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A 111 ‑turn circular coil of radius 2.11 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicular to the plane of the coil. The coil is connected to a 14.1 Ω resistor to create a closed circuit. During a time interval of 0.125 s, the magnetic field strength decreases uniformly from 0.669 T to zero. Find the energy, in millijoules, that is dissipated in the resistor during this time interval.

1 Answer

5 votes

Answer:

0.0061 J

Step-by-step explanation:

Parameters given:

Number of turns, N = 111

Radius of turn, r = 2.11 cm = 0.0211 m

Resistance, R = 14.1 ohms

Time taken, t = 0.125 s

Initial magnetic field, Bin = 0.669 T

Final magnetic field, Bfin = 0 T

The energy dissipated in the resistor is given as:

E = P * t

Where P = Power dissipated in the resistor

Power, P, is given as:

P = V² / R

Hence, energy will be:

E = (V² * t) / R

To find the induced voltage (EMF), V:

EMF = [-(Bfin - Bin) * N * A] / t

A is Area of coil

EMF = [-(0 - 0.669) * 111 * pi * 0.0211²] / 0.125

EMF = 0.83 V

Hence, the energy dissipated will be:

E = (0.83² * 0.125) / 14.1

E = 0.0061 J

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