Question

A 800 turn coil of effective area $0.05{{m}^{2}}$ is kept perpendicular to a magnetic field $5\times {{10}^{-5}}T$. When the plane of the coil is rotated by ${{90}^{\circ }}$ around any of its coplanar axis in 0.1 s, the emf induced in the coil will be :
$\begin{aligned} & \text{A}\text{. 2V} \\\ & \text{B}\text{. 0}\text{.2V} \\\ & \text{C}\text{. }2\times {{10}^{-3}}V \\\ & \text{D}\text{. 0}\text{.02V} \\\ \end{aligned}$

Answer 1

At first we have to look for the values that are given in the question, then we know the formula for finding the emf induced in the coil, we will see after writing the formula that only, one of the terms is available and we have to find the change in flux. Now to find the change in flux we will notice that the final flux is zero and we only have to calculate the initial flux.

Formula used: ${{e}_{\text{induced}}}=\dfrac{-d\phi }{dt}=\dfrac{-\Delta \phi }{dt}$
$\Delta \phi ={{\phi }_{f}}-{{\phi }_{i}}$
${{\phi }_{i}}=N(\vec{B}.\vec{A})$

Complete step by step answer:
According to the question given above we understand that,
The coil has 800 turns which is denoted as N.
The effective area of the coil is $0.05{{m}^{2}}$ or $5\times {{10}^{-2}}{{m}^{2}}$ which is denoted as A.
When the coil is kept perpendicular to the magnetic field of magnitude $5\times {{10}^{-5}}T$ which is denoted as B.
So, according to the question it says when the coil is rotated ${{90}^{\circ }}$ around any of the coplanar axis in 0.1s then what is the emf induced in the coil.
And the time will be denoted as $dt$
Now, according to the question to find the emf induced in the coil, we have to use the formula, ${{e}_{\text{induced}}}=\dfrac{-d\phi }{dt}=\dfrac{-\Delta \phi }{dt}$ …….Eq.1
Now, we know $dt$but we don’t know $\Delta \phi$, now to find $\Delta \phi$, we know that
$\Delta \phi ={{\phi }_{f}}-{{\phi }_{i}}$,
We know that ${{\phi }_{f}}$is equals to zero and we can find ${{\phi }_{i}}$ by the formula,
${{\phi }_{i}}=N(\vec{B}.\vec{A})$,
On placing all the terms we get,
${{\phi }_{i}}=800\times 5\times {{10}^{-5}}\times 5\times {{10}^{-2}}$.
So, on calculating the above equation we get, $2\times {{10}^{-3}}$
Now, taking Eq1,
${{e}_{\text{induced}}}=\dfrac{-d\phi }{dt}=\dfrac{-\Delta \phi }{dt}$,
Putting the given and derived values, we get,
${{e}_{\text{induced}}}=\dfrac{-d\phi }{dt}=\dfrac{-\Delta \phi }{dt}$
${{E}_{\text{induced}}}=-\dfrac{(-2\times {{10}^{-3}})}{0.1}$ ,
Which on solving gives us,
${{e}_{\text{induced}}}=0.02V$

So, the correct answer is “Option D”.

Note: In the equation ${{e}_{\text{induced}}}=\dfrac{-d\phi }{dt}=\dfrac{-\Delta \phi }{dt}$, $\Delta \phi$ is the change in the flux and $dt$ is the change in time. In the other formula, ${{\phi }_{i}}=N(\vec{B}.\vec{A})=0$, ‘N’ is the number of turns in the coil, ‘B’ is the magnetic field, and ‘A’ is the cross sectional area.