What is the magnetic flux through each turn of the coil at an instant when the current is 5.00 A?

+3 votes

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?
 µT · m2

asked Mar 6, 2012 in Physics by Joey33 ~Expert~ (1,216 points)
    

1 Answer

+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 * ΦB)/I)

Where ΦB 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 * ΦB)/I) * (DI/dt)

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!

answered Mar 6, 2012 by Hyperanthony ~Rookie~ (257 points)
selected Mar 7, 2012 by Joey33



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