A circular coil of radius R and N turns has n | vedclass

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

Question

(B) The torque $	au$ experienced by a current-carrying coil in a magnetic field is given by $\vec{	au} = \vec{M} 	imes \vec{B}_0$,where $\vec{M}$ i

Vedclass · Accepted Answer

$\pi N Q B_0 R^2$

Vedclass · Answer

(B) The torque $	au$ experienced by a current-carrying coil in a magnetic field is given by $\vec{	au} = \vec{M} 	imes \vec{B}_0$,where $\vec{M}$ is the magnetic moment of the coil.The magnitude of the torque is $	au = M B_0 \sin(	heta)$. Since the plane of the coil is vertical and the magnetic field $B_0$ is horizontal and parallel to the plane of the coil,the angle between the area vector (normal to the plane) and the magnetic field is $	heta = 90^\circ$. Thus,$\sin(90^\circ) = 1$.The magnetic moment of the coil is $M = N i A = N i (\pi R^2)$.Therefore,the torque is $	au = N i \pi R^2 B_0$.From the angular impulse-momentum theorem,the change in angular momentum $L$ is given by $\Delta L = \int 	au dt$.Substituting the expression for torque: $\Delta L = \int (N i \pi R^2 B_0) dt = N \pi R^2 B_0 \int i dt$.Since the total charge $Q$ discharged through the coil is $Q = \int i dt$,we get:$L = N \pi R^2 B_0 Q$.

Similar Questions

$A$ circular coil of $20$ turns and radius $10\, cm$ is placed in a uniform magnetic field of $0.10\, T$ normal to the plane of the coil. If the current in the coil is $5\, A$,then the torque acting on the coil will be...... $Nm$.

The figure represents four positions of a current-carrying coil in a magnetic field directed towards the right. $\hat{n}$ represents the direction of the area vector of the coil. The correct order of potential energy is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module