The magnetic flux through one turn of a 60-turn coil of wire is reduced from 35 Wb to 5.0 Wb in 0.80 s, which results in an induced current of 360 A. QUESTION CONTINUED IN DESCRIPTION

Question

The magnetic flux through one turn of a 60-turn coil of wire is reduced from 35 Wb to 5.0 Wb in 0.80 s, which results in an induced current of 360 A. What is the total resistance of the wire in the coil?

Answer 1

We are given N = 60 turns, Φ_i = 35 Wb, Φ_f = 5.0 Wb, t = 0.80 s, and I = 360 A.

Faraday's Law tells us that a change in magnetic flux through a closed loop will induce an electromotive force, or emf (ε):

ε = -N*(ΔΦ/Δt) = -(60)*([35 Wb - 5.0 Wb]/0.80s) = -2250 V.

The negative sign gives information about the direction of the current, and since the question does not ask about the direction, we will ignore the negative sign.

From here, we will use Ohm's Law to solve for the resistance in the configuration: ε = IR.

R = ε/I = (2250 V)/(360 A) = 6.25 Ω.

Hope this helps!