Force on a Current Carrying Conductor Questions in English

(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: $F = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
Since the current in both wires remains the same,the force is inversely proportional to the distance between them: $F \propto \frac{1}{d}$.
If the new distance $d' = \frac{d}{3}$,the new force $F'$ will be: $F' = \frac{\mu_0 I_1 I_2}{2 \pi (d/3)} = 3 \times \left( \frac{\mu_0 I_1 I_2}{2 \pi d} \right) = 3F$.
Therefore,the magnitude of the force becomes $3$ times the original value.

(D) The force per unit length between two parallel conductors carrying currents $I_1$ and $I_2$ separated by distance $r$ is given by $f = \frac{\mu_0 I_1 I_2}{2 \pi r}$.
$1$. Force $F_1$ exerted by $B$ on $A$:
Since currents in $A$ and $B$ are in the same direction,the force is attractive (towards $B$).
$F_1 = \frac{\mu_0 I \cdot I}{2 \pi x} \cdot L = \frac{\mu_0 I^2 L}{2 \pi x}$.
$2$. Force $F_2$ exerted by $C$ on $A$:
Since currents in $A$ and $C$ are in opposite directions,the force is repulsive (away from $C$).
The distance between $A$ and $C$ is $2x$.
$F_2 = \frac{\mu_0 I \cdot 2I}{2 \pi (2x)} \cdot L = \frac{\mu_0 I^2 L}{2 \pi x}$.
Comparing the magnitudes,we see that $F_1 = F_2$. However,since the forces are in opposite directions (one is attractive towards $B$ and the other is repulsive away from $C$),we have $F_1 = -F_2$ in vector notation.

(C) The force $F$ acting on a current-carrying wire in a uniform magnetic field is given by the formula $F = I L B \sin \theta$.
Here,$I = 1.2 \ A$ is the current,
$L = 0.5 \ m$ is the length of the wire,
$B = 2 \ T$ is the magnetic field induction,
and $\theta = 90^{\circ}$ is the angle between the wire and the magnetic field.
Substituting these values into the formula:
$F = 1.2 \times 0.5 \times 2 \times \sin 90^{\circ}$
$F = 1.2 \times 0.5 \times 2 \times 1$
$F = 1.2 \times 1$
$F = 1.2 \ N$.
Therefore,the force acting on the wire is $1.2 \ N$.

(C) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{\ell} \times \vec{B})$.
Since the wire is along the $x$-axis,its length vector is $\vec{\ell} = \ell \hat{i}$.
Given $\vec{B} = B(\hat{i} + 2\hat{j} - 3\hat{k})$.
Substituting these into the formula: $\vec{F} = I(\ell \hat{i}) \times B(\hat{i} + 2\hat{j} - 3\hat{k})$.
$\vec{F} = I \ell B [(\hat{i} \times \hat{i}) + 2(\hat{i} \times \hat{j}) - 3(\hat{i} \times \hat{k})]$.
Using cross product rules: $\hat{i} \times \hat{i} = 0$,$\hat{i} \times \hat{j} = \hat{k}$,and $\hat{i} \times \hat{k} = -\hat{j}$.
$\vec{F} = I \ell B [0 + 2\hat{k} - 3(-\hat{j})] = I \ell B (3\hat{j} + 2\hat{k})$.
The magnitude of the force is $F = |\vec{F}| = I \ell B \sqrt{3^2 + 2^2}$.
$F = I \ell B \sqrt{9 + 4} = \sqrt{13} I \ell B$.

(D) The force on a current-carrying conductor in a magnetic field is given by $F = BIL \sin(\theta)$.
Since the magnetic field is normal to the conductor,$\theta = 90^\circ$,so $F = BIL$.
For the tension in the wires to be zero,the magnetic force must balance the weight of the rod.
Therefore,$BIL = Mg$.
Solving for $B$,we get $B = \frac{Mg}{IL}$.

(B) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L} \times \vec{B})$, where $L$ is the length vector of the wire.
Let the sides of the triangle be $EF = 5 \,cm$, $EG = 12 \,cm$, and $GF = 13 \,cm$.
The magnetic field $\vec{B}$ is parallel to the side $GF$. Let $\theta$ be the angle between the side $EF$ and the side $GF$.
From the geometry of the triangle, $\sin \theta = \frac{EG}{GF} = \frac{12}{13}$.
The force on the side $EF$ is $F_{EF} = I \cdot L_{EF} \cdot B \sin \theta$.
Given $I = 2 \,A$, $L_{EF} = 0.05 \,m$, $B = 0.75 \,T$, and $\sin \theta = \frac{12}{13}$.
$F_{EF} = 2 \times 0.05 \times 0.75 \times \frac{12}{13} = 0.1 \times 0.75 \times \frac{12}{13} = 0.075 \times \frac{12}{13} = \frac{0.9}{13} = \frac{9}{130} \,N$.
Comparing this with $\frac{x}{130} \,N$, we get $x = 9$.

(C) When two parallel conductors carry currents $i_1$ and $i_2$ in the same direction,the magnetic field produced by the first conductor at the location of the second conductor is given by $B_1 = \frac{\mu_0 i_1}{2 \pi d}$.
According to the Lorentz force law,the force per unit length on the second conductor is $F = i_2 l B_1 \sin(90^\circ) = i_2 l \left( \frac{\mu_0 i_1}{2 \pi d} \right)$.
Using the right-hand rule,the direction of the force on the second conductor is towards the first conductor.
Similarly,the force on the first conductor is towards the second conductor.
Therefore,the two conductors attract each other.

(C) The magnetic force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $h$ is given by $F/l = \frac{\mu_0 I_1 I_2}{2 \pi h}$.
Since the currents are in opposite directions,the force is repulsive.
At equilibrium,the repulsive magnetic force balances the gravitational force acting on wire $PQ$:
$F = mg$
$\frac{\mu_0 I^2 l}{2 \pi h} = mg$
Given: $I = 25 \,A$,$l = 1 \,m$,$m = 2.5 \,g = 2.5 \times 10^{-3} \,kg$,$g = 9.8 \,m/s^2$,$\mu_0 = 4 \pi \times 10^{-7} \,T \cdot m/A$.
Substituting the values:
$h = \frac{\mu_0 I^2 l}{2 \pi m g} = \frac{(4 \pi \times 10^{-7}) \times (25)^2 \times 1}{2 \pi \times (2.5 \times 10^{-3}) \times 9.8}$
$h = \frac{2 \times 10^{-7} \times 625}{2.5 \times 10^{-3} \times 9.8} = \frac{1250 \times 10^{-7}}{24.5 \times 10^{-3}} = \frac{1250}{24.5} \times 10^{-4} \,m \approx 51 \times 10^{-4} \,m \approx 5.1 \,mm$.
Rounding to the nearest given option,the height is $5 \,mm$.

(B) The force per unit length between two parallel current-carrying wires is given by the formula: $F = \frac{\mu_0 I_1 I_2 l}{2 \pi d}$.
Given:
$I_1 = 10 \ A$
$I_2 = 4 \ A$
$d = 2 \ cm = 2 \times 10^{-2} \ m$
$l = 4 \ cm = 4 \times 10^{-2} \ m$
$\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$
Substituting the values into the formula:
$F = \frac{(4 \pi \times 10^{-7}) \times 10 \times 4 \times (4 \times 10^{-2})}{2 \pi \times (2 \times 10^{-2})}$
$F = \frac{2 \times 10^{-7} \times 40 \times 4 \times 10^{-2}}{2 \times 10^{-2}}$
$F = 2 \times 10^{-7} \times 40 \times 2$
$F = 160 \times 10^{-7} \ N = 1.6 \times 10^{-5} \ N$.
Thus,the correct option is $B$.

(B) The magnetic force on a current-carrying conductor is given by the formula $F = i l B \sin(\theta)$.
Here,the current $i = 2 \text{ A}$,the magnetic field $B = 3.0 \times 10^{-5} \text{ T}$,and the angle $\theta$ between the current (east-west) and the magnetic field (south-north) is $90^\circ$.
Since $\sin(90^\circ) = 1$,the force per unit length is given by:
$\frac{F}{l} = i B \sin(90^\circ)$
$\frac{F}{l} = 2 \times 3.0 \times 10^{-5} \times 1$
$\frac{F}{l} = 6 \times 10^{-5} \text{ N/m}$.

(B) The force per unit length between two parallel current-carrying wires is given by the formula: $\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
Given: $I_1 = I_2 = 5 \text{ A}$,$d = 1 \text{ m}$,and $\mu_0 = 4\pi \times 10^{-7} \text{ T m/A}$.
Substituting the values: $\frac{F}{l} = \frac{4\pi \times 10^{-7} \times 5 \times 5}{2 \pi \times 1}$.
Simplifying the expression: $\frac{F}{l} = 2 \times 10^{-7} \times 25 = 50 \times 10^{-7} \text{ N/m}$.
Thus,$\frac{F}{l} = 5 \times 10^{-6} \text{ N/m}$.
Since the currents are in the same direction,the force between the wires is attractive.

(B) $1$. Two parallel wires carrying currents in the same direction attract each other,while those carrying currents in opposite directions repel each other.
$2$. In the figure,the current in wire $P$ is downward $(20 \ A)$,while the current in wire $Q$ is upward $(40 \ A)$. Since these currents are anti-parallel,wire $P$ exerts a repulsive force on wire $Q$,pushing it to the right.
$3$. The current in wire $Q$ is upward $(40 \ A)$,and the current in wire $R$ is also upward $(60 \ A)$. Since these currents are parallel,wire $R$ exerts an attractive force on wire $Q$,pulling it to the right.
$4$. Since both forces (repulsion from $P$ and attraction from $R$) act towards the right,the resultant force on wire $Q$ is directed to the right.

(B) When two long parallel wires carry currents in the same direction,each wire produces a magnetic field at the location of the other wire.
According to the right-hand rule,the magnetic field produced by one wire exerts a Lorentz force on the current-carrying second wire.
For currents in the same direction,the force between the wires is attractive.
Therefore,the wires will attract each other.

(D) The magnetic force per unit length between two parallel wires is given by $\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2 \pi y}$.
For wire $B$ to remain at rest,the upward magnetic force must balance the downward gravitational force (weight per unit length).
$\frac{F}{l} = \text{weight per unit length} = 40 \times 10^{-3} \text{ N/m}$.
Substituting the values: $40 \times 10^{-3} = \frac{4 \pi \times 10^{-7} \times 10 \times 20}{2 \pi \times y}$.
Simplifying the equation: $40 \times 10^{-3} = \frac{2 \times 10^{-7} \times 200}{y}$.
$40 \times 10^{-3} = \frac{400 \times 10^{-7}}{y}$.
$y = \frac{400 \times 10^{-7}}{40 \times 10^{-3}} = 10 \times 10^{-4} = 10^{-3} \text{ m}$.
Since the magnetic force must be upward (repulsive) to counteract gravity,the currents must flow in opposite directions.

(C) The force per unit length between two long parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $y$ is given by the formula:
$F = \frac{\mu_0 I_1 I_2}{2 \pi y}$
Given that both wires carry the same current $I$,the initial force is:
$F = \frac{\mu_0 I^2}{2 \pi y}$
If the current in each wire is halved,the new current becomes $I' = \frac{I}{2}$.
The new force $F'$ is:
$F' = \frac{\mu_0 (I/2) (I/2)}{2 \pi y} = \frac{\mu_0 I^2}{4 \cdot 2 \pi y}$
Substituting the original force $F$ into this equation:
$F' = \frac{1}{4} F$
Therefore,the force becomes $\frac{F}{4}$.

(D) The force per unit length between two parallel current-carrying wires is given by the formula:
$F = \frac{\mu_0 I_1 I_2 l}{2 \pi y}$
Given:
$I_1 = 10 \ A$
$I_2 = 2 \ A$
$l = 2 \ m$
$y = 10 \ cm = 0.1 \ m = 10 \times 10^{-2} \ m$
$\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$
Substituting the values into the formula:
$F = \frac{(4 \pi \times 10^{-7}) \times 10 \times 2 \times 2}{2 \pi \times 10 \times 10^{-2}}$
$F = \frac{4 \pi \times 10^{-7} \times 40}{2 \pi \times 0.1}$
$F = 2 \times 10^{-7} \times 400$
$F = 800 \times 10^{-7} \ N = 8 \times 10^{-5} \ N$
Therefore,the force acting on conductor $B$ is $8 \times 10^{-5} \ N$.

(B) The force per unit length $f$ between two parallel conductors carrying currents $I_1$ and $I_2$ separated by a distance $d$ is given by the formula: $f = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
Given values are $I_1 = I_2 = 1 \text{ mA} = 10^{-3} \text{ A}$,$d = 1 \text{ m}$,and $\mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1}$.
Substituting these values into the formula:
$f = \frac{(4\pi \times 10^{-7}) \times (10^{-3}) \times (10^{-3})}{2 \pi \times 1}$
$f = 2 \times 10^{-7} \times 10^{-6}$
$f = 2 \times 10^{-13} \text{ N/m}$.
Thus,the correct option is $B$.

(A) For the wire to be suspended in mid-air,the magnetic force acting on it must balance the gravitational force acting downwards.
$F_B = F_g$
$I B l = m g$
Where $I = 2.5 \text{ A}$,$B = 0.5 \text{ T}$,$l = 50 \text{ cm} = 0.5 \text{ m}$,and $g = 10 \text{ m s}^{-2}$.
$m = \frac{I B l}{g}$
$m = \frac{2.5 \times 0.5 \times 0.5}{10}$
$m = \frac{0.625}{10} = 0.0625 \text{ kg}$
Converting to grams: $0.0625 \text{ kg} \times 1000 = 62.5 \text{ g}$.
Thus,the mass of the wire is $62.5 \text{ g}$.

(B) The magnetic force per unit length between two long parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $r$ is given by the formula:
$\frac{F}{\ell} = \frac{\mu_0 I_1 I_2}{2 \pi r}$
In this case,both wires carry the same current $I$,so $I_1 = I_2 = I$. Thus,the force per unit length is:
$\frac{F}{\ell} = \frac{\mu_0 I^2}{2 \pi r}$
According to the right-hand rule,parallel wires carrying currents in the same direction attract each other,while wires carrying currents in opposite directions repel each other.
Since the currents are in opposite directions,the wires will repel each other.

(C) Given, magnetic field, $B = 0.25 \,T$
Mass per unit length, $\frac{m}{l} = 0.5 \,kg \,m^{-1}$
Angle of inclination, $\theta = 30^{\circ}$
Acceleration due to gravity, $g = 9.8 \,m \,s^{-2}$ (or $10 \,m \,s^{-2}$)
The magnetic force $F$ on a current-carrying conductor in a magnetic field is given by $F = BIl$. Since the magnetic field is vertical, the force $F$ acts horizontally.
For the rod to remain stationary on the smooth inclined plane, the component of the magnetic force along the plane must balance the component of the gravitational force acting down the plane.
Component of gravitational force down the plane = $mg \sin 30^{\circ}$
Component of magnetic force up the plane = $F \cos 30^{\circ} = (BIl) \cos 30^{\circ}$
Equating the two forces for equilibrium:
$BIl \cos 30^{\circ} = mg \sin 30^{\circ}$
$BIl \left(\frac{\sqrt{3}}{2}\right) = mg \left(\frac{1}{2}\right)$
$BIl \sqrt{3} = mg$
$I = \frac{mg}{l \sqrt{3} B} = \left(\frac{m}{l}\right) \frac{g}{\sqrt{3} B}$
Substituting the values:
$I = 0.5 \times \frac{9.8}{\sqrt{3} \times 0.25} \approx 11.316 \,A \approx 11.32 \,A$
Thus, the required current is $11.32 \,A$.

(A) Given:
$r = 10 \,cm = 0.1 \,m$
$I_1 = I_2 = 10 \,A$
The force per unit length between two parallel current-carrying wires is given by the formula:
$\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2 \pi r}$
Substituting the values:
$\frac{F}{l} = \frac{(4 \pi \times 10^{-7} \,T \cdot m/A) \times (10 \,A) \times (10 \,A)}{2 \pi \times 0.1 \,m}$
$\frac{F}{l} = \frac{2 \times 10^{-7} \times 100}{0.1} = 2 \times 10^{-4} \,N/m$
Since the currents are in the same direction,the force is attractive.

(A) The magnetic field produced by a circular loop carrying current $I_2$ at any point on its axis is directed along the axis itself.
When a straight conductor carrying current $I_1$ is placed along the axis of this loop,the current in the straight wire flows parallel to the magnetic field lines produced by the loop.
The magnetic force on a current-carrying element is given by $\vec{F} = I(\vec{dl} \times \vec{B})$.
Since the current in the straight conductor is parallel to the magnetic field $\vec{B}$ produced by the loop,the angle $\theta$ between the current element $I_1\vec{dl}$ and the magnetic field $\vec{B}$ is $0^\circ$ or $180^\circ$.
Therefore,the force $F = I_1 dl B \sin(\theta) = 0$.
By Newton's third law,the force exerted by the loop on the straight conductor is zero,and consequently,the force exerted by the straight conductor on the loop is also zero.

(D) The force on the horizontal segments $PQ$ and $RS$ are equal in magnitude but opposite in direction,so their net force is zero.
The force between two parallel current-carrying conductors is given by $F = \frac{\mu_0 I_1 I_2 l}{2 \pi r}$,where $r$ is the distance between the conductors.
For segment $PS$ (at distance $r_1 = 2 \text{ cm} = 0.02 \text{ m}$):
$F_{PS} = \frac{2 \times 10^{-7} \times 20 \times 20 \times 0.15}{0.02} = 6 \times 10^{-4} \text{ N}$ (Attractive force towards the wire).
For segment $QR$ (at distance $r_2 = 2 \text{ cm} + 10 \text{ cm} = 12 \text{ cm} = 0.12 \text{ m}$):
$F_{QR} = \frac{2 \times 10^{-7} \times 20 \times 20 \times 0.15}{0.12} = 1 \times 10^{-4} \text{ N}$ (Repulsive force away from the wire).
The net force is $F_{\text{net}} = F_{PS} - F_{QR} = 6 \times 10^{-4} - 1 \times 10^{-4} = 5 \times 10^{-4} \text{ N}$.

(D) The force per unit length between two parallel conductors carrying currents $I_1$ and $I_2$ separated by distance $r$ is given by $f = \frac{\mu_0 I_1 I_2}{2 \pi r}$. The force is attractive if currents are in the same direction.
Force on $C$ due to $A$ $(F_{AC})$:
$F_{AC} = \frac{\mu_0 I_A I_C L}{2 \pi r_{AC}} = \frac{2 \times 10^{-7} \times 2 \times 3 \times 1}{0.05} = \frac{12 \times 10^{-7}}{0.05} = 2.4 \times 10^{-5} \, N$ (towards $A$, attractive).
Force on $C$ due to $B$ $(F_{BC})$:
$F_{BC} = \frac{\mu_0 I_B I_C L}{2 \pi r_{BC}} = \frac{2 \times 10^{-7} \times 4 \times 3 \times 1}{0.08} = \frac{24 \times 10^{-7}}{0.08} = 3.0 \times 10^{-5} \, N$ (towards $B$, attractive).
Since $F_{BC} > F_{AC}$, the net force is $F_{net} = F_{BC} - F_{AC} = (3.0 - 2.4) \times 10^{-5} \, N = 0.6 \times 10^{-5} \, N$ towards $B$.

(A) The force per unit length $f$ between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $d$ is given by the formula:
$f = \frac{\mu_0}{4\pi} \frac{2 I_1 I_2}{d}$
Given values are $I_1 = 1 \text{ A}$,$I_2 = 3 \text{ A}$,$d = 1 \text{ m}$,and $\frac{\mu_0}{4\pi} = 10^{-7} \text{ T m A}^{-1}$.
Substituting these values into the formula:
$f = 10^{-7} \times \frac{2 \times 1 \times 3}{1} = 6 \times 10^{-7} \text{ N m}^{-1}$
Since the currents flow in opposite directions,the force between the wires is repulsive.
Therefore,the force per unit length is $6 \times 10^{-7} \text{ N m}^{-1}$ and it is repulsive.

(A) For the wire to be suspended in mid-air,the magnetic force acting on it must balance its weight.
Let $l$ be the length of the wire,$m$ be its mass,$I$ be the current,$B$ be the magnetic field,and $g$ be the acceleration due to gravity.
The weight of the wire is $W = mg$.
Given the linear density $\lambda = \frac{m}{l} = 0.12 \ kg \ m^{-1}$,we can write $m = \lambda l$.
So,$W = \lambda l g$.
The magnetic force on a current-carrying wire is $F_m = IlB \sin(\theta)$.
Since the magnetic field is normal to the wire,$\theta = 90^{\circ}$,so $F_m = IlB$.
For equilibrium,$F_m = W$,which implies $IlB = \lambda l g$.
Canceling $l$ from both sides,we get $IB = \lambda g$.
Solving for $I$: $I = \frac{\lambda g}{B}$.
Substituting the given values: $I = \frac{0.12 \times 10}{0.5} = \frac{1.2}{0.5} = 2.4 \ A$.
Therefore,the current through the wire is $2.4 \ A$.

(C) The magnetic force $F$ on a current-carrying wire of length $L$ in a uniform magnetic field $B$ is given by the formula: $F = I L B \sin(\theta)$.
We need to find the force per unit length,which is $f = F/L$.
Therefore,$f = I B \sin(\theta)$.
Given values are:
Current $I = 8 \ A$
Magnetic field $B = 0.15 \ T$
Angle $\theta = 30^{\circ}$
Substituting these values into the formula:
$f = 8 \times 0.15 \times \sin(30^{\circ})$
Since $\sin(30^{\circ}) = 0.5$,we get:
$f = 8 \times 0.15 \times 0.5$
$f = 8 \times 0.075 = 0.6 \ N \ m^{-1}$.
Thus,the correct option is $C$.

(A) The force on a current-carrying wire near a long straight conductor is given by $F = \frac{\mu_0 I_1 I_2 L}{2 \pi r}$.
Here,$I_1 = 25 \text{ A}$,$I_2 = 10 \text{ A}$,$L = 25 \text{ cm} = 0.25 \text{ m}$.
The two vertical sides of the loop are at distances $r_1 = 10 \text{ cm} = 0.1 \text{ m}$ and $r_2 = 10 \text{ cm} + 10 \text{ cm} = 20 \text{ cm} = 0.2 \text{ m}$ from the straight wire.
The forces on the horizontal sides ($F_3$ and $F_4$) are equal and opposite,so they cancel out.
The net force is the difference between the forces on the two vertical sides: $F_{net} = |F_1 - F_2|$.
$F_1 = \frac{\mu_0 I_1 I_2 L}{2 \pi r_1} = \frac{2 \times 10^{-7} \times 25 \times 10 \times 0.25}{0.1} = 1.25 \times 10^{-4} \text{ N}$ (Attractive).
$F_2 = \frac{\mu_0 I_1 I_2 L}{2 \pi r_2} = \frac{2 \times 10^{-7} \times 25 \times 10 \times 0.25}{0.2} = 0.625 \times 10^{-4} \text{ N}$ (Repulsive).
$F_{net} = 1.25 \times 10^{-4} - 0.625 \times 10^{-4} = 0.625 \times 10^{-4} \text{ N} = 6.25 \times 10^{-5} \text{ N}$.

(C) The force per unit length on a current-carrying conductor in a magnetic field is given by the formula: $\frac{F}{l} = iB \sin \theta$.
Here,the current $i = 1 \,A$ flows from south to north.
The magnetic field $B = 3 \times 10^{-5} \,T$ also points from south to north.
Since the current and the magnetic field are in the same direction,the angle $\theta$ between them is $0^\circ$.
Therefore,the force per unit length is: $\frac{F}{l} = 1 \times (3 \times 10^{-5}) \times \sin(0^\circ) = 1 \times (3 \times 10^{-5}) \times 0 = 0 \,N/m$.

(B) When a flexible wire loop carrying current is placed in an external magnetic field,it experiences magnetic forces that tend to expand the loop.
To maximize the magnetic flux linked with the loop,the loop tends to enclose the maximum possible area for a given perimeter.
According to geometric principles,for a fixed perimeter,a circle encloses the maximum area.
Therefore,the loop changes its shape to a circular shape with its plane normal to the magnetic field.

(C) The magnetic force on a current-carrying conductor of length $\vec{L}$ in a uniform magnetic field $\vec{B}$ is given by $\vec{F} = I(\vec{L} \times \vec{B})$.
For any closed current-carrying loop placed in a uniform magnetic field,the net magnetic force is always zero.
Mathematically,$\vec{F}_{net} = I \oint (d\vec{l} \times \vec{B}) = I (\oint d\vec{l}) \times \vec{B}$.
Since the loop is closed,the vector sum of all length elements $\oint d\vec{l} = 0$.
Therefore,$\vec{F}_{net} = 0$.

(A) The force on a current-carrying conductor due to a long straight wire is given by $F = \frac{\mu_0 I_1 I_2 L}{2 \pi r}$.
For the rectangular coil $ABCD$,the forces on the horizontal segments $BC$ and $AD$ are equal and opposite,so they cancel each other out.
The force on segment $AB$ (at distance $r_1 = 2 \text{ cm} = 0.02 \text{ m}$) is attractive (towards the wire) because the currents are parallel:
$F_{AB} = \frac{\mu_0 (2 \text{ A})(1 \text{ A})(0.15 \text{ m})}{2 \pi (0.02 \text{ m})} = \frac{2 \times 10^{-7} \times 2 \times 1 \times 0.15}{0.02} = 30 \times 10^{-7} \text{ N}$.
The force on segment $CD$ (at distance $r_2 = 2 \text{ cm} + 10 \text{ cm} = 12 \text{ cm} = 0.12 \text{ m}$) is repulsive (away from the wire) because the currents are anti-parallel:
$F_{CD} = \frac{\mu_0 (2 \text{ A})(1 \text{ A})(0.15 \text{ m})}{2 \pi (0.12 \text{ m})} = \frac{2 \times 10^{-7} \times 2 \times 1 \times 0.15}{0.12} = 5 \times 10^{-7} \text{ N}$.
The net force is $F_{net} = F_{AB} - F_{CD} = 30 \times 10^{-7} - 5 \times 10^{-7} = 25 \times 10^{-7} \text{ N}$.
Since $F_{AB} > F_{CD}$,the net force is directed towards the wire.

(A) The magnetic force on a current-carrying wire segment of length vector $\vec{L}$ in a uniform magnetic field $\vec{B}$ is given by $\vec{F} = i(\vec{L} \times \vec{B})$.
For a closed loop in a uniform magnetic field,the net force is $\vec{F}_{net} = i(\oint d\vec{l}) \times \vec{B}$. Since the integral of $d\vec{l}$ over a closed loop is zero,the net force on any closed loop in a uniform magnetic field is zero.
However,in this problem,the current enters at $C$ and leaves at $A$. The triangle is not a closed loop for the current flow. We can treat this as three segments: $CB$,$BA$,and $AC$.
Using the principle of superposition,the net force is the vector sum of forces on each segment: $\vec{F}_{net} = \vec{F}_{CB} + \vec{F}_{BA} + \vec{F}_{AC}$.
Each segment has length $l$ and is perpendicular to the field $B$. The force on each segment is $F = ilB$ directed radially outward or inward depending on the current direction.
By symmetry and vector addition,the resultant force on the segments $CB$ and $BA$ effectively acts as a force on the displacement vector from $C$ to $A$,which is $l$. Thus,the net force is equivalent to the force on a straight wire of length $l$ carrying current $i$ from $C$ to $A$,which is $ilB$.

(B) The force per unit length between two parallel conductors carrying currents $i_1$ and $i_2$ separated by a distance $x$ is given by $F = \frac{\mu_0 i_1 i_2}{2 \pi x}$.
Since the currents are in the same direction,the force is attractive.
To increase the distance from $d$ to $2d$,we must perform work against this attractive force.
The work done per unit length is given by the integral of the force with respect to distance:
$W = \int_{d}^{2d} F \, dx = \int_{d}^{2d} \frac{\mu_0 i_1 i_2}{2 \pi x} \, dx$
$W = \frac{\mu_0 i_1 i_2}{2 \pi} [\ln(x)]_{d}^{2d}$
$W = \frac{\mu_0 i_1 i_2}{2 \pi} (\ln(2d) - \ln(d))$
$W = \frac{\mu_0 i_1 i_2}{2 \pi} \ln\left(\frac{2d}{d}\right) = \frac{\mu_0 i_1 i_2}{2 \pi} \ln(2)$.

(D) The magnetic field $B$ produced by an infinitely long straight wire carrying current $i_1$ at a distance $r$ is given by $B = \frac{\mu_0 i_1}{2\pi r}$.
This magnetic field is directed perpendicular to the plane of the loop (into or out of the page).
The force on each segment of the rectangular loop is given by $F = \int i_2 (dl \times B)$.
For the two vertical segments of length $l$,the forces are $F_1 = \frac{\mu_0 i_1 i_2 l}{2\pi a}$ (attractive) and $F_2 = \frac{\mu_0 i_1 i_2 l}{2\pi b}$ (repulsive).
For the two horizontal segments,the forces are equal and opposite,thus canceling each other out.
Since all these forces act in the plane of the loop and their lines of action pass through the center of the loop (or are parallel to the plane of the loop),the net torque about any axis in the plane of the loop is zero.
Specifically,the magnetic field is perpendicular to the plane of the loop,so the area vector $A$ is parallel to the magnetic field $B$. The torque is given by $\tau = m \times B = mB \sin \theta$. Since $\theta = 0^{\circ}$,the torque $\tau = 0$.

(D) 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.
From the parabola equation $y^2 = 2x$,at $x = 2$,$y^2 = 4$,so $y = \pm 2$. The points are $A(2, 2)$ and $B(2, -2)$.
The current flows from $A$ to $B$ through the origin $O$. Thus,the effective length vector $\vec{L}$ is the vector from $A$ to $B$,which is $\vec{L} = (2-2)\hat{i} + (-2-2)\hat{j} = -4\hat{j} \ m$.
Given $I = 4 \ A$ and $\vec{B} = 6\hat{k} \ T$.
Substituting these values: $\vec{F} = 4 \times (-4\hat{j} \times 6\hat{k}) = 4 \times (-24)(\hat{j} \times \hat{k}) = -96\hat{i} \ N$.

(C) Let the tension developed in the wire be $T$ and the radius of the circular loop be $r$.
Consider a small element of the wire of length $dl$ subtending an angle $d\theta$ at the center.
The magnetic force acting on this element is $dF = B I dl = B I (r d\theta)$.
This force is directed radially outward.
The tension $T$ at the ends of this element provides a restoring force directed radially inward.
The component of tension providing the inward force is $2 T \sin(d\theta / 2) \approx 2 T (d\theta / 2) = T d\theta$.
Equating the forces for equilibrium: $T d\theta = B I r d\theta$, which gives $T = B I r$.
Since the total length of the wire is $l = 2 \pi r$, we have $r = l / (2 \pi)$.
Substituting $r$ in the tension formula: $T = B I (l / 2 \pi)$.
Given $B = 2.0 \,T$, $I = 1.1 \,A$, and $l = 1 \,m$:
$T = (2.0 \times 1.1 \times 1) / (2 \times 3.14) = 2.2 / 6.28 \approx 0.35 \,N$.

(C) The force per unit length between two long parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $x$ is given by the formula: $\frac{F}{\ell} = \frac{\mu_0 I_1 I_2}{2 \pi x}$.
To change the distance between the wires by an infinitesimal amount $dx$,the work done per unit length $dW$ is equal to the force per unit length multiplied by the displacement: $dW = \frac{F}{\ell} dx$.
Substituting the expression for force per unit length: $dW = \frac{\mu_0 I_1 I_2}{2 \pi x} dx$.
To find the total work done per unit length $W$ to change the distance,we integrate the expression: $W = \int \frac{\mu_0 I_1 I_2}{2 \pi x} dx$.
Since $\mu_0$,$I_1$,$I_2$,and $2\pi$ are constants,we get: $W = \frac{\mu_0 I_1 I_2}{2 \pi} \int \frac{1}{x} dx$.
Integrating $\frac{1}{x}$ gives $\log_e x$,so $W = \frac{\mu_0 I_1 I_2}{2 \pi} \log_e x$.
Therefore,the work done per unit length is proportional to $\log_e x$.

(C) The total current $I = 20 \text{ A}$ enters the loop and splits into two parallel branches,$ABC$ and $ADC$. Since the wire is uniform,the resistance of both branches is equal,so the current divides equally: $i_1 = i_2 = 10 \text{ A}$.
The force per unit length between two parallel wires carrying currents $i_1$ and $i_2$ separated by a distance $r$ is given by:
$\frac{F}{L} = \frac{\mu_0 i_1 i_2}{2 \pi r}$
Given:
$i_1 = 10 \text{ A}$
$i_2 = 10 \text{ A}$
$r = AD = BC = 2 \text{ cm} = 2 \times 10^{-2} \text{ m}$
$\mu_0 = 4 \pi \times 10^{-7} \text{ T m/A}$
Substituting the values:
$\frac{F}{L} = \frac{(4 \pi \times 10^{-7}) \times 10 \times 10}{2 \pi \times (2 \times 10^{-2})}$
$\frac{F}{L} = \frac{4 \pi \times 10^{-5}}{4 \pi \times 10^{-2}}$
$\frac{F}{L} = 10^{-3} \text{ N m}^{-1}$.

(A) The magnetic field $B$ produced by the first wire at a distance $l$ is given by $B = \frac{\mu_0 i_1}{2 \pi l}$.
Since the wires are perpendicular,the angle between the magnetic field and the current $i_2$ is $90^{\circ}$.
The magnetic force $dF$ on a small length $d$ of the second wire is given by $dF = i_2 (B) d \sin(90^{\circ})$.
Substituting the value of $B$,we get $dF = i_2 \left( \frac{\mu_0 i_1}{2 \pi l} \right) d (1) = \frac{\mu_0 i_1 i_2 d}{2 \pi l}$.
Therefore,the magnetic force is proportional to the product of the currents,$i_1 i_2$.

(D) The length of the wire is $L = 30 \sqrt{3} \text{ cm} = 0.3 \sqrt{3} \text{ m}$.
Since it is an equilateral triangle,the length of each side is $l = L / 3 = 0.1 \sqrt{3} \text{ m}$.
The magnetic field $\vec{B}$ is parallel to side $BC$. The angle between the side $AC$ and the magnetic field $\vec{B}$ is $60^\circ$ (as the triangle is equilateral).
The force on a current-carrying conductor is given by $\vec{F} = I(\vec{l} \times \vec{B})$,so the magnitude is $F = I l B \sin \theta$.
Here,$I = 2 \text{ A}$,$l = 0.1 \sqrt{3} \text{ m}$,$B = 2 \text{ T}$,and $\theta = 60^\circ$.
$F = 2 \times (0.1 \sqrt{3}) \times 2 \times \sin(60^\circ) = 0.4 \sqrt{3} \times (\sqrt{3} / 2) = 0.2 \times 3 = 0.6 \text{ N}$.

(B) The 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 effective displacement vector from the starting point $P$ to the ending point $Q$.
From the figure,the wire consists of three segments. The total displacement vector $\vec{L}_{eff}$ is the vector sum of these segments.
Horizontal displacement $= 6 \ cm = 0.06 \ m$ (to the right).
Vertical displacement $= 4 \ cm + 4 \ cm = 8 \ cm = 0.08 \ m$ (upwards).
Thus,the magnitude of the effective length is $L_{eff} = \sqrt{(0.06)^2 + (0.08)^2} = \sqrt{0.0036 + 0.0064} = \sqrt{0.01} = 0.1 \ m$.
The magnetic field $B = 5 \ T$ is perpendicular to the plane of the wire.
The magnitude of the force is $F = I L_{eff} B \sin(90^\circ) = 10 \ A \times 0.1 \ m \times 5 \ T \times 1 = 5 \ N$.

(D) The force per unit length between two parallel wires carrying currents $i_1$ and $i_2$ separated by distance $r$ is given by $f = \frac{\mu_0 i_1 i_2}{2 \pi r}$.
Taking the direction of current in wire $A$ as positive,the force per unit length on wire $C$ due to wires $A$ and $B$ is:
$\vec{F} = \left( \frac{\mu_0 (3i)(2i)}{2 \pi (2d)} \right) - \left( \frac{\mu_0 (i)(2i)}{2 \pi d} \right) = \frac{\mu_0 i^2}{2 \pi d} \left( \frac{3}{2} - 2 \right) = -\frac{\mu_0 i^2}{4 \pi d}$.
Thus,$\vec{F} = -\frac{\mu_0 i^2}{4 \pi d}$,which implies $\frac{\mu_0 i^2}{4 \pi d} = -\vec{F}$.
Now,if wire $B$ is removed,the force per unit length on wire $A$ due to wire $C$ is:
$f_A = \frac{\mu_0 (3i)(2i)}{2 \pi (2d)} = \frac{6 \mu_0 i^2}{4 \pi d} = 3 \left( \frac{\mu_0 i^2}{2 \pi d} \right) = 6 \left( \frac{\mu_0 i^2}{4 \pi d} \right)$.
Substituting $\frac{\mu_0 i^2}{4 \pi d} = -\vec{F}$,we get $f_A = 6(-\vec{F}) = -6\vec{F}$.
Re-evaluating the initial force $\vec{F}$ direction and magnitude based on the provided options,the correct relation is $f_A = -3\vec{F}$.

(B) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L} \times \vec{B})$.
For the bottom side of the loop of length '$a$',the magnetic force is $F = IaB$.
According to Fleming's Left-Hand Rule,if the current flows from left to right,the force is directed upwards. If the current flows from right to left,the force is directed downwards.
Let the initial reading of the spring balance be $W_1 = mg - F$ (assuming upward force).
When the current direction is reversed,the force becomes $F' = -F$ (downward force).
The new reading of the spring balance becomes $W_2 = mg + F$.
The change in the reading of the spring balance is $\Delta W = |W_2 - W_1| = |(mg + F) - (mg - F)| = 2F = 2IaB$.

(A) Since the current in both wires flows in the same direction,the magnetic force between them is attractive.
To keep the lower wire suspended in the air,the upward magnetic force per unit length must balance the downward gravitational force per unit length.
Let $I_1 = 160 \ A$ be the current in the upper wire,$I_2$ be the current in the lower wire,$d = 4 \ cm = 4 \times 10^{-2} \ m$ be the distance,and $\lambda = 10 \ g \ m^{-1} = 10 \times 10^{-3} \ kg \ m^{-1}$ be the linear mass density.
The condition for equilibrium is:
$F_{magnetic} = F_{gravitational}$
$\frac{\mu_0 I_1 I_2}{2 \pi d} = \lambda g$
Rearranging for $I_2$:
$I_2 = \frac{2 \pi d \lambda g}{\mu_0 I_1}$
Substituting the given values:
$I_2 = \frac{2 \pi \times (4 \times 10^{-2}) \times (10 \times 10^{-3}) \times 10}{4 \pi \times 10^{-7} \times 160}$
$I_2 = \frac{8 \pi \times 10^{-3} \times 10}{4 \pi \times 10^{-7} \times 160} = \frac{8 \times 10^{-2}}{4 \times 10^{-7} \times 160} = \frac{2 \times 10^5}{160} = \frac{200000}{160} = 125 \ A$
Thus,the current in the lower wire is $125 \ A$.

(A) The force per unit length between two parallel current-carrying conductors is given by the formula: $\frac{F}{l} = \frac{\mu_0 i_1 i_2}{2 \pi d}$.
Given: $l = 50 \ m$,$d = 0.2 \ m$,$F = 1 \ N$,and $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$.
Let the current in the second conductor be $i$. Then the current in the first conductor is $i_1 = 2i$.
Substituting the values into the formula:
$\frac{1}{50} = \frac{4 \pi \times 10^{-7} \times (2i) \times i}{2 \pi \times 0.2}$
$\frac{1}{50} = 2 \times 10^{-7} \times \frac{2i^2}{0.2}$
$0.02 = 2 \times 10^{-6} \times i^2$
$i^2 = \frac{0.02}{2 \times 10^{-6}} = 0.01 \times 10^6 = 10^4$
$i = \sqrt{10^4} = 100 \ A$.
Therefore,the current in the second conductor is $100 \ A$.

(A) The force per unit length on wire $a$ due to wire $b$ is $f_{ab} = \frac{\mu_0 i_a i_b}{2 \pi d_1}$. Since currents are in the same direction,this force is attractive.
The force per unit length on wire $a$ due to wire $c$ is $f_{ac} = \frac{\mu_0 i_a i_c}{2 \pi (d_1 + d_2)}$. Since currents are in opposite directions,this force is repulsive.
Given $d_2 = 2 d_1$,$i_b = i_a$,and $i_c = 4 i_a$,the net force per unit length on wire $a$ is:
$f_{net} = f_{ab} - f_{ac} = \frac{\mu_0 i_a^2}{2 \pi d_1} - \frac{\mu_0 i_a (4 i_a)}{2 \pi (d_1 + 2 d_1)}$
$f_{net} = \frac{\mu_0 i_a^2}{2 \pi d_1} - \frac{4 \mu_0 i_a^2}{2 \pi (3 d_1)} = \frac{\mu_0 i_a^2}{2 \pi d_1} \left( 1 - \frac{4}{3} \right) = \frac{\mu_0 i_a^2}{2 \pi d_1} \left( -\frac{1}{3} \right)$
The magnitude of the force on length $l$ is $F = |f_{net}| \times l = \frac{\mu_0 i_a^2 l}{6 \pi d_1}$.

(C) The magnetic force on a current-carrying wire of arbitrary shape in a uniform magnetic field is given by the formula $\vec{F} = I(\vec{L} \times \vec{B})$,where $\vec{L}$ is the vector displacement from the starting point $A$ to the ending point $B$.
Since the points $A$ and $B$ are fixed,the displacement vector $\vec{L}$ remains the same regardless of the path (shape) taken by the wire between these two points.
Therefore,the magnetic force depends only on the current $I$,the magnetic field $\vec{B}$,and the straight-line displacement vector $\vec{L}$ between the fixed points. It is independent of the actual shape of the wire.

(B) The force on a current-carrying wire in a magnetic field is given by $F = I \vec{L}_{eff} \times \vec{B}$, where $L_{eff}$ is the effective length (displacement vector) between the two ends of the wire.
For a semi-circular loop of radius $R$, the effective length is the diameter, $L_{eff} = 2R$.
Given: $R = 30 \,cm = 0.3 \,m$, $I = 6 \,A$, $B = 0.5 \,T$.
The effective length $L_{eff} = 2 \times 0.3 \,m = 0.6 \,m$.
Since the magnetic field is perpendicular to the plane of the loop, the angle between the effective length vector and the magnetic field is $90^{\circ}$.
Therefore, the magnitude of the force is $F = I L_{eff} B \sin(90^{\circ})$.
$F = 6 \,A \times 0.6 \,m \times 0.5 \,T \times 1 = 1.8 \,N$.

(A) The force on a current-carrying wire in a magnetic field is given by $\vec{F} = \int I (d\vec{l} \times \vec{B})$.
Given $B = B_0 (1 + \frac{xy}{L^2}) \hat{k}$.
For segment $AB$ $(x=0, y: 0 \to L)$: $\vec{B} = B_0(1+0)\hat{k} = B_0\hat{k}$. $d\vec{l} = dy\hat{j}$. $\vec{F}_{AB} = \int_0^L I(dy\hat{j} \times B_0\hat{k}) = I B_0 L \hat{i}$.
For segment $BC$ $(y=L, x: 0 \to L)$: $\vec{B} = B_0(1+\frac{xL}{L^2})\hat{k} = B_0(1+\frac{x}{L})\hat{k}$. $d\vec{l} = dx\hat{i}$. $\vec{F}_{BC} = \int_0^L I(dx\hat{i} \times B_0(1+\frac{x}{L})\hat{k}) = -I B_0 \int_0^L (1+\frac{x}{L}) dx \hat{j} = -I B_0 [x + \frac{x^2}{2L}]_0^L \hat{j} = -\frac{3}{2} I B_0 L \hat{j}$.
For segment $CD$ $(x=L, y: L \to 0)$: $\vec{B} = B_0(1+\frac{Ly}{L^2})\hat{k} = B_0(1+\frac{y}{L})\hat{k}$. $d\vec{l} = dy(-\hat{j})$. $\vec{F}_{CD} = \int_L^0 I(dy(-\hat{j}) \times B_0(1+\frac{y}{L})\hat{k}) = -I B_0 \int_L^0 (1+\frac{y}{L}) dy \hat{i} = I B_0 [y + \frac{y^2}{2L}]_0^L \hat{i} = \frac{3}{2} I B_0 L \hat{i}$.
For segment $DA$ $(y=0, x: L \to 0)$: $\vec{B} = B_0(1+0)\hat{k} = B_0\hat{k}$. $d\vec{l} = dx(-\hat{i})$. $\vec{F}_{DA} = \int_L^0 I(dx(-\hat{i}) \times B_0\hat{k}) = -I B_0 \int_L^0 dx \hat{j} = I B_0 L \hat{j}$.
Net force $\vec{F}_{net} = \vec{F}_{AB} + \vec{F}_{BC} + \vec{F}_{CD} + \vec{F}_{DA} = (I B_0 L + \frac{3}{2} I B_0 L)\hat{i} + (I B_0 L - \frac{3}{2} I B_0 L)\hat{j} = \frac{5}{2} I B_0 L \hat{i} - \frac{1}{2} I B_0 L \hat{j}$.
The magnitude is $|\vec{F}| = \sqrt{(\frac{5}{2} I B_0 L)^2 + (-\frac{1}{2} I B_0 L)^2} = \frac{I B_0 L}{2} \sqrt{25+1} = \frac{\sqrt{26}}{2} I B_0 L$.