Question

An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner's 1.80 T field with his fingers pointing in the direction of the field. His wedding ring has a diameter of 2.23 cm, and it takes 0.320 s to move it into the field.
1. What average current is induced in the ring in A if its resistance is 0.0100 Ω?

2. What average power is dissipated in mW?

3. What magnetic field is induced at the center of the ring in T?

Answer 1

Given that,

Magnetic field strength is

B = 1.8T

The wedding ring has a diameter of

d = 2.23 cm = 0.023m

Time take t = 0.32 secs

A. Current induced

From ohms law

V= iR, given that R = 0.01Ω

So, we need to get the induced emf

Using

ε = -NdΦ / dt

Where Φ = BA

ε = -A ∆B / ∆t

ε = -¼πd²(B2-B1) / (t2-t1)

ε = -¼ × π × 0.023² × -1.8 / 0.32

ε = 0.0023371 V

Then

I = ε / R

I = 0.002337 / 0.01

I =0.2337 A

B. Power discippated?

Power is given as

P = iV

P = 0.2337 × 0.002337

P = 0.0005462 W

P = 0.56 mW

C. The magnetic field at the centre of the ring.

The electric field at the centre of the ring is zero because each part of the ring will cause a symmetrical opposite magnitude at every point,

Then, B = 0 T