An emf of 13.0 mV is induced in a 482-turn coil when the current is changing at the rate of 10.0 A/s. What is the magnetic flux through each turn of the coil at an instant when the current is 5.00 A?

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+3 votes

Best answer

Self-induced emf = 13.0 mV

Turns: 482-turn

DI/dt: 10.0 A/s

Current: 5.00 A

We just need to manipulate a few equations for inductance to solve this. We know that

**self-induced emf = -L * (DI/dt)**

Where L represents inductance. We know that inductance is:

**L = ((N * Φ _{B})/I)**

Where Φ_{B }is magnetic flux, N is the number of turns in the coil, and I is the current. With this information, we can substitute the equation for inductance into the equation for self-induced EMF.

**self-induced emf = -****((N * Φ _{B})/I) **

From the prompt, we have the values for self-induced emf, N, I, and DI/dt. This leaves the value for magnetic flux as the only value, which only requires some quick algebraic manipulation to solve!

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