Question

In a coil of area $10c{m^2}$ and $10$ turns with magnetic field directed perpendicular to the plane and is changing with the rate of ${10^8}Gauss{s^{ - 1}}$ . The resistance of the coil is $20\Omega$ . The current in the coil will be :
(A) $0.5A$
(B) $5A$
(C) $50A$
(D) $5 \times {10^8}A$

Answer 1

Hint : Here, in the above question we have to use the concept of magnetic flux, magnetic field intensity. Just take care of changing the CGS units to SI units. Use $1Guass = {10^{ - 4}}T$ .

Complete Step By Step Answer:
We have the given data as follows,
Area of the coil, $A = {10^2}c{m^2} = {10^{ - 4}}{m^2}$
No. of turns of the coil, $N = 10$
Resistance of the Coil, $R = 20\Omega$
Rate of change of magnetic field is ${10^8}Gauss{s^{ - 1}}$ $= {10^{ - 8}}T$
The Emf of the coil can be written as
$\varepsilon = N\dfrac{{d\phi }}{{dt}}$ ……(for $N$ no. of turns) $(1)$
We know that the magnetic flux is the product of the total area of coil and magnetic field intensity $B$
$\phi = A \times B$
Let us put this value in equation $(1)$
$\varepsilon = NA\dfrac{{dB}}{{dt}}$ ………(here area is constant and magnetic field is changing)
Now putting the value from the given data,
$\varepsilon = 10 \times 10 \times {10^{ - 4}} \times {10^4} = 100V$
Since, we have written all the terms in SI units.
Now, according to the definition of current, we have
$I = \dfrac{\varepsilon }{R}$
$I = \dfrac{\varepsilon }{R} = \dfrac{{100}}{{20}}$
$\Rightarrow I = 5A$
Thus, we have obtained current flowing in the coil is $5A$ .
Correct answer is option B.

Note :
Here we used the formula of magnetic flux and emf to calculate the emf, magnetic flux depends on the area of exposure and the magnetic field. Thus the emf is nothing but the voltage applied to the coil. If the terms are given and you see the need to change them from CGS to SI or vice versa please do so.