Question

Suppose we are given a square coil that is 3.5 cm on a side containing 100 loops of very fine wire. The total resistance of this coil is 1.5 ✕ 10−3 Ω. This coil is initially centered on the origin with its sides parallel to the x- and y-axes. It is being rotated about the y-axis in a uniform magnetic field of 1.5 T along the z-axis, with an angular velocity of 2.0 rad/sec. Find the magnitude of the peak value of the induced current flowing in this coil under these conditions.

Answer 1

Answer:

Magnitude of the peak value of the induced current flowing in this coil under these conditions = 245 A

Explanation:

To calculate the magnitude of the peak induced current, we first need to calculate the peak emf in the loop.

The induced emf from Faraday's law of electromagnetic induction is given as

E = NABw (sin wt)

The peak value of the Emf will be when (sin wt) = 1.

Peak emf = E = NABw

where

N = number of turns = 100 loops of wire

A = Cross sectional Area of the loop = 0.035 × 0.035 = 0.001225 m² ( a square)

B = magnetic field = 1.5 T

w = angular velocity = 2.0 rad/s

Peak emf = E = NABw = 100 × 0.001225 × 1.5 × 2 = 0.3675 V

But Ohm's law explains that voltage is related to current through V = IR, hence,

Emf = (current) × (resistance of the coil)

0.3675 = I × (1.5 ✕ 10⁻³)

I = (0.3675/0.0015) = 245 A

Hope this Helps!!!

Answer 2

Answer:

The magnitude of the peak value of the induced current is 245 A

Explanation:

The magnetic flux is

Φ = B*A*cosθ

If an emp is induced opposed the change in flux is

E = -dΦ/dt

|E| = dBAcosθ/dt

if θ = wt

sinwt = 1The magnitu

|E| = B*A*w*sin(wt)

fon n = 100 loops,

E = 100 * 1.5 * 0.001225 * 2 = 0.3675 V

I = 0.3675/1.5x10^-3 = 245 A