Question

A single loop of wire with an area of 0.0900m2 is in a uniform magnetic field that has an
initial value of 3.8 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 0.600O, find the current induced in the loop.

Answer 1

Concepts and reason

The concept to be used in this problem is of Faraday’s Law of Electromagnetic Induction.

First, apply Faraday’s Law of Electromagnetic Induction law to find the emf induced in the loop. Then, using the voltage-current relation find the current flowing in the loop.

Fundamentals

The Law of Electromagnetic Induction states that the emf induced in the closed circuit is equal to the change in the magnetic flux through the circuit.

$\varepsilon = - N\frac{{d{\phi _B}}}{{dt}}$

Here $N$ is the number of turns of the coil and ${\phi _B}$ is the magnetic flux.

The expression of magnetic flux is given by,

${\phi _B} = BA$

Here $B$ is the magnitude of the magnetic field and $A$ is the area of the surface.

The voltage-current relation is given by the Ohm's law:

$I = \frac{V}{R}$

Here $V$ is the emf induced in the loop and $R$ is the resistance of the loop.

(a)

The Law of Electromagnetic Induction states that the emf induced in the closed circuit is equal to the change in the magnetic flux through the circuit.

$\varepsilon = - N\frac{{d{\phi _B}}}{{dt}}$

Here $N$ is the number of turns of the coil and ${\phi _B}$ is the magnetic flux.

Substitute $BA$ for ${\phi _m}$ ,

$\begin{array}{c}\\\varepsilon = - N\frac{d}{{dt}}\left( {BA} \right)\\\\ = - NA\frac{{dB}}{{dt}}\\\end{array}$

Substitute 1 for $N$ , $0.09{\rm{ }}{{\rm{m}}^2}$ for $A$ and $0.19{\rm{ T}} \cdot {{\rm{s}}^{ - 1}}$ for $\frac{{dB}}{{dt}}$ ,

$\begin{array}{c}\\\varepsilon = - \left( 1 \right)\left( {0.09{\rm{ }}{{\rm{m}}^2}} \right)\left( {0.19{\rm{ T}} \cdot {{\rm{s}}^{ - 1}}} \right)\\\\ = - 0.0171{\rm{ V}}\\\\{\rm{ = }} - 1.71 \times {10^{ - 2}}{\rm{ V}}\\\end{array}$

Hence the magnitude of the induced emf is $1.71 \times {10^{ - 2}}{\rm{ V}}$ .

(b)

The current flowing in the loop is given by,

$I = \frac{\varepsilon }{R}$

Here $\varepsilon$ is emf induced in the loop and $R$ is the resistance of the loop.

Substitute $1.71 \times {10^{ - 2}}{\rm{ V}}$ for $\varepsilon$ and $0.6\Omega$ for $R$ ,

$\begin{array}{c}\\I = \frac{{1.71 \times {{10}^{ - 2}}{\rm{ V}}}}{{0.6\Omega }}\\\\ = 2.85 \times {10^{ - 2}}{\rm{ A}}\\\end{array}$

Ans: Part a

The emf induced in the loop is $1.71 \times {10^{ - 2}}{\rm{ V}}$ .

Part b

The Current induced in the loop is $2.85 \times {10^{ - 2}}{\rm{ A}}$ .