A square coil of wire with side 8.0 cm and 50 turns sits in a uniform magnetic field that is perpendicular to the plane of the coil.The coil is pulled quickly out of the magnetic field in 0.2 s.If the resistance of the coil is 15 ohm and a current of 12 mA is induced in the coil,calculate the value of the magnetic field. A)5.6 T B)0.11 T C)7.5 $\times$ 10^-3 T D)1.4 T E)9.1 T

Question

Quizplus · Accepted Answer

We can use Faraday's Law to solve this problem:
EMF = -N (ΔΦ/Δt)
where EMF is the induced electromotive force, N is the number of turns in the coil, and ΔΦ/Δt is the rate of change of magnetic flux.
Since the magnetic field is perpendicular to the plane of the coil, the magnetic flux through the coil is given by:
Φ = BA
where B is the magnetic field and A is the area of the coil.
We can substitute this expression into Faraday's Law and solve for B:
EMF = -N Δ(BA)/Δt
EMF = -NAB(Δt/Δt)
B = -EMF/NAΔt
Substituting the given values, we get:
B = -(12 mA)(15 Ω)/(50)(0.08 m) (0.2 s)
B = 0.11 T
Therefore, the correct answer is B).

A Square Coil of Wire with Side 8 ×\times× 10-3 T D)1

A Square Coil of Wire with Side 8 $\times$ 10^-3 T
D)1