Question

A single-turn loop of wire has a resistance of 6.00 Ω and a cross-sectional area 300 cm2 and is perpendicular to a uniform magnetic field which increases at a constant rate from 0.200 T to 3.60 T in 0.500 seconds. What is the magnitude of the induced current in the loop?
A. 0.34 AB. 34 mAC. 2.04 AD. 36 mAE. 0.36 A

Answer 1

Given that,

Resistance of wire

R = 6.00 Ω

Cross sectional area

A = 300cm² = 0.03m²

Initial magnetic field Bi = 0.2T

Final magnetic field Bf = 3.6T

Time taken t = 0.5s

Current?

From ohms law

V=iR

Then, I = V/R

Therefore, we need to find the induced EMF in the coil

ε = -dΦ/dt

Where Φ =BA

Where the Area is constant

ε = -dΦ/dt = -d(BA)/dt

Since A is constant

ε = -A•dB/dt

ε = -A•(∆B)/∆t

ε = -0.03(3.6—0.2)/0.5

ε = -0.204 V

Then, the magnitude of the EMF is 0.204 V

So,

I = V/R

I = 0.204/6

I = 0.034 Amps

I = 34 mA

Option B is correct.