Question

If two coils with a mutual inductance of 5.00 mH have a 2.00 A current switched off in 30.0 ms, what voltage is induced in one coil?

a) 0.300 V
b) 1.500 V
c) 3.000 V
d) 15.000 V

Answer 1

Using Faraday's law of induction and the provided mutual inductance, current change, and time interval, the induced voltage is calculated to be approximately 0.333 V, making 0.300 V the closest option to the calculated value.

Step-by-step explanation:

When a current in one coil is turned off, the change in current induces a voltage in a nearby coil through mutual inductance. The voltage induced in one coil by the change in current through the other is given by Faraday's law of induction:

V = -M (ΔI/Δt)

Where:

• V is the induced voltage in volts (V)
• M is the mutual inductance between the two coils in henrys (H)
• ΔI is the change in current in amperes (A)
• Δt is the time in which the current change occurs in seconds (s)

Given the mutual inductance (M) of 5.00 mH, a current change (ΔI) of 2.00 A, and a time interval (Δt) of 30.0 ms, we can calculate the induced voltage (V):

V = -5.00 x 10^-3 H * (2.00 A / 30.0 x 10^-3 s)

V = -0.333... V

The negative sign indicates the induced voltage is in a direction to oppose the change in current, according to Lenz's law. The magnitude of the induced voltage is approximately 0.333 V. Hence, the closest answer choice provided is 0.300 V (a).