Force on a Current Carrying Conductor Questions in English

(C) The magnetic field produced by one wire at the position of the other wire is given by $B = \frac{\mu_0 I}{2 \pi d}$.
According to the Lorentz force law,the force per unit length $f$ on a wire carrying current $I$ in a magnetic field $B$ is $f = I B \sin(90^\circ) = I B$.
Substituting the value of $B$,we get $f = I \left( \frac{\mu_0 I}{2 \pi d} \right) = \frac{\mu_0 I^2}{2 \pi d}$.
Since the currents are in the same direction,the wires attract each other.

(A) The magnetic force on a current-carrying wire is given by $\vec{F} = i(\vec{L} \times \vec{B})$,where $\vec{L}$ is the vector length from the start to the end of the wire segment.
For the loop,we can consider the segments $AC$,$CD$,$AE$,$ED$,and the diagonal $AD$.
However,the total force on a closed loop in a uniform magnetic field is zero. But here,the current flows in specific branches. We calculate the force on each segment:
$1$. For segment $AC$: $\vec{L}_{AC} = 2\hat{j}$,$\vec{B} = -2\hat{k}$. $\vec{F}_{AC} = 3(2\hat{j} \times -2\hat{k}) = -12\hat{i}\,N$.
$2$. For segment $CD$: $\vec{L}_{CD} = 2\hat{i}$,$\vec{B} = -2\hat{k}$. $\vec{F}_{CD} = 3(2\hat{i} \times -2\hat{k}) = 12\hat{j}\,N$.
$3$. For segment $AE$: $\vec{L}_{AE} = 2\hat{i}$,$\vec{B} = -2\hat{k}$. $\vec{F}_{AE} = 3(2\hat{i} \times -2\hat{k}) = 12\hat{j}\,N$.
$4$. For segment $ED$: $\vec{L}_{ED} = 2\hat{j}$,$\vec{B} = -2\hat{k}$. $\vec{F}_{ED} = 3(2\hat{j} \times -2\hat{k}) = -12\hat{i}\,N$.
$5$. For segment $AD$: $\vec{L}_{AD} = 2\hat{i} + 2\hat{j}$,$\vec{B} = -2\hat{k}$. $\vec{F}_{AD} = 3((2\hat{i} + 2\hat{j}) \times -2\hat{k}) = 3(-4\hat{j} + 4\hat{i}) = 12\hat{i} - 12\hat{j}\,N$.
Summing all forces: $\vec{F}_{net} = \vec{F}_{AC} + \vec{F}_{CD} + \vec{F}_{AE} + \vec{F}_{ED} + \vec{F}_{AD} = (-12\hat{i}) + (12\hat{j}) + (12\hat{j}) + (-12\hat{i}) + (12\hat{i} - 12\hat{j}) = -12\hat{i} + 12\hat{j}\,N$.
The magnitude is $|\vec{F}_{net}| = \sqrt{(-12)^2 + (12)^2} = \sqrt{144 + 144} = \sqrt{288} = 12\sqrt{2}\,N$.
Wait,re-evaluating the path: The current splits at $A$ and recombines at $D$. The force is the vector sum of forces on all segments. The calculation above yields $12\sqrt{2}\,N$. Given the options,the intended logic likely treats the paths $ACD$ and $AED$ as parallel paths and $AD$ as a third path,but the standard vector sum is $12\sqrt{2}\,N$. Since $12\sqrt{2}$ is not an option,let's re-read the diagram. If the current $i$ flows through $ACD$,$AED$,and $AD$ independently,the total force is the sum of forces on these three paths: $\vec{F}_{ACD} = \vec{F}_{AC} + \vec{F}_{CD} = -12\hat{i} + 12\hat{j}$. $\vec{F}_{AED} = \vec{F}_{AE} + \vec{F}_{ED} = 12\hat{j} - 12\hat{i}$. $\vec{F}_{AD} = 12\hat{i} - 12\hat{j}$. Total $\vec{F} = -12\hat{i} + 12\hat{j} + 12\hat{j} - 12\hat{i} + 12\hat{i} - 12\hat{j} = -12\hat{i} + 12\hat{j}$. Magnitude $12\sqrt{2}$. None match. If the question implies $3i$ total,then $36\sqrt{2}$ is the result.

(B) The magnetic force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = I(\vec{L} \times \vec{B})$,where $\vec{L}$ is the vector displacement from the starting point to the end point of the wire.
For any current-carrying path between two points $A$ and $C$,the net magnetic force depends only on the straight-line displacement vector $\vec{AC}$ connecting $A$ and $C$.
In the given circuit,the current $I$ splits into two paths between $A$ and $C$: path $ABC$ and path $ADC$.
For path $ABC$,the displacement vector is $\vec{AC}$ directed from $A$ to $C$ with magnitude $l$. The force is $\vec{F}_{ABC} = I_1 (\vec{l} \times \vec{B})$.
For path $ADC$,the displacement vector is also $\vec{AC}$ directed from $A$ to $C$ with magnitude $l$. The force is $\vec{F}_{ADC} = I_2 (\vec{l} \times \vec{B})$.
Since the total current $I = I_1 + I_2$,the total force on the loop is $\vec{F} = \vec{F}_{ABC} + \vec{F}_{ADC} = (I_1 + I_2) (\vec{l} \times \vec{B}) = I (\vec{l} \times \vec{B})$.
Since the magnetic field is perpendicular to the plane of the loop,the magnitude of the force is $F = I l B$.

(C) Consider a small element of the wire subtending an angle $d\theta$ at the center. The magnetic force on this element is $dF = I(R d\theta)B = IBR d\theta$,acting radially outward.
Let $T$ be the tension in the wire. The components of tension at the ends of this small element,$T \sin(d\theta/2)$ at each end,act radially inward to balance the magnetic force.
For small angles,$\sin(d\theta/2) \approx d\theta/2$. Thus,the net inward force is $2T \sin(d\theta/2) \approx 2T(d\theta/2) = T d\theta$.
Equating the inward force to the outward magnetic force:
$T d\theta = IBR d\theta$
Therefore,the tension in the wire is $T = IBR$.

(A) The force per unit length between two long parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $d$ is given by $F = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
According to the problem,if the currents are in the same direction,$I_1 I_2 > 0$,the wires attract each other,and the force $F$ is defined as 'negative'.
If the currents are in opposite directions,$I_1 I_2 < 0$,the wires repel each other,and the force $F$ is defined as 'positive'.
Thus,the relationship is $F = -k(I_1 I_2)$,where $k = \frac{\mu_0}{2 \pi d}$ is a positive constant.
This represents a straight line passing through the origin with a negative slope. Therefore,the graph showing the dependence of $F$ on $I_1 I_2$ is a straight line in the second and fourth quadrants,which corresponds to graph $A$.

(A) Given,length of wire $Q$,$L = 25\, cm = 0.25\, m$.
The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $r$ is given by $f = \frac{\mu_0 I_1 I_2}{2\pi r}$.
Force on wire $Q$ due to wire $R$ $(F_{QR})$:
$I_Q = 10\, A$,$I_R = 20\, A$,$r = 0.05\, m$.
Since currents are in opposite directions,the force is repulsive (towards left).
$F_{QR} = \frac{\mu_0 I_Q I_R}{2\pi r} \times L = 2 \times 10^{-7} \times \frac{10 \times 20}{0.05} \times 0.25 = 20 \times 10^{-5}\, N$ (towards left).
Force on wire $Q$ due to wire $P$ $(F_{QP})$:
$I_Q = 10\, A$,$I_P = 30\, A$,$r = 0.03\, m$.
Since currents are in opposite directions,the force is repulsive (towards right).
$F_{QP} = \frac{\mu_0 I_Q I_P}{2\pi r} \times L = 2 \times 10^{-7} \times \frac{10 \times 30}{0.03} \times 0.25 = 50 \times 10^{-5}\, N$ (towards right).
Net force $F_{\text{net}} = F_{QP} - F_{QR} = 50 \times 10^{-5} - 20 \times 10^{-5} = 30 \times 10^{-5}\, N = 3 \times 10^{-4}\, N$ towards right.

(A) The magnetic force on the top segment of the loop (length $a$) is given by $\overrightarrow{F} = i(\overrightarrow{a} \times \overrightarrow{B})$. Since the current $i$ flows from left to right in the top segment and the magnetic field $\overrightarrow{B}$ is directed into the plane,the force $\overrightarrow{F}$ acts vertically upwards.
Magnitude of the magnetic force is $F = iBa$.
Given that $i > mg/Ba$,we have $iBa > mg$,which means the upward magnetic force is greater than the downward gravitational force $mg$.
As a result,the net force on the loop is upward,causing the mass $m$ to rise.
Since the loop moves against the gravitational force,work is done on the system by the external source (the battery driving the current $i$).
Therefore,the weight rises and work is done on the system.

(C) The current in the circuit is $I = \frac{V}{R} = \frac{24}{12} = 2 \, A$.
The initial equilibrium condition for the wire of mass $m$ supported by two springs of constant $k$ is $2kx_1 = mg$,where $x_1 = 0.5 \, cm = 0.5 \times 10^{-2} \, m$.
When the magnetic field $B$ is applied,the additional force $F_m = IlB$ causes an additional stretch $x_2 = 0.3 \, cm = 0.3 \times 10^{-2} \, m$. The new equilibrium condition is $2k(x_1 + x_2) = mg + IlB$.
Subtracting the first equation from the second gives $2kx_2 = IlB$.
Substituting $2k = \frac{mg}{x_1}$ into the equation,we get $\frac{mg}{x_1} x_2 = IlB$.
Solving for $B$: $B = \frac{mgx_2}{Ilx_1} = \frac{10 \times 10^{-3} \times 10 \times 0.3 \times 10^{-2}}{2 \times 5 \times 10^{-2} \times 0.5 \times 10^{-2}} = \frac{0.3}{0.5} = 0.6 \, T$.
Since the springs stretch further,the magnetic force must be directed downwards. By Fleming's Left-Hand Rule,for a current flowing in the wire,the magnetic field must be directed into the plane of the page to produce a downward force.

(A) The magnetic field $B_A$ produced by the infinitely long wire $A$ at a distance $r$ is given by $B_A = \frac{\mu_0 I_A}{2 \pi r}$.
Here,$I_A = 10\, A$ and $r = 10\, cm = 0.1\, m$.
So,$B_A = \frac{4\pi \times 10^{-7} \times 10}{2\pi \times 0.1} = 2 \times 10^{-5}\, T$.
The force $F$ on a current-carrying wire $B$ of length $L$ in a magnetic field $B_A$ is given by $F = I_B L B_A \sin(\theta)$.
Since the wires are parallel,the angle $\theta = 90^\circ$ (between the current direction and the magnetic field lines),so $\sin(90^\circ) = 1$.
Given $I_B = 2\, A$ and $L = 2\, m$,we have $F = 2 \times 2 \times (2 \times 10^{-5}) = 8 \times 10^{-5}\, N$.

(D) The magnetic field $B$ produced by an infinitely long wire at a distance $r$ is $B = \frac{\mu_0 I}{2\pi r}$.
For a small current loop with magnetic moment $M = I_L \pi a^2$ (where $I_L$ is the loop current) placed in a non-uniform magnetic field,the force is given by $F = \nabla (M \cdot B)$.
Since the magnetic field $B$ from the wire varies as $1/r$,the gradient of the field varies as $1/r^2$.
Thus,the force $F$ on the loop due to the wire (or vice-versa) is proportional to $M \times (\text{gradient of } B)$.
$F \propto M \times \frac{1}{d^2} \propto (I_L a^2) \times \frac{1}{d^2}$.
Given the options,we look for the dependence on $a$ and $d$. The force is proportional to $a^2/d^2$,which is equivalent to $(a/d)^2$.

(B) Let the square loop have vertices $P, Q, R, S$ such that $PQ$ is the side closest to the wire at distance $a$. The magnetic field $B$ due to the long wire at a distance $r$ is $B = \frac{\mu_0 I_1}{2\pi r}$.
$1$. For the side $PQ$ (length $a$,distance $a$): The current flows downwards. The magnetic field is directed into the page. By Fleming's Left-Hand Rule,the force $F_1$ is directed away from the wire (repulsive) with magnitude $F_1 = I_2 B_1 a = I_2 \left( \frac{\mu_0 I_1}{2\pi a} \right) a = \frac{\mu_0 I_1 I_2}{2\pi}$.
$2$. For the side $RS$ (length $a$,distance $2a$): The current flows upwards. The magnetic field is directed into the page. By Fleming's Left-Hand Rule,the force $F_2$ is directed towards the wire (attractive) with magnitude $F_2 = I_2 B_2 a = I_2 \left( \frac{\mu_0 I_1}{2\pi (2a)} \right) a = \frac{\mu_0 I_1 I_2}{4\pi}$.
$3$. For the top and bottom sides: The forces on these segments are equal and opposite,thus they cancel each other out.
$4$. The net force is $F_{net} = F_1 - F_2 = \frac{\mu_0 I_1 I_2}{2\pi} - \frac{\mu_0 I_1 I_2}{4\pi} = \frac{\mu_0 I_1 I_2}{4\pi}$. Since $F_1 > F_2$,the net force is repulsive.

(B) The force per unit length on a wire carrying current $I$ due to another parallel wire carrying current $I'$ at a distance $r$ is given by $F = \frac{\mu_0 I I'}{2 \pi r}$.
For the net force on wire $C$ to be zero,the forces exerted by wires $A$ and $B$ must be equal in magnitude and opposite in direction.
Let wire $C$ be at a distance $x$ from $A$ and $(d-x)$ from $B$. The force per unit length due to $A$ is $F_A = \frac{\mu_0 I_1 I}{2 \pi x}$ (repulsive if currents are in the same direction,attractive if opposite).
The force per unit length due to $B$ is $F_B = \frac{\mu_0 I_2 I}{2 \pi (d-x)}$.
Since the currents $I_1$ and $I_2$ are in opposite directions (as shown in the figure),the forces will be in the same direction if $C$ is between $A$ and $B$. Thus,the equilibrium point must lie outside the region between $A$ and $B$.
Equating the magnitudes of the magnetic fields produced by $A$ and $B$ at the position of $C$:
$\frac{\mu_0 I_1}{2 \pi x} = \frac{\mu_0 I_2}{2 \pi |x - d|}$
$\frac{I_1}{x} = \frac{I_2}{|x - d|}$
Case $1$: $x > d$,then $x - d = x - d$,so $I_1(x - d) = I_2 x \Rightarrow x(I_1 - I_2) = I_1 d \Rightarrow x = \frac{I_1 d}{I_1 - I_2}$.
Case $2$: $x < 0$,then $|x - d| = d - x$,so $I_1(d - x) = I_2 x \Rightarrow I_1 d = x(I_1 + I_2) \Rightarrow x = \frac{I_1 d}{I_1 + I_2}$ (This is not possible as $x$ is distance from $A$).
Re-evaluating the geometry for equilibrium: The point where the net magnetic field is zero is $x = \frac{I_1 d}{I_1 - I_2}$.

(B) Given:
Current $I = 8\,A$
Angle $\theta = 30^{\circ}$
Magnetic field $B = 0.15\,T$
Length per unit length $\ell = 1\,m$
The magnetic force on a current-carrying conductor is given by $F = BI\ell \sin \theta$.
The magnetic force per unit length is $f = \frac{F}{\ell} = BI \sin \theta$.
Substituting the values:
$f = 0.15 \times 8 \times \sin 30^{\circ}$
$f = 1.2 \times 0.5$
$f = 0.6\,N\,m^{-1}$.

(C) In the absence of a magnetic field,the weight added in one pan balances the rectangular coil in the other pan of the balance.
Let $M = 500 \, g = 0.5 \, kg$. The balance condition is $Mgl = W_{coil}l$,where $l$ is the arm length of the balance.
When a current $I = 9.8 \, A$ is passed through the coil and a magnetic field $B = 0.4 \, T$ is switched on,a magnetic force $F_m = IBL$ acts on the arm $CD$ of length $L = 1.5 \, cm = 1.5 \times 10^{-2} \, m$.
According to Fleming's left-hand rule,the magnetic force acts vertically downwards on the arm $CD$.
To regain the balance,an additional mass $m$ must be added to the pan containing the $500 \, g$ mass.
The new balance condition is $(M + m)gl = W_{coil}l + F_m l$.
Since $Mgl = W_{coil}l$,we have $mgl = F_m l = (IBL)l$.
Therefore,$m = \frac{IBL}{g} = \frac{9.8 \times 0.4 \times 1.5 \times 10^{-2}}{9.8}$.
$m = 0.4 \times 1.5 \times 10^{-2} \, kg = 0.6 \times 10^{-2} \, kg = 6 \times 10^{-3} \, kg = 6 \, g$.

(C) $1$. Two parallel wires carrying currents in the same direction attract each other, while wires carrying currents in opposite directions repel each other.
$2$. Wire $P$ carries current into the paper $(\otimes)$, wire $Q$ carries current out of the paper $(\odot)$, and wire $R$ carries current into the paper $(\otimes)$.
$3$. Since $P$ and $Q$ have currents in opposite directions, wire $Q$ exerts a repulsive force on $P$, pushing it to the left (away from $Q$). Let this force be $F_Q$.
$4$. Since $P$ and $R$ have currents in the same direction, wire $R$ exerts an attractive force on $P$, pulling it downwards (towards $R$). Let this force be $F_R$.
$5$. The resultant force $F$ on wire $P$ is the vector sum of $F_Q$ and $F_R$. Since both forces have the same magnitude, the resultant force $F$ will point in the direction of arrow $C$, which bisects the angle between the horizontal and vertical directions.

(B) $1$. For conductors $A$ and $B$: When two parallel conductors carry current in the same direction,they produce magnetic fields that result in an attractive force between them.
$2$. For electron beams $x$ and $y$: Electron beams consist of moving negative charges. Since they are moving in the same direction,they constitute parallel currents in the same direction. However,they also possess a net negative charge density,which results in a strong electrostatic repulsive force. In the case of electron beams,the electrostatic repulsion dominates over the magnetic attraction,leading to a net repulsion between the beams.

(A) The magnetic force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $d$ is given by $f = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
Wires carrying current in the same direction attract each other,while those carrying current in opposite directions repel each other.
For wire $R$:
$1$. Force due to wire $Q$ $(f_{RQ})$: Currents are in opposite directions (one into the page,one out of the page),so they repel. The force is directed away from $Q$ (to the right).
$f_{RQ} = \frac{\mu_0 (4)(2)}{2 \pi (1)} = 2 \times 10^{-7} \times \frac{8}{1} = 16 \times 10^{-7} \, N/m$.
$2$. Force due to wire $P$ $(f_{RP})$: Currents are in the same direction (both out of the page),so they attract. The force is directed towards $P$ (to the left).
$f_{RP} = \frac{\mu_0 (1)(2)}{2 \pi (3)} = 2 \times 10^{-7} \times \frac{2}{3} = \frac{4}{3} \times 10^{-7} \, N/m$.
Net force per unit length on $R$ is $f_{net} = f_{RQ} - f_{RP} = (16 - \frac{4}{3}) \times 10^{-7} = \frac{48-4}{3} \times 10^{-7} = \frac{44}{3} \times 10^{-7} \, N/m$.

(B) The magnetic force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = i(\vec{L}_{eff} \times \vec{B})$,where $\vec{L}_{eff}$ is the vector displacement from the starting point to the end point of the wire.
The wire starts at $x = 0$ and ends at $x = 2L$.
At $x = 0$,$y = a \sin(0) = 0$.
At $x = 2L$,$y = a \sin\left(\frac{\pi(2L)}{L}\right) = a \sin(2\pi) = 0$.
Thus,the starting point is $(0, 0)$ and the ending point is $(2L, 0)$.
The effective length vector $\vec{L}_{eff}$ is the displacement vector from $(0, 0)$ to $(2L, 0)$,which is $\vec{L}_{eff} = 2L \hat{i}$.
The magnetic field is uniform and directed into the plane,so $\vec{B} = -B \hat{k}$ (assuming the plane is the $xy$-plane).
The magnetic force is $\vec{F} = i(2L \hat{i} \times -B \hat{k}) = -2iLB(\hat{i} \times \hat{k}) = -2iLB(-\hat{j}) = 2iLB \hat{j}$.
The magnitude of the force is $2iLB$.

(C) The magnetic field is $\vec B = B_0 \left( 5 + \frac{x}{l} \right) \hat k$. Let the vertices of the square loop be $P(0,0)$,$Q(l,0)$,$R(l,l)$,and $S(0,l)$.
The force on a current-carrying wire is given by $\vec F = i \int d\vec l \times \vec B$.
$1$. For segment $PQ$ (along $x$-axis,$y=0$,$x$ from $0$ to $l$): $d\vec l = dx \hat i$,$\vec B = B_0 (5 + x/l) \hat k$. $\vec F_{PQ} = i \int_0^l (dx \hat i) \times B_0 (5 + x/l) \hat k = -i B_0 \int_0^l (5 + x/l) dx \hat j = -i B_0 [5x + x^2/(2l)]_0^l \hat j = -i B_0 (5l + l/2) \hat j = -5.5 B_0 il \hat j$.
$2$. For segment $RS$ (parallel to $x$-axis,$y=l$,$x$ from $l$ to $0$): $d\vec l = dx \hat i$,$\vec B = B_0 (5 + x/l) \hat k$. $\vec F_{RS} = i \int_l^0 (dx \hat i) \times B_0 (5 + x/l) \hat k = -i B_0 \int_l^0 (5 + x/l) dx \hat j = i B_0 [5x + x^2/(2l)]_0^l \hat j = 5.5 B_0 il \hat j$.
$3$. For segment $QR$ (parallel to $y$-axis,$x=l$,$y$ from $0$ to $l$): $d\vec l = dy \hat j$,$\vec B = B_0 (5 + l/l) \hat k = 6 B_0 \hat k$. $\vec F_{QR} = i \int_0^l (dy \hat j) \times 6 B_0 \hat k = 6 B_0 il \hat i$.
$4$. For segment $SP$ (parallel to $y$-axis,$x=0$,$y$ from $l$ to $0$): $d\vec l = dy \hat j$,$\vec B = B_0 (5 + 0/l) \hat k = 5 B_0 \hat k$. $\vec F_{SP} = i \int_l^0 (dy \hat j) \times 5 B_0 \hat k = -5 B_0 il \hat i$.
Net force $\vec F_{net} = \vec F_{PQ} + \vec F_{RS} + \vec F_{QR} + \vec F_{SP} = (-5.5 + 5.5) B_0 il \hat j + (6 - 5) B_0 il \hat i = B_0 il \hat i$.
The magnitude is $B_0 il$.

(A) The magnetic force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = i(\vec{L} \times \vec{B})$,where $\vec{L}$ is the displacement vector from the starting point to the ending point of the wire.
For a semicircular ring of radius $R$,the displacement vector $\vec{L}$ connects the two ends of the semicircle. The length of this straight-line displacement is the diameter of the semicircle,which is $2R$.
Since the plane of the wire is perpendicular to the magnetic field $\vec{B}$,the magnitude of the force is $F = iLB \sin(90^\circ) = i(2R)B$.
Therefore,the net force acting on the ring is $F = 2BiR$.

(C) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $d$ is given by $f = \frac{{\mu _0}{I_1}{I_2}}{2\pi d}$.
Since all wires carry the same current $I$,the force per unit length exerted by wire $A$ on $B$ is $f_A = \frac{{\mu _0}{I^2}}{2\pi d}$ (directed towards $A$).
The force per unit length exerted by wire $C$ on $B$ is $f_C = \frac{{\mu _0}{I^2}}{2\pi d}$ (directed towards $C$).
Since the wires $A$,$B$,and $C$ are arranged such that the angle between the directions of forces $f_A$ and $f_C$ is $90^\circ$,the net force per unit length on wire $B$ is:
$f_{net} = \sqrt{f_A^2 + f_C^2} = \sqrt{\left(\frac{{\mu _0}{I^2}}{2\pi d}\right)^2 + \left(\frac{{\mu _0}{I^2}}{2\pi d}\right)^2}$
$f_{net} = \sqrt{2 \left(\frac{{\mu _0}{I^2}}{2\pi d}\right)^2} = \sqrt{2} \frac{{\mu _0}{I^2}}{2\pi d} = \frac{{\mu _0}{I^2}}{\sqrt{2}\pi d}$.

(B) The force per unit length on a current-carrying conductor in a magnetic field is given by the formula $f = \frac{F}{L} = I B \sin \theta$.
Here,the magnetic field $B = 3 \times 10^{-5} \, T$ is directed from south to north.
The current $I = 1 \, A$ flows from east to west.
Since the direction of the current (east-west) is perpendicular to the direction of the magnetic field (south-north),the angle $\theta = 90^{\circ}$.
Substituting the values into the formula:
$f = 1 \, A \times (3 \times 10^{-5} \, T) \times \sin(90^{\circ})$
$f = 1 \times 3 \times 10^{-5} \times 1$
$f = 3 \times 10^{-5} \, N/m$.

(B) When a flexible current-carrying loop is placed in an external magnetic field,the magnetic force acts on the wire to maximize the magnetic flux through the loop.
According to the principle of maximum area for a given perimeter,the loop will expand to enclose the maximum possible area.
For a fixed perimeter,a circle is the geometric shape that encloses the maximum area.
Therefore,the flexible wire will take the shape of a circle.

(A) Current in wire $A$,$I_{A} = 8.0 \, A$.
Current in wire $B$,$I_{B} = 5.0 \, A$.
Distance between the wires,$r = 4.0 \, cm = 0.04 \, m$.
Length of the section of wire $A$,$l = 10 \, cm = 0.1 \, m$.
The force per unit length between two parallel wires is given by $f = \frac{\mu_{0} I_{A} I_{B}}{2 \pi r}$.
The total force $F$ on a length $l$ is $F = \frac{\mu_{0} I_{A} I_{B} l}{2 \pi r}$.
Substituting the values:
$F = \frac{(4 \pi \times 10^{-7} \, T \cdot m/A) \times 8.0 \, A \times 5.0 \, A \times 0.1 \, m}{2 \pi \times 0.04 \, m}$.
$F = \frac{2 \times 10^{-7} \times 40 \times 0.1}{0.04} \, N$.
$F = \frac{8 \times 10^{-7}}{0.04} \, N = 2 \times 10^{-5} \, N$.
Since the currents are in the same direction,the force is attractive.

(B) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L}_{eff} \times \vec{B})$,where $\vec{L}_{eff}$ is the vector displacement from the start point $A$ to the end point $C$.
The wire consists of a straight segment $AB$ of length $3 \, m$ and a curved segment $BC$ which is a quarter circle of radius $R = 4 \, m$.
The coordinates are: $A = (0, 0)$,$B = (3, 0)$,and $C = (3, 4)$.
The effective length vector $\vec{L}_{eff} = \vec{AC} = (3 \hat{i} + 4 \hat{j}) \, m$.
The magnetic field is directed into the page,so $\vec{B} = -2 \hat{k} \, T$.
The force is $\vec{F} = I(\vec{L}_{eff} \times \vec{B}) = 2 \, A \times [(3 \hat{i} + 4 \hat{j}) \times (-2 \hat{k})] \, T$.
Using the cross product rules ($\hat{i} \times \hat{k} = -\hat{j}$ and $\hat{j} \times \hat{k} = \hat{i}$):
$\vec{F} = 2 \times [3(\hat{i} \times -2\hat{k}) + 4(\hat{j} \times -2\hat{k})] = 2 \times [-6(-\hat{j}) + -8(\hat{i})] = 2 \times [6\hat{j} - 8\hat{i}] = (12\hat{j} - 16\hat{i}) \, N$.
The magnitude of the force is $F = \sqrt{(12)^2 + (-16)^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \, N$.

(A) The magnetic field $B$ is directed outwards (represented by dots) and is uniform.
According to Lenz's Law, the induced current in the loop will create a magnetic field that opposes the change in magnetic flux.
However, in this scenario, the magnetic field is constant.
If we consider the force on a current-carrying element $d\vec{l}$ in a magnetic field $\vec{B}$, the force is given by $d\vec{F} = i(d\vec{l} \times \vec{B})$.
For a circular loop with current $i$ flowing in a uniform magnetic field $\vec{B}$ directed outwards, the force on every small element $d\vec{l}$ will be directed radially outwards.
This is because the cross product of the tangential current element and the outward magnetic field results in a force directed away from the center.
Therefore, the loop will experience a net outward force, causing it to expand.

(A) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $d$ is given by $f = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
Parallel currents attract each other,while anti-parallel currents repel each other.
For wire $R$:
$1$. Force due to wire $Q$ $(F_{RQ})$: Currents in $Q$ $(4 \text{ A})$ and $R$ $(6 \text{ A})$ are in the same direction,so they attract. Thus,$F_{RQ}$ is towards the left (towards $Q$).
$2$. Force due to wire $P$ $(F_{RP})$: Currents in $P$ $(2 \text{ A})$ and $R$ $(6 \text{ A})$ are in the same direction,so they attract. Thus,$F_{RP}$ is towards the left (towards $P$).
Since both forces $F_{RQ}$ and $F_{RP}$ act towards the left,the resultant force on $R$ is towards the left.

(C) The magnetic force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = i(\vec{L}_{eff} \times \vec{B})$,where $\vec{L}_{eff}$ is the vector displacement from the starting point to the ending point of the wire.
Here,the wire starts at $(0, 0)$ and ends at $(2L, 0)$.
Therefore,the effective length vector is $\vec{L}_{eff} = (2L - 0)\hat{i} = 2L\hat{i}$.
The magnetic field is uniform and directed into the page,so $\vec{B} = -B\hat{k}$ (assuming the $xy$-plane is the plane of the wire).
The force is $\vec{F} = i(2L\hat{i} \times -B\hat{k}) = i(2LB)(\hat{j}) = 2iLB\hat{j}$.
The magnitude of the force is $F = 2iLB$.

(D) The magnetic field $B$ produced by the long straight wire $1$ at a perpendicular distance $r$ is given by $B = \frac{\mu_0 i_1}{2\pi r}$.
The force $dF$ on a current element $i_2 \, dl$ placed in a magnetic field $B$ is given by $dF = i_2 (dl \times B) = i_2 \, dl \, B \sin \alpha$,where $\alpha$ is the angle between the current element $dl$ and the magnetic field $B$.
The magnetic field $B$ is perpendicular to the plane containing the wires (directed inwards). The current element $dl$ lies in the plane of the wires. Therefore,the angle between $dl$ and $B$ is $90^{\circ}$.
Thus,$dF = i_2 \, dl \left( \frac{\mu_0 i_1}{2\pi r} \right) \sin 90^{\circ} = \frac{\mu_0 i_1 i_2}{2\pi r} dl$.

(A) The magnetic force $F$ acting on the current-carrying wire is given by $F = IlB$,which acts in the upward direction to counteract gravity.
For the wire to be suspended in mid-air,the magnetic force must balance the gravitational force:
$mg = IlB$
Rearranging for the magnetic field $B$:
$B = \frac{mg}{Il}$
Given:
$m = 200 \;g = 0.2 \;kg$
$l = 1.5 \;m$
$I = 2 \;A$
$g = 9.8 \;m/s^2$
Substituting the values:
$B = \frac{0.2 \times 9.8}{2 \times 1.5} = \frac{1.96}{3} \approx 0.653 \;T$
Thus,the magnitude of the magnetic field is approximately $0.65 \;T$.

(A) The magnetic force on a current-carrying conductor is given by $\vec{F} = I(\vec{l} \times \vec{B})$.
The magnitude of the force is $F = I L B \sin \theta$.
The force per unit length is $f = F/L = I B \sin \theta$.
$(a)$ When the current flows from east to west, the angle between the current direction and the magnetic field (south to north) is $\theta = 90^{\circ}$.
Thus, $f = I B \sin 90^{\circ} = I B = (1 \; A) \times (3.0 \times 10^{-5} \; T) = 3.0 \times 10^{-5} \; N/m$.
The direction of the force is downwards (using the right-hand rule).
$(b)$ When the current flows from south to north, the angle between the current direction and the magnetic field is $\theta = 0^{\circ}$.
Thus, $f = I B \sin 0^{\circ} = 0 \; N/m$.
There is no force on the conductor in this case.

(B) The magnetic force per unit length $f$ on a current-carrying wire is given by the formula $f = B I \sin \theta$.
Given values:
$I = 8 \; A$
$B = 0.15 \; T$
$\theta = 30^{\circ}$
Substituting these values into the formula:
$f = 0.15 \times 8 \times \sin 30^{\circ}$
$f = 1.2 \times 0.5$
$f = 0.6 \; N \; m^{-1}$
Therefore,the magnitude of the magnetic force per unit length is $0.6 \; N \; m^{-1}$.

(C) Length of the wire,$l = 3 \; cm = 0.03 \; m$.
Current flowing in the wire,$I = 10 \; A$.
Magnetic field,$B = 0.27 \; T$.
Angle between the current and magnetic field,$\theta = 90^{\circ}$ (Because the magnetic field produced by a solenoid is along its axis and the current-carrying wire is kept perpendicular to the axis).
Magnetic force exerted on the wire is given by the formula:
$F = B I l \sin \theta$
Substituting the values:
$F = 0.27 \times 10 \times 0.03 \times \sin 90^{\circ}$
$F = 0.27 \times 10 \times 0.03 \times 1$
$F = 8.1 \times 10^{-2} \; N$.
Hence,the magnetic force on the wire is $8.1 \times 10^{-2} \; N$. The direction of the force can be obtained from Fleming's left-hand rule.

(D) Current in wire $A$,$I_A = 8.0 \; A$.
Current in wire $B$,$I_B = 5.0 \; A$.
Distance between the wires,$r = 4.0 \; cm = 0.04 \; m$.
Length of the section of wire $A$,$l = 10 \; cm = 0.1 \; m$.
The force per unit length between two parallel wires is given by $f = \frac{\mu_0 I_A I_B}{2 \pi r}$.
The total force $F$ on a length $l$ is $F = \frac{\mu_0 I_A I_B l}{2 \pi r} = \frac{\mu_0}{4 \pi} \cdot \frac{2 I_A I_B l}{r}$.
Substituting the values:
$F = 10^{-7} \times \frac{2 \times 8.0 \times 5.0 \times 0.1}{0.04}$.
$F = 10^{-7} \times \frac{8.0}{0.04} = 10^{-7} \times 200 = 2 \times 10^{-5} \; N$.
Since the currents are in the same direction,the force is attractive.

(A) Given:
Length of the rod,$l = 0.45\; m$
Mass of the rod,$m = 60\; g = 60 \times 10^{-3}\; kg$
Acceleration due to gravity,$g = 9.8\; m s^{-2}$
Current in the rod,$I = 5.0\; A$
$(a)$ For the tension in the wires to be zero,the upward magnetic force must balance the downward weight of the rod.
$F_B = mg$
$BIl = mg$
$B = \frac{mg}{Il} = \frac{60 \times 10^{-3} \times 9.8}{5.0 \times 0.45} = 0.26\; T$
Thus,a magnetic field of $0.26\; T$ directed normal to the conductor is required.
$(b)$ When the current direction is reversed,the magnetic force $F_B$ acts downwards along with the weight $mg$.
Total tension $T = F_B + mg = BIl + mg$
$T = (0.26 \times 5.0 \times 0.45) + (60 \times 10^{-3} \times 9.8)$
$T = 0.588 + 0.588 = 1.176\; N$

(A) Current in both wires,$I = 300\; A$.
Distance between the wires,$r = 1.5\; cm = 0.015\; m$.
The force per unit length between two parallel current-carrying wires is given by the formula:
$f = \frac{F}{l} = \frac{\mu_0 I^2}{2 \pi r}$
Where $\mu_0 = 4 \pi \times 10^{-7}\; T\cdot m/A$ is the permeability of free space.
Substituting the values:
$f = \frac{4 \pi \times 10^{-7} \times (300)^2}{2 \pi \times 0.015}$
$f = \frac{2 \times 10^{-7} \times 90000}{0.015}$
$f = \frac{0.018}{0.015} = 1.2\; N/m$.
Since the current in the wires flows in opposite directions (one to the motor and one returning),the force between them is repulsive.

(A) Magnetic field strength,$B = 1.5 \; T$
Radius of the cylindrical region,$r = 10 \; cm = 0.1 \; m$
Current in the wire,$I = 7.0 \; A$
$(a)$ If the wire intersects the axis,the length of the wire inside the field is the diameter,$l = 2r = 0.2 \; m$. The angle between the magnetic field (East-West) and current (North-South) is $\theta = 90^{\circ}$.
Force $F = B I l \sin \theta = 1.5 \times 7.0 \times 0.2 \times \sin 90^{\circ} = 2.1 \; N$. By Fleming's Left-Hand Rule,the direction is vertically downward.
$(b)$ When the wire is turned,the length $l'$ inside the field becomes $l' = l / \sin \theta$,where $\theta = 45^{\circ}$.
Force $F = B I l' \sin \theta = B I (l / \sin \theta) \sin \theta = B I l = 1.5 \times 7.0 \times 0.2 = 2.1 \; N$. The force remains $2.1 \; N$ vertically downward.
$(c)$ If the wire is lowered by $d = 6.0 \; cm$,the length of the chord is $l'' = 2 \sqrt{r^2 - d^2} = 2 \sqrt{10^2 - 6^2} = 2 \sqrt{64} = 16 \; cm = 0.16 \; m$.
Force $F = B I l'' = 1.5 \times 7.0 \times 0.16 = 1.68 \; N$. The direction is vertically downward.

(N/A) As shown in the figure,consider a conductor $PQ$ of length $l$ and cross-sectional area $A$,carrying current $I$ along the positive $y$-direction. The magnetic field $\overrightarrow{B}$ acts along the positive $z$-direction.
The electrons drift towards the left with drift velocity $\overrightarrow{v_{d}}$.
Each electron experiences a force along the positive $x$-axis,which is given by:
$\vec{f} = -e(\overrightarrow{v_{d}} \times \overrightarrow{B})$
If $n$ is the number of free electrons per unit volume,then the total number of electrons in the conductor is:
$N = n \times \text{Volume} = nAl$
The total force on the conductor is:
$\overrightarrow{F} = N \vec{f} = nAl[-e(\overrightarrow{v_{d}} \times \overrightarrow{B})]$
$= nAe[-(l \overrightarrow{v_{d}} \times \overrightarrow{B})]$
Since $I\vec{l}$ represents a current element vector in the direction of current,we can write:
$\vec{v}_{d} = v_{d} \vec{l}$
$\therefore \overrightarrow{F} = nAe(v_{d} \vec{l} \times \overrightarrow{B}) = nAev_{d}(\vec{l} \times \overrightarrow{B})$
Since $nAev_{d} = I$ (current),
$\therefore \vec{F} = I(\vec{l} \times \overrightarrow{B})$
The magnitude is $F = IlB \sin \theta$,where $\theta$ is the angle between $\overrightarrow{B}$ and $\vec{l}$.
For a wire of arbitrary shape,we calculate the force by considering it as a collection of linear strips $d\vec{l}$ and summing:
$\overrightarrow{F} = \sum I(d\vec{l} \times \overrightarrow{B})$

(N/A) The magnetic force $F$ acting on a straight current-carrying wire of length $L$ carrying current $I$ placed in a uniform magnetic field $B$ is given by the vector product:
$F = I(\vec{L} \times \vec{B})$
Where:
$I$ is the current in the wire (in Amperes),
$\vec{L}$ is the length vector of the wire (in meters),directed along the flow of current,
$\vec{B}$ is the uniform magnetic field (in Tesla),
$F$ is the magnetic force (in Newtons).
In terms of magnitude,this can be written as $F = BIL \sin(\theta)$,where $\theta$ is the angle between the direction of the current and the magnetic field.

(N/A) Consider two long parallel wires $a$ and $b$ separated by a distance $d$,carrying currents $I_{a}$ and $I_{b}$ respectively.
The conductor $a$ produces a magnetic field $\overrightarrow{B}_{a}$ at all points along the conductor $b$. According to the right-hand rule,the direction of this field is perpendicular to the plane containing the wires.
The magnitude of the magnetic field produced by wire $a$ at distance $d$ is given by:
$B_{a} = \frac{\mu_{0} I_{a}}{2 \pi d}$ ... $(1)$
The conductor $b$ carrying current $I_{b}$ experiences a magnetic force due to the field $B_{a}$. The force on a segment of length $L$ of wire $b$ is given by:
$\overrightarrow{F} = I \vec{L} \times \overrightarrow{B}$
Since the current and magnetic field are perpendicular,the magnitude of the force $F_{ba}$ on wire $b$ due to wire $a$ is:
$F_{ba} = I_{b} L B_{a} = \frac{\mu_{0} I_{a} I_{b} L}{2 \pi d}$ ... $(2)$
Similarly,the force $F_{ab}$ on a segment of length $L$ of wire $a$ due to the current in wire $b$ is equal in magnitude to $F_{ba}$ but acts in the opposite direction:
$\overrightarrow{F}_{ba} = -\overrightarrow{F}_{ab}$ ... $(3)$
This result is consistent with Newton's third law of motion.
It is observed that currents flowing in the same direction attract each other,while currents flowing in opposite directions repel each other.

(N/A) The magnetic force $d\vec{F}$ on a current-carrying element $I\vec{dl}$ placed in a magnetic field $\vec{B}$ is given by the vector product:
$d\vec{F} = I(d\vec{l} \times \vec{B})$
where $I$ is the current,$d\vec{l}$ is the infinitesimal length vector of the conductor,and $\vec{B}$ is the external magnetic field.
The direction of the magnetic force is determined by the Fleming's Left-Hand Rule.
According to this rule,if we stretch the thumb,forefinger,and middle finger of the left hand such that they are mutually perpendicular to each other,then:
$1$. The forefinger points in the direction of the magnetic field $(\vec{B})$.
$2$. The middle finger points in the direction of the current $(I)$.
$3$. The thumb will point in the direction of the magnetic force $(d\vec{F})$ acting on the conductor.

(A) According to the Biot-Savart law and the Lorentz force law,when two parallel wires carry current in the same direction,the magnetic field produced by one wire exerts an attractive force on the other wire.
Conversely,when the currents flow in opposite directions,the magnetic fields produced by the wires result in a repulsive force between them.
Therefore,the correct sequence is attraction followed by repulsion.

(N/A) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $d$ is given by the formula: $F/L = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
If $I_1 = I_2 = 1 \ A$ and $d = 1 \ m$,then the force per unit length is $F/L = \frac{4 \pi \times 10^{-7} \times 1 \times 1}{2 \pi \times 1} = 2 \times 10^{-7} \ N/m$.
Therefore,$1 \ A$ is defined as the constant current which,if maintained in two straight parallel conductors of infinite length and negligible circular cross-section,placed $1 \ m$ apart in a vacuum,would produce between these conductors a force equal to $2 \times 10^{-7} \ N$ per meter of length.

(N/A) Electromagnets are used in various devices due to their ability to turn magnetism on and off. Common uses include:
$1$. Electric bells: Used to create the ringing sound.
$2$. Loudspeakers: Used to convert electrical signals into sound waves.
$3$. Telephones: Used in the diaphragms to convert electrical signals into sound.
$4$. Cranes: Used in industrial settings to lift and move heavy magnetic materials like iron and steel.

(D) The magnetic field $B_1$ produced by the long wire carrying current $I_1$ along the $x$-axis at a point $(0, 0, d)$ is given by the Biot-Savart Law. Using the right-hand rule,the direction of the magnetic field at $(0, 0, d)$ due to current $I_1$ along the $x$-axis is in the negative $y$-direction ($-j$ direction).
The magnetic force $dF$ on a current element $I_2 dl$ is given by $dF = I_2 (dl \times B_1)$.
Here,the current element $I_2 dl$ is directed along the $y$-axis (let's say $dl = dl \hat{j}$).
The magnetic field $B_1$ at $(0, 0, d)$ is directed along the negative $y$-axis $(B_1 = -B_1 \hat{j})$.
Since the current element $dl$ and the magnetic field $B_1$ are both parallel to the $y$-axis,the cross product $dl \times B_1$ is zero because $\hat{j} \times \hat{j} = 0$.
Therefore,the magnetic force $dF$ exerted on the segment of the second wire at $O_2$ is zero.

(0.51 CM) The magnetic force on the wire $PQ$ due to the current-carrying wire on the table must balance the gravitational force acting on $PQ$ for it to remain in equilibrium at height $h$.
The magnetic field $B$ at a distance $h$ from the long straight wire carrying current $I = 25 \ A$ is given by:
$B = \frac{\mu_0 I}{2 \pi h}$
The magnetic force $F_m$ on the wire $PQ$ of length $l = 1 \ m$ carrying current $I$ is:
$F_m = I l B = I l \left( \frac{\mu_0 I}{2 \pi h} \right) = \frac{\mu_0 I^2 l}{2 \pi h}$
Since the currents are in opposite directions,the magnetic force is repulsive and acts upwards. At equilibrium,this force balances the weight $mg$ of the wire $PQ$:
$mg = \frac{\mu_0 I^2 l}{2 \pi h}$
Rearranging for $h$:
$h = \frac{\mu_0 I^2 l}{2 \pi m g}$
Given: $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$,$I = 25 \ A$,$l = 1 \ m$,$m = 2.5 \times 10^{-3} \ kg$,$g = 9.8 \ m/s^2$:
$h = \frac{(4 \pi \times 10^{-7}) \times (25)^2 \times 1}{2 \pi \times (2.5 \times 10^{-3}) \times 9.8}$
$h = \frac{2 \times 10^{-7} \times 625}{2.5 \times 10^{-3} \times 9.8} = \frac{1250 \times 10^{-7}}{24.5 \times 10^{-3}} \approx 51.02 \times 10^{-4} \ m$
$h \approx 0.51 \times 10^{-2} \ m = 0.51 \ cm$

(10 G) The magnetic force acting on the arm $CD$ creates a torque that disturbs the balance. To regain equilibrium,we must add an additional mass $m$ to the side of the coil.
The magnetic force $F_m$ on the arm $CD$ is given by $F_m = N I L B \sin(90^\circ) = N I L B$,where $N = 100$ is the number of turns,$I = 4.9 \text{ A}$ is the current,$L = 1 \text{ cm} = 0.01 \text{ m}$ is the length of the arm,and $B = 0.2 \text{ T}$ is the magnetic field.
$F_m = 100 \times 4.9 \times 0.01 \times 0.2 = 0.098 \text{ N}$.
This force acts downwards (by Fleming's Left-Hand Rule). To balance this,we add a mass $m$ such that the weight $mg$ equals the magnetic force $F_m$.
$mg = F_m$
$m \times 9.8 = 0.098$
$m = \frac{0.098}{9.8} = 0.01 \text{ kg} = 10 \text{ g}$.
Therefore,an additional mass of $10 \text{ g}$ must be added.

(B) The magnetic field at point $P$ is produced by the two semi-infinite segments of the wire.
Using the formula for the magnetic field due to a semi-infinite wire,$B = \frac{\mu_0 i}{4 \pi r}$.
At point $P$,the distance $r = 5 \, cm = 5 \times 10^{-2} \, m$.
The total magnetic field $B$ at $P$ is the sum of the fields from the two semi-infinite segments:
$B = \frac{\mu_0 i}{4 \pi r} + \frac{\mu_0 i}{4 \pi r} = 2 \times \frac{\mu_0 i}{4 \pi r}$
Substituting the values: $B = 2 \times \frac{10^{-7} \times 5}{5 \times 10^{-2}} = 2 \times 10^{-5} \, T$.
The force per unit length on a current-carrying wire is given by $f = iB$.
$f = 5 \, A \times 2 \times 10^{-5} \, T = 10^{-4} \, N/m$.

(A) The net magnetic force on any closed current-carrying loop placed in a uniform magnetic field is always zero.
This is because the magnetic force on a current element $I d\vec{l}$ is given by $d\vec{F} = I(d\vec{l} \times \vec{B})$.
For a closed loop,the net force is the integral $\oint I(d\vec{l} \times \vec{B}) = I(\oint d\vec{l}) \times \vec{B}$.
Since the vector sum of all displacement elements $d\vec{l}$ in a closed loop is zero (i.e.,$\oint d\vec{l} = 0$),the net force $\vec{F}$ is zero.

(A) When a flexible current-carrying loop is placed in an external magnetic field,every infinitesimal segment $(d\ell)$ of the wire experiences a magnetic force given by $dF = i(d\ell \times B)$.
This force acts perpendicular to both the current direction and the magnetic field.
Due to the symmetry of the magnetic force acting radially outward (or inward depending on the current direction),the wire experiences a uniform tension that forces it to expand into a circular shape to maximize the area enclosed by the loop,with its plane oriented normal to the magnetic field lines.