A rigid square loop of side a carrying curren | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 l

Vedclass · Accepted Answer

Repulsive and equal to $\frac{\mu_0 I_1 I_2}{4\pi}$

Vedclass · Answer

(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.

$A$ rigid square loop of side $a$ carrying current $I_2$ is lying on a horizontal surface near a long wire carrying current $I_1$ in the same plane as shown in the figure. The net force on the loop due to the wire will be:

Similar Questions

$A$ square loop of length $L$ is placed with its edges parallel to the $XY$-axes. The loop carries a current $I$. If the magnetic field in the region varies as $B = B_0 \left(1 + \frac{xy}{L^2}\right) \hat{k}$, then the magnitude of the net force on the loop will be:

The figure shows a conducting loop $ADCA$ carrying current $i$ and placed in a region of uniform magnetic field $B_0$. The part $ADC$ forms a semicircle of radius $R$. The magnitude of the force on the semicircular part of the loop is equal to

$A$ conducting loop carrying a current $I$ is placed in a uniform magnetic field pointing into the plane of the paper as shown. The loop will have a tendency to

$A$ wire carrying current '$I$' along the $x$-axis has length '$L$' and is kept in a magnetic field $\vec{B} = B(\hat{i} + 2\hat{j} - 2\hat{k}) \text{ T}$. The magnitude of the magnetic force acting on the wire is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module