A round flat metal coil has 180 turns and negligible resistance. It is connected in a series circuit with a 17-Ω resistor, with nothing else in the circuit. You measure that a 6.0-A current flows through the resistor when a magnetic field through the coil, perpendicular to its area, is changing at 3.0 T/s. What is the radius of the coil?

(A) 0.043 m

(B) 0.014 m

(C) 0.25 m

(D) 0.54 m

Question

Physics

Lane Arnold

3 July, 05:52

A round flat metal coil has 180 turns and negligible resistance. It is connected in a series circuit with a 17-Ω resistor, with nothing else in the circuit. You measure that a 6.0-A current flows through the resistor when a magnetic field through the coil, perpendicular to its area, is changing at 3.0 T/s. What is the radius of the coil?

(A) 0.043 m

(B) 0.014 m

(C) 0.25 m

(D) 0.54 m

+5

Answers (1)

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Frisky · Answer 1

Answer: C) 0.25 m

Explanation: In order to explain this problem we have to consider the Faraday law, we have:

ε=-dФ/dt where ε the emf induced by the change of the magnetic field given by dФ/dt.

then ε=I*R=17*6=102 V

We have a coil then we have the magnetic flux as follow:

Ф=N*A*B then we have

dФ/dt = N*A*dB/dt where A and N is the area and number of turn of the coil.

A=π*R^2 where R is the radius of teh coil.

Finally we have;

dФ/dt = N*π*R^2*dB/dt then

R = [dФ/dt / (N*π*dB/dt) ]^1/2 = [102 / (180*π*3) ]1/2=245.2*10^-3=≅0.25m