What is the radius of a tightly wound solenoid of circular cross-section that has 180 turns if a change in its internal magnetic field of 3.0 T/s causes a 6.0 A current to flow? The resistance of the circuit that contains the solenoid is 17 Ω. The only emf source for the circuit is the induced emf. A. 0.54 m B. 0.043 m C. 0.25 m D. 0.014 m

Respuesta :

Answer:

C. 0.25 m

Explanation:

Given;

current flow in the solenoid, I = 6.0 A

number of turns of the solenoid, N = 180 turns

the resistance of the circuit, R = 17 Ω

change in the magnetic field, dB/dt = 3.0 T/s

The emf of the circuit is given by;

V = IR

V = 6 x 17

V = 102 Volts

Magnitude of induced emf is given by;

[tex]E = N(\frac{dB}{dt} )A[/tex]

Where;

A is area of the solenoid

[tex]E = N(\frac{dB}{dt} )A\\\\A = \frac{E}{N(\frac{dB}{dt} )} \\\\A = \frac{102}{180 *3} \\\\A = 0.18889 \ m^2[/tex]

Area of the circular solenoid is given by;

A = πr²

where;

r is radius of the solenoid

[tex]r = \sqrt{\frac{A}{\pi} }\\\\r = \sqrt{\frac{0.18889}{\pi} }\\\\r = 0.25 \ m[/tex]

Therefore, the correct option is C. 0.25 m