Question

A coil of mean area $500c{{m}^{2}}$ and having 1000 turns is held perpendicular to a uniform field of 0.4 gauss. The coil is turned through ${{180}^{0}}$ in $\dfrac{1}{10}$ second. The average induced emf is
A.0.02V
B. 0.04V
C. 1.4V
D. 0.08V

Answer 1

Hint: To find the answer to this question, first find the initial flux and the final flux after the rotation of the coil in the magnetic field. Emf can be given in terms of the change in flux per unit time. Obtain the change in flux and time interval and put it in the equation to find the induced emf.

Complete step by step solution:
Given, the area of the coil is, $A=500c{{m}^{2}}$
Converting it into SI unit i.e. in ${{m}^{2}}$
$A=500c{{m}^{2}}=500\times {{10}^{-4}}{{m}^{2}}=5\times {{10}^{-2}}{{m}^{2}}$
The uniform magnetic field is,
$B=0.4G=0.4\times {{10}^{-4}}T$
Time taken to turn the coil by an angle of ${{180}^{0}}$ is,
$t=\dfrac{1}{10}s=0.1s$
Now, the magnetic flux through a coil placed in a uniform magnetic field is given by,
$\phi =NBA\cos \theta $
Where, N is the number of turns in the coil, B is the magnetic field and A is the area of the coil.
$\theta $ is the angle between the coil and the magnetic field.
Now, initially the magnetic flux through the coil is,
$\begin{align}
  & {{\phi }_{1}}=NBA\cos {{0}^{0}} \\
 & {{\phi }_{1}}=NBA \\
\end{align}$
Finally, the coils turn through an angle ${{180}^{0}}$. So, the magnetic flux through the coil is given as,
 $\begin{align}
  & {{\phi }_{2}}=NBA\cos {{180}^{0}} \\
 & {{\phi }_{2}}=-NBA \\
\end{align}$
Now, the change in magnetic flux us given as,
$\begin{align}
  & \Delta \phi ={{\phi }_{2}}-{{\phi }_{1}} \\
 & \Delta \phi =-NBA-NBA \\
 & \Delta \phi =-2NBA \\
\end{align}$
Taking the magnitude only, the change in magnetic flux through the coil is,
$\Delta \phi =2NBA$
Now, the induced emf in a coil moving in a magnetic field is given by the expression.
$emf=\dfrac{\Delta \phi }{\Delta t}$
Putting the values on the above equation, we get,
$\begin{align}
  & emf=\dfrac{2NBA}{\Delta t} \\
 & emf=\dfrac{2\times 1000\times 0.4\times {{10}^{-4}}\times 5\times {{10}^{-2}}}{0.1} \\
 & emf=0.04V \\
\end{align}$
So, the induced emf on the coil is 0.04 V.

Note: The induced voltage in a circuit due to the change in magnetic flux is directly proportional to this rate of change of the magnetic flux through the circuit. It is known as the phenomenon of electromagnetic induction.