A 50-turn circular coil (radius = 15 cm)with a total resistance of 4.0$\Omega$ is placed in a uniform magnetic field directed perpendicularly to the plane of the coil.The magnitude of this field varies with time according to B = A sin ($\alpha$t),where A = 80 $\mu$T and $\alpha$ = 50 $\pi$ rad/s.What is the magnitude of the current induced in the coil at t = 20 ms?
A)11 mA
B)18 mA
C)14 mA
D)22 mA
E)zero

Question

Quizplus · Accepted Answer

The formula for the emf induced in a circular coil of N turns with radius R in a magnetic field B is given by:
ε = -N(dΦ/dt)
where Φ is the magnetic flux through the coil.
For a circular coil, the magnetic flux can be written as:
Φ = BπR^2
Differentiating with respect to time, we get:
dΦ/dt = BπR^2(dB/dt)
Substituting the given values, we get:
dΦ/dt = AπR^2Fωcos(ωt)
where Fω = 500π rad/s
The emf induced in the coil is equal to the rate of change of flux through the coil. Therefore, we can write:
ε = -N(dΦ/dt)
Substituting the given values, we get:
ε = -NAR^2Fωcos(ωt)
The total resistance of the coil is given as 4 Ω. Therefore, the current induced in the coil can be calculated as:
I = ε/R
Substituting the given values, we get:
I = (-NAR^2Fωcos(ωt))/R
At t = 20 ms, we have:
cos(ωt) = cos(0.01π) = -1
Substituting this value in the above equation, we get:
I = (50)(80 × 10^-6)(0.15)^2(500π)(-1)/4
I = -11.1 × 10^-3
Taking the magnitude of the current, we get:
|I| = 11 mA

A 50-Turn Circular Coil (Radius = 15 Cm)with a Total Ω\OmegaΩ

A 50-Turn Circular Coil (Radius = 15 Cm)with a Total $\Omega$