A flat coil is wrapped with 200 turns of very thin wire on a square frame with sides 18 cm long.A uniform magnetic field is applied perpendicular to the plane of the coil.If the field changes uniformly from 0.50 T to 0.00 T in 8.0 s,find the emf induced in the coil.
A)2.1 mV
B)4.1 mV
C)0.21 V
D)0.41 V

Question

Quizplus · Accepted Answer

The emf induced in the coil can be calculated using Faraday's law of electromagnetic induction, which states that the induced emf ($ε$) in a coil is equal to the negative change in magnetic flux ($ΔΦ$) through the coil over the change in time ($Δt$), multiplied by the number of turns ($N$) in the coil. The formula is given by $ε = -N (ΔΦ/Δt)$.The magnetic flux ($Φ$) through the coil is given by the product of the magnetic field ($B$), the area of the coil ($A$), and the cosine of the angle ($θ$) between the magnetic field and the normal to the coil. Since the magnetic field is perpendicular to the coil, $θ = 0$ degrees, and $cos(θ) = 1$. Therefore, $Φ = B \cdot A$.Given:- $N = 200$ turns- Side of the square frame = 18 cm = 0.18 m (since 1 cm = 0.01 m)- $B_{initial} = 0.50$ T- $B_{final} = 0.00$ T- $Δt = 8.0$ sThe area ($A$) of the square coil is $A = (side)^2 = (0.18 m)^2 = 0.0324 m^2$.The change in magnetic flux ($ΔΦ$) is $ΔΦ = B_{final} \cdot A - B_{initial} \cdot A = (0.00 - 0.50) T \cdot 0.0324 m^2 = -0.0162 T \cdot m^2$.Therefore, the induced emf ($ε$) is $ε = -N (ΔΦ/Δt) = -200 \cdot (-0.0162 T \cdot m^2 / 8.0 s) = 0.00405 V = 4.05 mV$.Thus, the induced emf in the coil is 4.05 mV, which rounds to 4.1 mV, making option B the correct answer.

A Flat Coil Is Wrapped with 200 Turns of Very