A solenoid with 100 turns and a cross sectional area of 20 cm² has a magnetic field of 0.60 T along its axis.If the field is confined within the solenoid and changes at a rate of 0.20 T/s,the magnitude of the induced EMF in the solenoid will be A)0.0080 V. B)0.080 V. C)0.040 V. D)0.40 V.

Question

Quizplus · Accepted Answer

The induced EMF in a solenoid can be calculated using the equation:
EMF = -N*dΦ/dt
where N is the number of turns, Φ is the magnetic flux through the solenoid, and dt is the change in time.

Since the magnetic field is confined within the solenoid, the magnetic flux is simply the product of the magnetic field, the number of turns, and the cross-sectional area:
Φ = B*A*N

Substituting the given values and differentiating, we get:
EMF = -N*dΦ/dt
EMF = -N*B*A*dN/dt
EMF = -(100)*(0.60)*(0.20)*(20*10^-4)
EMF = 0.040 V

Therefore, the magnitude of the induced EMF in the solenoid is 0.040 V, which corresponds to choice (C) as the correct answer.

A Solenoid with 100 Turns and a Cross Sectional Area