Force on a Current Carrying Conductor Questions in English

(B) When two parallel wires carry currents in the same direction,they attract each other due to the magnetic force.
Conversely,when two parallel wires carry currents in opposite directions,the magnetic fields produced by the wires result in a force that pushes them apart.
Therefore,the wires repel each other.

(C) The magnetic field produced by the long straight wire at a distance $r$ is given by $B = \frac{\mu_0 i}{2\pi r}$.
For the side of the loop closer to the wire at distance $r_1$,the current flows in the same direction as the wire. According to Fleming's left-hand rule,the force $F_1$ on this side is attractive (towards the wire).
For the side of the loop farther from the wire at distance $r_2$,the current flows in the opposite direction to the wire. The force $F_2$ on this side is repulsive (away from the wire).
Since $r_1 < r_2$,the magnetic field $B_1$ at the closer side is stronger than the magnetic field $B_2$ at the farther side $(B_1 > B_2)$.
Consequently,the attractive force $F_1$ is greater than the repulsive force $F_2$ $(F_1 > F_2)$.
Therefore,the net force $F_{net} = F_1 - F_2$ acts towards the wire,and the loop will move towards the wire.

(C) The magnetic force on a small current element $d\overrightarrow{l}$ in a magnetic field $\overrightarrow{B}$ is given by $d\overrightarrow{F} = i(d\overrightarrow{l} \times \overrightarrow{B})$.
For a closed loop,the net magnetic force is $\overrightarrow{F} = \oint i(d\overrightarrow{l} \times \overrightarrow{B})$.
Since the magnetic field $\overrightarrow{B}$ is uniform,it can be taken out of the integral: $\overrightarrow{F} = i(\oint d\overrightarrow{l}) \times \overrightarrow{B}$.
For any closed loop,the vector sum of all infinitesimal length elements $\oint d\overrightarrow{l}$ is zero.
Therefore,the net magnetic force $\overrightarrow{F} = i(0) \times \overrightarrow{B} = 0$.

(B) The magnetic field $B$ produced by one wire at a distance $b$ is given by the formula: $B = \frac{\mu_0 i}{2\pi b}$.
According to the Lorentz force law,the force $F$ on a length $L$ of the second wire carrying current $i$ in this magnetic field is $F = i L B \sin(\theta)$.
Since the wires are parallel,the angle $\theta = 90^\circ$,so $\sin(90^\circ) = 1$.
Therefore,the force per unit length $f = \frac{F}{L} = i B$.
Substituting the value of $B$,we get $f = i \left( \frac{\mu_0 i}{2\pi b} \right) = \frac{\mu_0 i^2}{2\pi b}$.

(A) The force per unit length between two parallel current-carrying wires is given by $F = \frac{\mu_0 i_1 i_2}{2\pi r}$.
For a wire of length $l$,the total force is $F = \frac{\mu_0 i_1 i_2 l}{2\pi r} = \frac{\mu_0}{4\pi} \cdot \frac{2 i_1 i_2 l}{r}$.
Given: $i_1 = 10 \, A$,$i_2 = 2 \, A$,$r = 10 \, cm = 0.1 \, m$,$l = 2 \, m$,and $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$.
Substituting the values:
$F = 10^{-7} \times \frac{2 \times 10 \times 2 \times 2}{0.1} = 10^{-7} \times \frac{80}{0.1} = 10^{-7} \times 800 = 8 \times 10^{-5} \, N$.

(B) When two streams of protons move parallel to each other in the same direction,two types of forces act between them: the electrostatic force and the magnetic force.
$1$. The electrostatic force between two streams of protons is repulsive because both streams consist of positively charged particles.
$2$. The magnetic force between two parallel currents moving in the same direction is attractive.
$3$. However,for charged particles moving in space,the electrostatic repulsive force is significantly stronger than the magnetic attractive force.
$4$. Therefore,the net force between the two streams is repulsive,causing them to repel each other.

(B) The magnetic field produced by a circular current-carrying loop at any point on its axis is directed along the axis itself.
As shown in the figure,the straight wire is placed along the axis of the circular loop.
Therefore,the current in the straight wire flows parallel (or anti-parallel) to the magnetic field produced by the circular loop.
The magnetic force on a current-carrying conductor is given by $\vec{F} = i(\vec{L} \times \vec{B})$.
Since the current element $\vec{L}$ is parallel to the magnetic field $\vec{B}$,the angle $\theta$ between them is $0^\circ$ or $180^\circ$.
Thus,$\vec{F} = iLB \sin(0^\circ) = 0$.
Hence,the force of interaction between the two conductors is zero.

(A) When two parallel wires carry electric 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 at the position of the second wire is perpendicular to the direction of the current in the second wire.
By applying the Lorentz force law $(F = I(L \times B))$,it is determined that the magnetic force acting on each wire is directed towards the other wire.
Therefore,two parallel wires carrying currents in the same direction exert an attractive force on each other.

(A) The force per unit length between two parallel current-carrying wires is given by the formula:
$f = \frac{\mu_0}{4\pi} \frac{2 i_1 i_2}{r}$
Given:
$i_1 = i_2 = 5\, A$
$r = 0.1\, m$
$\frac{\mu_0}{4\pi} = 10^{-7}\, T\cdot m/A$
Substituting the values:
$f = 10^{-7} \times \frac{2 \times 5 \times 5}{0.1}$
$f = 10^{-7} \times \frac{50}{0.1}$
$f = 10^{-7} \times 500 = 5 \times 10^{-5}\, N/m$
Thus,the correct option is $A$.

(C) The force per unit length between two parallel wires carrying currents $i_1$ and $i_2$ separated by distance $a$ is given by $F = \frac{\mu_0}{4\pi} \frac{2 i_1 i_2}{a}$.
Given,$F = 1 \times 10^{-3} \ N$ for $i_1 = 10 \ A$ and $i_2 = 10 \ A$.
When both currents are doubled,the new currents are $i_1' = 2 i_1$ and $i_2' = 2 i_2$.
The new force $F'$ is given by $F' = \frac{\mu_0}{4\pi} \frac{2 (2 i_1) (2 i_2)}{a} = 4 \times \left( \frac{\mu_0}{4\pi} \frac{2 i_1 i_2}{a} \right) = 4 \times F$.
Substituting the value of $F$,we get $F' = 4 \times 1 \times 10^{-3} \ N = 4 \times 10^{-3} \ N$.

(C) The force $F$ experienced by a current-carrying conductor in a magnetic field is given by the formula: $F = B I L \sin \theta$.
Given:
Current $I = 3\, A$,
Length $L = 40\, cm = 0.4\, m$,
Magnetic field $B = 500\, G = 500 \times 10^{-4}\, T = 0.05\, T$,
Angle $\theta = 30^\circ$.
Substituting the values:
$F = (0.05) \times 3 \times 0.4 \times \sin(30^\circ)$
$F = 0.05 \times 3 \times 0.4 \times 0.5$
$F = 0.03\, N = 3 \times 10^{-2}\, N$.

(B) The force per unit length between two long parallel current-carrying wires is given by the formula: $\frac{F}{l} = \frac{\mu_0}{2\pi} \frac{i_1 i_2}{r}$.
Given: $i_1 = 1 \ A$,$i_2 = 1 \ A$,$r = 1 \ m$,and $\frac{\mu_0}{4\pi} = 10^{-7} \ T \cdot m/A$.
Substituting these values into the formula:
$\frac{F}{l} = 2 \times 10^{-7} \times \frac{1 \times 1}{1} = 2 \times 10^{-7} \ N/m$.

(A) The force $F$ on a current-carrying conductor in a magnetic field is given by the formula: $F = BIl \sin \theta$.
Here,$F = 15\, N$,$B = 2\, T$,$I = 10\, A$,and $l = 1.5\, m$.
Substituting these values into the formula:
$15 = 2 \times 10 \times 1.5 \times \sin \theta$
$15 = 30 \times \sin \theta$
$\sin \theta = \frac{15}{30} = \frac{1}{2}$
Since $\sin \theta = \frac{1}{2}$,the angle $\theta = 30^\circ$.

(B) According to the law of magnetic force between two parallel current-carrying conductors,the force per unit length is given by $f = \frac{\mu_0 I_1 I_2}{2\pi d}$.
Since the currents $I_1 = I$ and $I_2 = 10I$ are in the same direction,the force is attractive.
According to Newton's third law of motion,the force exerted by conductor $A$ on $B$ must be equal in magnitude and opposite in direction to the force exerted by conductor $B$ on $A$.
Therefore,both conductors attract each other with the same force.

(A) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $r$ is given by $f = \frac{\mu_0 I_1 I_2}{2\pi r}$. The total force on a length $L$ is $F = \frac{\mu_0 I_1 I_2 L}{2\pi r} = 2 \times 10^{-7} \frac{I_1 I_2 L}{r}$.
$1$. Force on wire $Q$ due to wire $P$ $(I_P = 30\,A, I_Q = 10\,A, r = 0.1\,m, L = 0.1\,m)$:
Since currents are in opposite directions,the force is repulsive (towards the left).
$F_P = 2 \times 10^{-7} \times \frac{30 \times 10 \times 0.1}{0.1} = 6 \times 10^{-5}\,N$ (towards the left).
$2$. Force on wire $Q$ due to wire $R$ $(I_R = 20\,A, I_Q = 10\,A, r = 0.02\,m, L = 0.1\,m)$:
Since currents are in opposite directions,the force is repulsive (towards the right).
$F_R = 2 \times 10^{-7} \times \frac{20 \times 10 \times 0.1}{0.02} = 20 \times 10^{-5}\,N$ (towards the right).
$3$. Net force on wire $Q$:
$F_{net} = F_R - F_P = 20 \times 10^{-5} - 6 \times 10^{-5} = 14 \times 10^{-5}\,N = 1.4 \times 10^{-4}\,N$ (towards the right).

(A) The force $F$ on a current-carrying conductor in a uniform magnetic field is given by the formula: $F = B I L \sin \theta$,where $B$ is the magnetic field,$I$ is the current,$L$ is the length of the wire,and $\theta$ is the angle between the magnetic field and the current direction.
Given values are: $F = 7.5 \ N$,$B = 2 \ T$,$I = 5 \ A$,and $L = 1.5 \ m$.
Substituting these values into the formula:
$7.5 = 2 \times 5 \times 1.5 \times \sin \theta$
$7.5 = 15 \times \sin \theta$
$\sin \theta = \frac{7.5}{15} = 0.5$
Since $\sin \theta = 0.5$,we have $\theta = 30^\circ$.

(C) The force on a current-carrying conductor in a uniform magnetic field is given by $\vec{F} = I(\vec{L} \times \vec{B})$.
For a bent wire,the total force is equivalent to the force on a straight wire connecting the initial and final points,which is the displacement vector $\vec{L}_{AC}$.
Given $AB = 3\, cm$ and $BC = 4\, cm$,the length of the hypotenuse $AC$ is $\sqrt{AB^2 + BC^2} = \sqrt{3^2 + 4^2} = 5\, cm = 0.05\, m$.
The magnitude of the force is $F = I L_{AC} B \sin(\theta)$. Since the magnetic field is perpendicular to the plane,$\theta = 90^\circ$ and $\sin(90^\circ) = 1$.
Substituting the values: $F = 10\, A \times 0.05\, m \times 5\, T = 2.5\, N$.

(A) When a current is passed through a spring,each turn of the spring acts as a circular loop carrying current.
Since the current flows in the same direction in all adjacent turns,these turns attract each other due to the magnetic force between parallel currents.
As a result,the spring experiences an attractive force between its coils,causing it to compress.

(A) For wire $C$ to experience no net force,the magnetic force exerted on it by wire $D$ must be equal in magnitude and opposite in direction to the force exerted by wire $B$.
Since all currents are in the same direction,the force between parallel wires is attractive.
Let $x$ be the distance of wire $C$ from wire $D$. Then the distance of wire $C$ from wire $B$ is $(15 - x)\, cm$.
The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $r$ is given by $f = \frac{\mu_0 I_1 I_2}{2\pi r}$.
Equating the forces on wire $C$:
$\frac{\mu_0 I_D I_C}{2\pi x} = \frac{\mu_0 I_B I_C}{2\pi (15 - x)}$
$\frac{15 \times 5}{x} = \frac{10 \times 5}{15 - x}$
$\frac{15}{x} = \frac{10}{15 - x}$
$15(15 - x) = 10x$
$225 - 15x = 10x$
$25x = 225$
$x = 9\, cm$.

(D) According to Fleming's Left-Hand Rule,if we stretch the thumb,forefinger,and middle finger of the left hand such that they are mutually perpendicular to each other:
$1$. The forefinger points in the direction of the magnetic field (North).
$2$. The middle finger points in the direction of the current (Upward).
$3$. The thumb will point in the direction of the force.
Using the right-hand coordinate system where North is $+y$,Up is $+z$,and East is $+x$:
- Current $\vec{I} = I \hat{k}$
- Magnetic field $\vec{B} = B \hat{j}$
- Force $\vec{F} = I(\vec{L} \times \vec{B}) = I(L \hat{k} \times B \hat{j}) = ILB(\hat{k} \times \hat{j}) = -ILB \hat{i}$
Since the negative $x$-direction corresponds to West,the force is directed towards the West.

(C) The force on a current-carrying conductor in a magnetic field is given by the formula $F = BIl \sin(\theta)$.
Here,the current $I = 10\, A$ and the magnetic field $B = {10^{ - 4}}\, T$.
Since the power line is along the east-west direction and the earth's magnetic field is generally considered to be perpendicular to the wire in this context for maximum force calculation,we take $\theta = 90^\circ$,so $\sin(90^\circ) = 1$.
The force per unit length is given by $f = \frac{F}{l} = BI$.
Substituting the values: $f = {10^{ - 4}} \times 10 = {10^{ - 3}}\, N/m$.
Therefore,the correct option is $C$.

(B) The force $F$ on a current-carrying wire in a magnetic field is given by the formula $F = BIl \sin(\theta)$.
Here,$B = 2\,T$ (magnetic field induction),
$I = 1.2\,A$ (current),
$l = 0.5\,m$ (length of the wire),
and $\theta = 90^\circ$ (since the magnetic field is perpendicular to the wire),so $\sin(90^\circ) = 1$.
Substituting the values into the formula:
$F = 2 \times 1.2 \times 0.5 \times 1$
$F = 1.2\,N$.
Therefore,the force on the wire is $1.2\,N$.

(A) The force per unit length between two parallel current-carrying wires is given by the formula: $F/L = \frac{\mu_0}{4\pi} \frac{2i_1 i_2}{d}$.
Given: $\mu_0/4\pi = 10^{-7}\,T\cdot m/A$,$i_1 = i_2 = 10\,A$,and $d = 10\,cm = 0.1\,m$.
Substituting the values: $F/L = 10^{-7} \times \frac{2 \times 10 \times 10}{0.1} = 10^{-7} \times \frac{200}{0.1} = 10^{-7} \times 2000 = 2 \times 10^{-4}\,N/m$.
Since the currents are in the same direction,the force between the wires is attractive.

(C) The force per unit length between two long parallel wires carrying currents $i_1$ and $i_2$ separated by a distance $a$ is given by the formula:
$\frac{F}{l} = \frac{\mu_0}{4\pi} \times \frac{2i_1i_2}{a}$
Given that the currents are equal,$i_1 = i_2 = i$,and the separation $a = 1\,m$. The force per unit length is $\frac{F}{l} = 2 \times 10^{-7}\,N/m$.
Substituting the values into the formula:
$2 \times 10^{-7} = 10^{-7} \times \frac{2i^2}{1}$
$2 \times 10^{-7} = 2 \times 10^{-7} \times i^2$
$i^2 = 1$
$i = 1\,A$
Therefore,the current flowing through each wire is $1.0\,A$.

(A) The force on sides $BC$ and $AD$ are equal in magnitude but opposite in direction,so their net force is zero.
For side $AB$ (at distance $r_1 = 2 \, cm = 2 \times 10^{-2} \, m$):
$F_{AB} = \frac{\mu_0 I_1 I_2 L}{2 \pi r_1} = 2 \times 10^{-7} \times \frac{2 \times 1 \times 0.15}{2 \times 10^{-2}} = 3 \times 10^{-6} \, N$ (attractive,towards the wire).
For side $CD$ (at distance $r_2 = 2 \, cm + 10 \, cm = 12 \, cm = 12 \times 10^{-2} \, m$):
$F_{CD} = \frac{\mu_0 I_1 I_2 L}{2 \pi r_2} = 2 \times 10^{-7} \times \frac{2 \times 1 \times 0.15}{12 \times 10^{-2}} = 0.5 \times 10^{-6} \, N$ (repulsive,away from the wire).
The net force is $F_{net} = F_{AB} - F_{CD} = 3 \times 10^{-6} - 0.5 \times 10^{-6} = 2.5 \times 10^{-6} \, N = 25 \times 10^{-7} \, N$,directed towards the wire.

(B) The force per unit length between two parallel wires carrying currents $i_1$ and $i_2$ separated by a distance $d$ is given by $F/L = \frac{\mu_0 i_1 i_2}{2\pi d}$.
Given: $i_1 = i_2 = 5\,A$,$d = 0.5\,m$,and $\frac{\mu_0}{2\pi} = 2 \times 10^{-7}\,T\cdot m/A$.
Substituting the values: $F/L = \frac{2 \times 10^{-7} \times 5 \times 5}{0.5} = \frac{50 \times 10^{-7}}{0.5} = 100 \times 10^{-7} = 10^{-5}\,N/m$.
Since the currents are in opposite directions,the force between the wires is repulsive.

(A) The force per unit length between two parallel current-carrying wires is given by the formula: $f = \frac{\mu_0}{4\pi} \frac{2 i_1 i_2}{r}$.
Given values are $i_1 = 10 \, A$,$i_2 = 5 \, A$,$r = 0.1 \, m$,and $\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A$.
Substituting these values into the formula:
$f = 10^{-7} \times \frac{2 \times 10 \times 5}{0.1} = 10^{-7} \times \frac{100}{0.1} = 10^{-7} \times 1000 = 10^{-4} \, N/m$.
Since the currents are flowing in opposite directions,the force between the wires is repulsive.

(A) The force $F$ on a current-carrying conductor in a magnetic field is given by the formula: $F = B I L \sin \theta$.
Given:
Magnetic field $B = 1.5\, T$,
Current $I = 10\, A$,
Length $L = 1\, m$,
Angle $\theta = {30^\circ}$.
Substituting these values into the formula:
$F = 1.5 \times 10 \times 1 \times \sin({30^\circ})$
$F = 15 \times 0.5$
$F = 7.5\, N$.

(A) The magnetic force on a current-carrying wire segment 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 segment.
For any closed loop, the starting point and the ending point are the same. Therefore, the effective length vector $\vec{L}_{eff}$ for a closed loop is zero.
Since $\vec{L}_{eff} = 0$, the net magnetic force $\vec{F} = I(0 \times \vec{B}) = 0$.
Thus, the net force on the closed coil is $Zero$.

(B) The force per unit length between two parallel conductors carrying currents $i_1$ and $i_2$ separated by distance $r$ is given by $f = \frac{\mu_0}{2\pi} \frac{i_1 i_2}{r}$.
For conductor $A$ and $B$: Both carry current $I$ in the same direction,so the force $F_1$ exerted by $B$ on $A$ is attractive.
$F_1 = \frac{\mu_0}{2\pi} \frac{I \cdot I}{x} = \frac{\mu_0 I^2}{2\pi x}$.
For conductor $B$ and $C$: $B$ carries current $I$ and $C$ carries current $2I$ in opposite directions,so the force $F_2$ exerted by $C$ on $B$ is repulsive.
$F_2 = \frac{\mu_0}{2\pi} \frac{I \cdot 2I}{x} = \frac{2 \mu_0 I^2}{2\pi x} = 2 F_1$.
Since $F_1$ is attractive and $F_2$ is repulsive,they act in opposite directions. Thus,$F_2 = -2F_1$ or $F_1 = -0.5 F_2$. However,looking at the magnitudes and directions,the force $F_2$ is twice the magnitude of $F_1$ and acts in the opposite direction. Given the options,$F_2 = 2F_1$ represents the relationship between the magnitudes.

(A) The force per unit length between two parallel current-carrying conductors is given by the formula:
$f = \frac{\mu_0}{4\pi} \frac{2 I_1 I_2}{r}$
Given:
$I_1 = 5\,A$,$I_2 = 8\,A$,$r = 0.5\,m$,and $\frac{\mu_0}{4\pi} = 10^{-7}\,T\cdot m/A$.
Substituting the values:
$f = 10^{-7} \times \frac{2 \times 5 \times 8}{0.5}$
$f = 10^{-7} \times \frac{80}{0.5}$
$f = 10^{-7} \times 160$
$f = 1.6 \times 10^{-5}\,N/m$.
Since the currents are in the same direction,the force is attractive.

(B) The force per unit length between two parallel current-carrying wires is given by $f = \frac{\mu_0 i_1 i_2}{2\pi r}$.
Since the currents in all three wires are in the same direction,the force between any two wires is attractive.
Let $i_A = 1 \text{ A}$,$i_B = 2 \text{ A}$,and $i_C = 3 \text{ A}$. The distance between adjacent wires is $d$.
Force on $B$ due to $A$ $(F_{BA})$ is attractive,so it acts towards $A$: $F_{BA} = \frac{\mu_0 i_A i_B}{2\pi d} = \frac{\mu_0 (1)(2)}{2\pi d} = \frac{\mu_0}{\pi d}$.
Force on $B$ due to $C$ $(F_{BC})$ is attractive,so it acts towards $C$: $F_{BC} = \frac{\mu_0 i_B i_C}{2\pi d} = \frac{\mu_0 (2)(3)}{2\pi d} = \frac{3\mu_0}{\pi d}$.
Since $F_{BC} > F_{BA}$,the net force on $B$ is $F_{net} = F_{BC} - F_{BA} = \frac{2\mu_0}{\pi d}$ directed towards $C$.

(C) The force per unit length between two long parallel conductors is given by $F = \frac{\mu_0 I_1 I_2}{2 \pi d}$.
Initially,the force is $F = k \frac{I_1 I_2}{d}$,where $k = \frac{\mu_0}{2 \pi}$.
In the new scenario,the current $I_1$ becomes $2I_1$ and the direction is reversed,so the new current is $-2I_1$. The distance $d$ becomes $3d$.
The new force $F'$ is given by $F' = k \frac{(-2I_1) I_2}{3d} = -\frac{2}{3} \left( k \frac{I_1 I_2}{d} \right)$.
Substituting the initial force $F$,we get $F' = -\frac{2}{3} F$.

(B) The force on a current-carrying conductor in a magnetic field is given by $F = B i l \sin(\theta)$.
Given:
Length $l = 1\, m$
Magnetic field $B = 0.98\, T$
Force $F = 1\, kg\text{-wt} = 1 \times 9.8\, N = 9.8\, N$
Angle $\theta = 90^\circ$ (since it is at right angles),so $\sin(90^\circ) = 1$.
Substituting the values:
$9.8 = 0.98 \times i \times 1$
$i = \frac{9.8}{0.98} = 10\, A$.
Therefore,the current flowing through the wire is $10\, A$.

(D) The force per unit length between two parallel wires carrying currents $i_1$ and $i_2$ separated by distance $d$ is given by $F = \frac{\mu_0 i_1 i_2 l}{2 \pi d}$.
Given that $i_1 = i_2 = i$,the formula becomes $F = \frac{\mu_0 i^2 l}{2 \pi d} = 2 \times 10^{-7} \times \frac{i^2 l}{d}$.
Substituting the given values: $30 \times 10^{-7} = 2 \times 10^{-7} \times \frac{i^2 \times 9}{0.15}$.
Simplifying the equation: $30 = 2 \times \frac{i^2 \times 9}{0.15}$.
$30 = \frac{18 i^2}{0.15}$.
$i^2 = \frac{30 \times 0.15}{18} = \frac{4.5}{18} = 0.25$.
Therefore,$i = \sqrt{0.25} = 0.5 \, A$.

(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 the formula: $\frac{F}{l} = \frac{\mu_0 i_1 i_2}{2\pi d}$.
Since both wires carry the same current $i$ in the same direction,$i_1 = i_2 = i$.
Substituting these values,we get: $\frac{F}{l} = \frac{\mu_0 i^2}{2\pi d}$.
According to the right-hand rule,parallel currents in the same direction exert an attractive force on each other.
Therefore,the wires will attract each other with a force of $\frac{\mu_0 i^2}{2\pi d}$ per unit length.

(C) The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $r$ is given by $f = \frac{\mu_0 I_1 I_2}{2\pi r}$.
For a length $L$,the force is $F = \frac{\mu_0 I_1 I_2 L}{2\pi r} = 2 \times 10^{-7} \times \frac{I_1 I_2 L}{r}$.
$1$. Force on wire $C$ due to wire $D$ $(F_D)$:
$I_D = 30\, A$,$I_C = 10\, A$,$r_{DC} = 3\, cm = 3 \times 10^{-2}\, m$,$L = 25\, cm = 0.25\, m$.
Since currents are in opposite directions,the force is repulsive (towards the right).
$F_D = 2 \times 10^{-7} \times \frac{30 \times 10 \times 0.25}{3 \times 10^{-2}} = 2 \times 10^{-7} \times \frac{75}{3 \times 10^{-2}} = 5 \times 10^{-4}\, N$ (towards right).
$2$. Force on wire $C$ due to wire $G$ $(F_G)$:
$I_G = 20\, A$,$I_C = 10\, A$,$r_{CG} = 2\, cm = 2 \times 10^{-2}\, m$,$L = 0.25\, m$.
Since currents are in opposite directions,the force is repulsive (towards the left).
$F_G = 2 \times 10^{-7} \times \frac{20 \times 10 \times 0.25}{2 \times 10^{-2}} = 2 \times 10^{-7} \times \frac{50}{2 \times 10^{-2}} = 5 \times 10^{-4}\, N$ (towards left).
$3$. Net force on wire $C$:
Since $F_D$ and $F_G$ are equal in magnitude and opposite in direction,the net force is $F_{net} = F_D - F_G = 5 \times 10^{-4} - 5 \times 10^{-4} = 0\, N$.

(C) The forces acting on the rod are the gravitational force $(mg)$ acting downwards and the magnetic force $(F_m = ilB)$ acting horizontally. For the rod to remain stationary on the smooth inclined plane,the component of the gravitational force down the plane must be balanced by the component of the magnetic force up the plane.
The component of gravitational force down the plane is $F_g = mg \sin 60^\circ$.
The component of magnetic force up the plane is $F_m \cos 60^\circ = (ilB) \cos 60^\circ$.
Equating these two forces for equilibrium:
$mg \sin 60^\circ = ilB \cos 60^\circ$
$B = \frac{mg \sin 60^\circ}{il \cos 60^\circ} = \frac{mg}{il} \tan 60^\circ$
Given: $m = 10 \, g = 0.01 \, kg$,$l = 10 \, cm = 0.1 \, m$,$i = 1.73 \, A$,$g = 10 \, m/s^2$,$\tan 60^\circ = \sqrt{3} \approx 1.73$.
$B = \frac{0.01 \times 10}{0.1 \times 1.73} \times 1.73 = \frac{0.1}{0.173} \times 1.73 = 1 \, T$.

(A) When two current-carrying wires are placed parallel to each other,the magnetic force between them depends on the direction of the current.
$1$. When the wires are connected in parallel,the current flows in the same direction in both wires. According to Ampere's law,currents flowing in the same direction attract each other.
$2$. When the wires are connected in series,the current flows in opposite directions in the two wires. Currents flowing in opposite directions repel each other.
Therefore,there is an attraction force when in parallel and a repulsion force when in series.

(B) The magnetic force on a small current element $d\vec{l}$ is given by $d\vec{F} = I(d\vec{l} \times \vec{B})$.
Given that the magnetic field $\vec{B}$ is directed out of the paper and the current $I$ flows in the circular wire,we apply Fleming's left-hand rule or the right-hand rule for the cross product.
For any small segment of the circular wire,the current $I$ flows tangentially,and the magnetic field $\vec{B}$ is perpendicular to the plane of the wire (outwards).
The cross product $d\vec{l} \times \vec{B}$ results in a force vector that points radially outward from the center of the circle at every point along the wire.
Since the wire is elastic,these radially outward forces acting on all parts of the wire will cause it to expand,resulting in a stretching force.

(C) The magnetic field $B$ produced by wire $1$ 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 containing the wires.
The force $dF$ on a current element $i_2 dl$ in a magnetic field $B$ is given by $dF = i_2 (dl \times B)$.
The magnitude of the force is $dF = i_2 B dl \sin \phi$,where $\phi$ is the angle between the current element $dl$ and the magnetic field $B$.
Since the magnetic field $B$ is perpendicular to the plane of the wires,it is perpendicular to the current element $dl$ as well. Thus,$\phi = 90^\circ$ and $\sin \phi = 1$.
However,the force between two current-carrying wires is specifically due to the component of the current element that is parallel to the other wire.
The component of $dl$ parallel to wire $1$ is $dl \cos \theta$.
The force on this component is $dF = B \cdot i_2 \cdot (dl \cos \theta) = \left( \frac{\mu_0 i_1}{2\pi r} \right) i_2 dl \cos \theta = \frac{\mu_0 i_1 i_2 dl \cos \theta}{2\pi r}$.

(B) The net magnetic force on a current-carrying closed loop placed in a uniform magnetic field is zero. Therefore,the loop cannot undergo translational motion. This eliminates options $(c)$ and $(d)$.
According to Fleming's left-hand rule,the magnetic force $\overrightarrow{F_m} = I(\overrightarrow{dl} \times \overrightarrow{B})$ acts on each small element of the loop. Given that the magnetic field $\overrightarrow{B}$ is directed into the plane of the paper and the current $I$ flows in a counter-clockwise direction (as shown in the figure),the force on each element is directed radially outwards. Consequently,the loop will have a tendency to expand.

(A) In equilibrium,the magnetic force on wire $PQ$ must balance its weight.
$mg = B i l$
Here,$B$ is the magnetic field produced by the long wire $AB$ at a distance $r$ from it,given by $B = \frac{\mu_0}{4\pi} \frac{2i}{r}$.
Substituting this into the equilibrium condition:
$mg = \left( \frac{\mu_0}{4\pi} \frac{2i}{r} \right) i l$
Given: $m = 1.0\, g = 10^{-3}\, kg$,$l = 50\, cm = 0.5\, m$,$i = 50\, A$,$g = 10\, m/s^2$.
$10^{-3} \times 10 = 10^{-7} \times \frac{2 \times (50)^2}{r} \times 0.5$
$10^{-2} = 10^{-7} \times \frac{5000}{r} \times 0.5$
$10^{-2} = \frac{2.5 \times 10^{-4}}{r}$
$r = \frac{2.5 \times 10^{-4}}{10^{-2}} = 2.5 \times 10^{-2}\, m = 25\, mm$.

(C) The magnetic field $B$ produced by the long wire $AB$ at a distance $r$ is given by $B = \frac{\mu_0 i_1}{2 \pi r}$.
According to the right-hand rule,the direction of this magnetic field is into the plane of the paper.
The force $dF$ on an element $dx$ of the wire $CD$ at distance $x$ from $AB$ is $dF = i_2 (dx) B = i_2 (dx) \frac{\mu_0 i_1}{2 \pi x}$.
Using Fleming's left-hand rule,the direction of this force is upward (away from $AB$ if $i_1$ and $i_2$ are directed as shown).
Since the magnetic field $B$ decreases as the distance $x$ increases,the force per unit length is greater near $C$ and smaller near $D$.
This non-uniform force distribution results in a net upward force and a torque that causes the rod to rotate clockwise.

(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 displacement vector from the start to the end of the wire segment.
For the upper triangle $ACD$,the path is $C \to A \to D$. The displacement vector $\vec{L}_{CAD}$ is the vector from $C$ to $D$,which is the same as the segment $CD$.
Similarly,for the lower triangle $CDE$,the path is $C \to E \to D$. The displacement vector $\vec{L}_{CED}$ is also the vector from $C$ to $D$.
There is also a wire segment directly from $C$ to $D$ with current $i$ flowing from $C$ to $D$.
Thus,the total force is the sum of forces on three parallel segments,each of length $l = 1 \, m$ carrying current $i = 2 \, A$ in the same direction from $C$ to $D$.
$F_{net} = F_{CAD} + F_{CED} + F_{CD} = 3 \times (i \cdot l \cdot B)$
$F_{net} = 3 \times (2 \, A \times 1 \, m \times 4 \, T) = 3 \times 8 \, N = 24 \, N$.

(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 starting point $A$ to the end point $C$.
From the figure,the displacement vector $\vec{AC}$ has a magnitude of $3\,cm = 3 \times 10^{-2}\,m$ and is directed along the positive $x$-axis.
The magnetic field $\vec{B}$ is directed into the plane (negative $z$-direction,$-2\hat{k}\,T$).
The effective length vector is $\vec{L}_{eff} = 3 \times 10^{-2}\hat{i}\,m$.
The force is $\vec{F} = I(\vec{L}_{eff} \times \vec{B}) = 2 \times (3 \times 10^{-2}\hat{i} \times -2\hat{k}) = 2 \times (6 \times 10^{-2}\hat{j}) = 12 \times 10^{-2}\,N$ along the positive $y$-axis.
Given mass $m = 10\,g = 10^{-2}\,kg$.
The acceleration $a = \frac{F}{m} = \frac{12 \times 10^{-2}}{10^{-2}} = 12\,m\,s^{-2}$ along the $y$-axis.
Note: The provided options contain a discrepancy regarding the direction. Based on the calculation,the force and acceleration are along the $y$-axis.

(C) The magnetic field is $\vec{B} = B\hat{j}$. The force on a current-carrying wire is given by $\vec{F} = i(\vec{L} \times \vec{B})$.
For segments $PQ$ and $TU$,the length vector is parallel to the magnetic field (along $Y$-axis),so $\vec{F}_{PQ} = 0$ and $\vec{F}_{TU} = 0$.
For segment $QR$,the length vector is along the $Z$-axis $(\vec{L} = L\hat{k})$,so $\vec{F}_{QR} = i(L\hat{k} \times B\hat{j}) = -iLB\hat{i}$.
For segment $RS$,the length vector is along the $-X$-axis $(\vec{L} = -L\hat{i})$,so $\vec{F}_{RS} = i(-L\hat{i} \times B\hat{j}) = -iLB\hat{k}$.
For segment $ST$,the length vector is along the $-Z$-axis $(\vec{L} = -L\hat{k})$,so $\vec{F}_{ST} = i(-L\hat{k} \times B\hat{j}) = iLB\hat{i}$.
The net force is $\vec{F}_{net} = \vec{F}_{PQ} + \vec{F}_{QR} + \vec{F}_{RS} + \vec{F}_{ST} + \vec{F}_{TU} = 0 + (-iLB\hat{i}) + (-iLB\hat{k}) + (iLB\hat{i}) + 0 = -iLB\hat{k}$.
The magnitude of the net force is $iLB$.

(B) The magnetic force on a current-carrying wire is given by $\overrightarrow{F} = i(\overrightarrow{l} \times \overrightarrow{B})$.
Given $\overrightarrow{l} = l\hat{i}$ and $\overrightarrow{B} = B_0(\hat{i} + \hat{j} + \hat{k})$.
Substituting these values: $\overrightarrow{F} = i[l\hat{i} \times B_0(\hat{i} + \hat{j} + \hat{k})] = B_0il[\hat{i} \times \hat{i} + \hat{i} \times \hat{j} + \hat{i} \times \hat{k}]$.
Using cross product rules $\hat{i} \times \hat{i} = 0$,$\hat{i} \times \hat{j} = \hat{k}$,and $\hat{i} \times \hat{k} = -\hat{j}$,we get:
$\overrightarrow{F} = B_0il(0 + \hat{k} - \hat{j}) = B_0il(\hat{k} - \hat{j})$.
The magnitude of the force is $F = |\overrightarrow{F}| = B_0il \sqrt{(1)^2 + (-1)^2} = \sqrt{2} B_0il$.