Question

An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner's 1.50 T field with his fingers pointing in the direction of the field. His wedding ring has a diameter of 2.13 cm, and it takes 0.390 s to move it into the field.

(a) What average current is induced in the ring if its resistance is 0.0100 Ω? (Enter the magnitude in amperes.)

__________ A

(b) What average power is dissipated (in W)?

____________ W

(c) What average magnetic field is induced at the center of the ring? (Enter the magnitude in teslas.)

_____________ T

(d) What is the direction of this induced magnetic field relative to the MRI's field?

o parallel

o antiparallel    

o The magnitude is zero

Answer 1

Emf Induced in magnetic field is given by:

|E| = N*d(phi)/dt

phi = magnetic flux = B*A

|E| = N*d(B*A)/dt = N*A*dB/dt

Given that N = 1,

A = Cross-sectional area of ring = pi*d^2/4 = pi*0.0213^2/4

dB = Change in magnetic field = 1.50 - 0 = 1.50 T

dt = time interval = 0.390 sec

Now average current induced in ring will be:

from Ohm's law:

I_avg = |E|/R = N*A*dB/(R*dt)

R = Resistance of ring = 0.0100 ohm

So,

I_avg = 1*pi*0.0213^2*1.50/(4*0.0100*0.390)

I_avg = 0.137 Amp

Part B.

Average Power dissipated will be:

P = I_avg^2*R

P = 0.137^2*0.0100

P = 0.000188 W = 1.88*10^-4 W

Part C.

Magnetic field at the center of ring due to induced current will be:

B = u0*I_avg/(2*r) = u0*I_avg/d

d = diameter = 2.13 cm = 0.0213 m

B = 4*pi*10^-7*0.137/0.0213

B = 8.08*10^-6 T

Part D.

From lorenz rule, ucrrent will oppose the increase in magnetic field in rong.

So, the direction of this induced magnetic field is antiparallel to the MRI's field.

Therefore correct option is B.