Question

A circular loop of radius 13 cm carries a current of 13 A. A flat coil of radius 0.94 cm, having 58 turns and a current of 1.9 A, is concentric with the loop. The plane of the loop is perpendicular to the plane of the coil. Assume the loop's magnetic field is uniform across the coil. What is the magnitude of (a) the magnetic field produced by the loop at its center and (b) the torque on the coil due to the loop

Answer 1

Answer:

$6.28\times 10^{-5}\ \text{T}$

$1.92\times 10^{-6}\ \text{Nm}$

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi 10^{-7}\ \text{H/m}$

$I_l$ = Current in circular loop = 13 A

$r_l$ = Radius of circular loop = 13 cm

$N$ = Number of turns = 58

$r_c$ = Radius of coil = 0.94 cm

$I_c$ = Current in coil = 1.9 A

$\theta$ = Angle between loop and coil = $90^{\circ}$

Magnitude of magnetic field in circular loop

$B_l=\dfrac{\mu_0I_l}{2r_l}\\\Rightarrow B_l=\dfrac{4\pi 10^{-7}\times 13}{2\times 13\times 10^{-2}}\\\Rightarrow B_l=6.28\times 10^{-5}\ \text{T}$

The magnetic field produced by the loop at its center is $6.28\times 10^{-5}\ \text{T}$.

Torque is given by

$\tau=\pi NI_cr_c^2B_l\sin\theta\\\Rightarrow \tau=\pi 58\times 1.9\times (0.94\times 10^{-2})^2\times 6.28\times 10^{-5}\sin90^{\circ}\\\Rightarrow \tau=1.92\times 10^{-6}\ \text{Nm}$

The torque on the coil due to the loop $1.92\times 10^{-6}\ \text{Nm}$.