Question

A 0.500 mm diameter copper wire makes up a 136 turns, 7.60 cm diameter coil. There is a magnetic field parallel to the axis of the coil. If the induced current in the coil is 2.80 A, what is the rate of change of the magnetic field? a) 0.0839 T/s b) 3.63×10-6 T/s c) 12.8 T/s d) 1.74×103 T/s

Answer 1

The induced emf ( in T/s) = NBA

Where

N= No of turns .

B = Magnetic field

A = Area of coil = R2

Where R = Radius of coil

This emf will be equal to IR

Where I = current and

R = resistance = L/a

Where 'a' is the cross section area of the coil = r2 where 'r' is the radius of the cross section of wire and

L = length of wire = 2RN

Now equating both induced emf and the V = IR

NBA = IR

Putting all the values of N ,B ,A, I and R

from above given expression the net result will come out -

B = (2I ) / Rr2

Put value of the = 1.68 x 10 -8 Ohm - m ( Standard value at NTP)

I = 2.8 A ( given )

R = 7.6 / 2 = 3.8 cm = 3.8 x 10-2 m.

r = 0.25 mm = 0.25 x 10 -3 m.

On putting all those values in the above expression we will get the value of B = 12.62 T/s which is almost equal to 12.8 T / s hence the option C will be correct.

There is minor differences in the decimal places about 0.18 which is due to rounding off the decimal values.

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