Question

A magnetic field in a certain region is given by `B = B_o cos (ωt)hatk` and a coil of radius a with resistance R is placed in the x-y plane with its centre at the origin in the magnetic field (Figure) . Find the magnitude and the direction of the current at (a, 0, 0) at t = π/2ω, t = π/ω and t = 3π/2ω.

Answer 1

1. First law: Whenever the number of magnetic lines of force (magnetic flux) passing through a circuit changes, an emf is produced in the circuit called induced emf. The induced emf persists only as long as there is a change or cutting of flux.
2. Second law: The induced emf is given by the rate of charge of magnetic flux linked with the circuit, i.e, `e = - (dphi)/(dt)`. For N turns `e = - (N dphi)/(dt)`; NEgative sign indicates that induced emf (e) opposes the change of flux.

First we need to find out the flux passing through the ring at any instant and that is given by

`phi_m = vecB * vecA = BA cos theta`

And as we know both `vecA` (area vector) and `vecB` (magnetic field vector) are directed along z-axis. So, angle between them is 0.

So, cos θ = 1   .....(∵ θ = 0)

⇒ `phi_m = BA`

Area of coil of radius a = πa²

ε = `B_0(pia^2) cos ωt`

By Faraday's law of electromagnetic induction,

Magnitude of induced emf is given by ε = `B_0(pia^2) ω sin ωt`

This causes flow of indeed current, which is given by `I = (B_0(pia^2) ω sin ωt)/R`

Now, the value of current at different instants,

(i) `t = pi/(2ω)`

`I = (B_0(pia^2)ω)/R` along `hatj`

Because sin ωt = `sin(ω pi/(2ω)) = sin  pi/2` = 1

(ii) `t = pi/ω, I = (B(pia^2)ω)/R` = 0

Because, sin ωt = `sin(ω pi/ω) = sin pi` = 0

(iii) `t = 3/2 pi/ω`

`I = (B(pia^2)ω)/R` along `-hatj`

sin ωt = `sin(ω * (3pi)/(2ω)) = sin  (3pi)/2` = – 1