Question

A magnetic field of flux density acts normal in a coil of turns having $100c{{m}^{2}}$ area. The emf induced if the coil is removed from the magnetic field in $0.1s$ is

$A)\text{ }50V$
$B)\text{ 6}0V$
$C)\text{ 8}0V$
$D)\text{ 4}0V$

Answer 1

Hint: This problem can be solved by using the direct formula for the magnetic flux through a coil in terms of the magnetic field, number of turns and the area of the coil and then using this value of magnetic flux to get the emf induced from the direct formula relating the emf induced to the change in magnetic flux per unit time.
Formula used:
$\phi =NAB\cos \theta $
$\left| \varepsilon \right|=\dfrac{\Delta \phi }{\Delta t}$

Complete answer:
We will first find the magnetic flux linking through the coil and then using this value find the emf induced in the coil when this magnetic flux is removed in a time period.
The magnetic flux $\phi $ linking through a coil of $N$ turns and surface area $A$ placed in a magnetic field of magnitude $B$ is given by
$\phi =NAB\cos \theta $ ---(1)
Where $\theta $ is the angle made by the direction of the magnetic field with the perpendicular to the area of the coil.
The magnitude $\left| \varepsilon \right|$ of the emf $\varepsilon $ induced in a coil due to a change $\Delta \phi $ in the magnetic flux linking through a coil in a time interval $\Delta t$ is given by
$\left| \varepsilon \right|=\left| \dfrac{\Delta \phi }{\Delta t} \right|$ --(2)
Now, let us analyze the question.
The magnitude of the magnetic field is $B=10T$.
The number of turns in the coil is $N=50$.
The surface area of the coil is $A=100c{{m}^{2}}=100\times {{10}^{-4}}{{m}^{2}}={{10}^{-2}}{{m}^{2}}$ $\left( \because 1c{{m}^{2}}={{10}^{-4}}{{m}^{2}} \right)$
Since, it is given that the magnetic field is normal to the coil, therefore the angle made by the magnetic field with the perpendicular to the area of the coil is $\theta ={{0}^{0}}$.
Let the flux linking through the coil be $\phi $.
Now, using (1), we get
$\phi =50\times {{10}^{-2}}\times 10\times \cos {{0}^{0}}=5\times 1=5Wb$ $\left( \because \cos {{0}^{0}}=1 \right)$ --(3)
Now, it is given that the flux is removed from the coil in $\Delta t=0.1s$.
Therefore, the change in the magnetic flux linking through the coil will be $\Delta \phi =0-\phi =-\phi $
Let the magnitude of the emf induced in the coil due to this change be $\left| \varepsilon \right|$.
Now, using (2), we get
$\left| \varepsilon \right|=\left| \dfrac{\Delta \phi }{\Delta t} \right|=\left| \dfrac{-\phi }{\Delta t} \right|=\dfrac{\phi }{\Delta t}$
Putting (3) in the above equation, we get
$\left| \varepsilon \right|=\dfrac{5}{0.1}=50V$
Hence, the magnitude of the emf induced in the coil is $50V$.

So, the correct answer is “Option A”.

Note:
Students must note that the direction of the emf induced in the coil will be such that it produces a current that can make up for or opposes the change in the magnetic flux linking through the coil by producing a magnetic flux of its own but in the opposite direction. This is also called as the Lens’ law and hence, the total picture of the emf induced in the coil is given by the formula,
$\varepsilon =-\dfrac{\Delta \phi }{\Delta t}$
Where the negative sign is implication of the fact that the emf induced is such that it opposes the change in the magnetic flux linking through the coil. However, while finding out the magnitude, the negative sign can be done away with.