The magnetic flux that passes through one turn of a 22-turn coil of wire changes to 2.5 Wb from 13.0 Wb in a time of 0.052 s. The average induced current in the coil is 260 A. What is the resistance of the wire

17.09 Ω

Explanation:

First we find the voltage induced in the coil, using

E = NdΦ/dt Equation 1

Where E = Induced voltage, N = Number of turns, dΦ = change in magnetic flux, dt = change in time/ Time interval

Give: N = 22 turns, dΦ = 13-2.5 = 10.5 Wb, dt = 0.052 s.

Substitute into equation 1

E = 22(10.5)/0.052

E = 4442.31 V.

Using Ohm's Law,

E = IR Equation 2

Where I = current, R = Resistance of the wire, E = Voltage.

make R the subject of the equation,

R = E/I Equation 3

Give: I = 260 A, E = 4442.31 V

Substitute into equation 3

R = 4442.31/260

R = 17.09 Ω

what are the options.

explanation:

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