Question

A single loop of wire with an area of 0.0900${\mathbf{m}}^{\mathbf{2}}$is in a uniform magnetic field that has an initial value of 3.80 T, is perpendicular to the plane of the loop, and is decreasing at a constant rate of 0.190 T/s. (a) What emf is induced in this loop? (b) If the loop has a resistance of 0.600 Ω, find the current induced in the loop.

Short answer

1. Emf induced in the loop is _{$\left|\epsilon \right|=0.0171V$}
2. Current induced in the loop is l = 0.0285A

Step-by-step solution

Magnetic flux

a. Consider a single loop of wire with an Area $A=0.0900m^2$ that is placed perpendicularly in a magnetic field with an initial value of B = 3.80 T and a constant rate of _{$\left|\raisebox{1ex}{$dB$}\!\left/ \!\raisebox{-1ex}{$dT$}\right.\right|=0.190\raisebox{1ex}{$T$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$} decrease. Consequently, the magnitude of the induced emf is given by

$\left|\epsilon \right|=\left|\frac{d{\Phi }_B}{dt}\right|$

Here,

_{${\Phi }_B=BA\cos \left(\varphi \right)$}

Putting $\varphi =0°$, so the equation is;

$\left|\epsilon \right|=A\frac{dB}{dt}$

On putting the values;

_{$$\begin{gathered} \left|\epsilon \right|=\frac{\left(0.0900m^2\right)}{\left(0.190T/s\right)} \\ \left|\epsilon \right|=0.0171V \end{gathered}$$}

Hence, the Emf induced in the loop is$\left|\epsilon \right|=0.0171V$

Induced current

b. If the resistance of the loop is $R=0.600\Omega$, the induced current is

$$\begin{gathered} l=\frac{\epsilon }{R}=\frac{0.0171V}{0.600\Omega } \\ l=0.0285A \end{gathered}$$

Hence, the current induced in the loop is$l=0.0285A$