Question

A flat coil of $500$ turns each of area $50c{m^2}$, rotates in a uniform magnetic field of $0.14Wb/{m^2}$ about an axis normal to the field at an angular speed of $150rad/s$. The coil has a resistance of $5\Omega $. The induced e.m.f is applied to an external resistance of $10\Omega $. The peak current through the resistance is,
A. $1.5A$
B. $2.5A$
C. $3.5A$
D. $4.5A$

Answer 1

Hint: The induced electromotive force is also called electromagnetic induction or electromotive induction force. This force is the generation of the potential difference in a coil due to the changes of the magnetic flux that is linked to it. To solve the question, consider the formula that links the number of turns, flux density, and area.

Formula used:
The peak current formula,
$ \Rightarrow {i_0} = \dfrac{{{E_0}}}{R}$
Where, ${i_0}$is the peak current, ${E_0}$ is the induced emf, $R$ is the resistance.
$ \Rightarrow {E_0} = NBA\omega $
Where, ${E_0}$ is the induced emf, $N$ is the number of turns, $B$ magnetic flux density, and $A$ is the area.
The total resistance,
$ \Rightarrow {R_{total}} = R + R'$
Where $R$ is the resistance.

Complete step by step solution:
Given a flat coil has $500$ turns each with the area of $50c{m^2}$. The coil rotates in the uniform magnetic field of $0.14Wb/{m^2}$. This axis is normal to the field at $150rad/s$. The resistance of the coil is $5\Omega $. An induced emf is applied to the external resistance of $10\Omega $.
To find the peak current through the resistance.
To solve the question, use the formula of the induced emf. The formula of induced is,
$ \Rightarrow {E_0} = NBA\omega $
Where, ${E_0}$is the induced emf, $N$is the number of turns, $B$ magnetic flux density, and $A$ is the area.
The values given are,
$ \Rightarrow N = 500$
$ \Rightarrow A = 50c{m^2}$ converting into meters $A = 50 \times {10^{ - 4}}{m^2}$
$ \Rightarrow B = 0.14Wb/{m^2}$
$ \Rightarrow \omega = 150rad/s$
Substitute in the formula,
$ \Rightarrow {E_0} = NBA\omega $
$ \Rightarrow {E_0} = 500 \times 50 \times {10^{ - 4}} \times 0.14 \times 150$
Multiply the values.
$ \Rightarrow {E_0} = 52.5v$
To find the total resistance,
$ \Rightarrow {R_{total}} = R + R'$
The values are,
$ \Rightarrow R = 5\Omega $
$ \Rightarrow R' = 10\Omega $
Substitute the values.
$ \Rightarrow {R_{total}} = 5\Omega + 10\Omega $
Add the values.
$ \Rightarrow {R_{total}} = 15\Omega $
The peak current value is,
$ \Rightarrow {i_0} = \dfrac{{{E_0}}}{R}$
Where, ${i_0}$is the peak current, ${E_0}$ is the induced emf, $R$ is the resistance.
Substitute the calculated values.
$ \Rightarrow {i_0} = \dfrac{{52.5}}{{15}}$
Divide the values.
$ \Rightarrow {i_0} = 3.5A$
The value of peak current through the resistance is $3.5A$.
Therefore, the correct option is $\left( C \right)$.

Note:
There are ways to induce an electromotive force. The first way involves the placement of an electric conductor in the magnetic field that is moving. The second way that involves is the placement of the constantly moving electric conductor in the static magnetic field. The induced emf is used in the working of galvanometers, generators, and transformers.