A round flat conducting loop is placed perpendicular to a uniform 0.50 T $\vec { B }$ field.If the area of the loop increases at a rate of 3.0 × $10 ^{- 3}$ $m ^ { 2 }$ /s,what is the induced emf in the loop?
A)4.3 mV
B)0 mV
C)1.5 mV
D)1.7 mV
E)5.5 mV

Question

Quizplus · Accepted Answer

The induced emf ($\epsilon$) in a loop due to a changing magnetic flux ($\Phi_B$) is given by Faraday's law of electromagnetic induction, $\epsilon = -\frac{d\Phi_B}{dt}$. The magnetic flux $\Phi_B$ through the loop is given by $\Phi_B = B \cdot A$, where $B$ is the magnetic field strength and $A$ is the area of the loop. Given that $B = 0.50 \, T$ and $\frac{dA}{dt} = 3.0 \times 10^{-3} \, m^2/s$, the rate of change of magnetic flux $\frac{d\Phi_B}{dt} = B \cdot \frac{dA}{dt} = 0.50 \cdot 3.0 \times 10^{-3} = 1.5 \times 10^{-3} \, T \cdot m^2/s$, which corresponds to 1.5 mV.

A Round Flat Conducting Loop Is Placed Perpendicular to a Uniform