Question

A coil of resistance $40\Omega$ is connected to a galvanometer of $160\Omega$ resistance. The coil has a radius $6mm$ and turns $100$. This coil is placed between the poles of a magnet such that the magnetic field is perpendicular to the coil. If the coil is dragged out then the charge through the galvanometer is $32\mu C$. The magnetic field is?
(A) $6.55{\text{ T}}$
(B) $5.66{\text{ T}}$
(C) $0.655{\text{ T}}$
(D) $0.566{\text{ T}}$

Answer 1

Hint
To solve this question, we need to use the Faraday’s law of electromagnetic induction which gives the value of the emf induced in a coil when the magnetic flux through the coil changes. Then on equating it with the emf given by the Ohm’s law will get the final answer.

Formula Used: The formula used in solving this question is given by
$\Rightarrow e = \dfrac{{d\varphi }}{{dt}}$, here $e$ is the emf induced in a conductor, $\varphi$ is the magnetic flux, and $t$ is the time.
$\varphi = BA\cos \theta$ here $\varphi$ is the magnetic flux through a coil of area $A$ through which magnetic field of $B$ is being passed, and $\theta$ is the angle between the magnetic field and the area vector of the coil.

Complete step by step answer
Let the magnetic field be $B$
The magnitude of the emf induced in a coil placed in a magnetic field is given by the Faraday’s law as
$\Rightarrow e = \dfrac{{d\varphi }}{{dt}}$ …..(1)
The flux is given by the formula
$\Rightarrow \varphi = BA\cos \theta$
As the coil is of $100$ turns, so on substituting we get the flux as,
$\varphi = 100BA\cos \theta$
The magnetic field is perpendicular to the coil, as given in the question, so the angle $\theta = {0^ \circ }$. Thus the flux is
$\Rightarrow \varphi = 100BA\cos {0^ \circ }$
Since cosine of 0 is 1
$\Rightarrow \varphi = 100BA$
Now, putting this value of flux in (1) we get
$\Rightarrow e = \dfrac{{d\left( {100BA} \right)}}{{dt}}$
As the area of the coil is constant, so we have
$\Rightarrow e = 100A\dfrac{{dB}}{{dt}}$ …...(2)
Now, from ohm’s law the induced emf can also be given as
$\Rightarrow e = IR$ ……...(3)
We know that the current is equal to the rate of flow of charge, that is
$\Rightarrow I = \dfrac{{dq}}{{dt}}$
Putting this in (3) we get
$\Rightarrow e = R\dfrac{{dq}}{{dt}}$ …….(4)
Therefore, on equating (2) and (4) we get
$100A\dfrac{{dB}}{{dt}} = R\dfrac{{dq}}{{dt}}$
Cancelling $dt$ from both the sides, we have
$\Rightarrow 100AdB = Rdq$
Dividing both sides by the area $A$ we get
$\Rightarrow 100dB = \dfrac{R}{A}dq$ …….(5)
According to the question, we have the resistance of the galvanometer $G = 160\Omega$ which is connected in series with the coil, which has the resistance $r = 40\Omega$. So the net resistance in the circuit becomes
$\Rightarrow R = G + r$
$\Rightarrow R = 160 + 40 = 200\Omega$
Also, the area of the coil is given as
$\Rightarrow A = \pi {r^2}$
According to the question, $r = 6mm = 6 \times {10^{ - 3}}m$. So the area of the coil becomes
$\Rightarrow A = \pi \times {\left( {6 \times {{10}^{ - 3}}} \right)^2}$
On solving we get
$\Rightarrow A = 1.13 \times {10^{ - 4}}{m^2}$
The charge through the galvanometer is
$\Rightarrow dq = 32\mu C$
$\Rightarrow dq = 32 \times {10^{ - 6}}C$
Substituting all these values in (5) we get
$\Rightarrow 100dB = \dfrac{{200}}{{1.13 \times {{10}^{ - 4}}}} \times 32 \times {10^{ - 6}}$
$\Rightarrow 100dB = 56.63$
Finally dividing RHS by $100$ we get the magnetic field as
$\Rightarrow dB = 0.5663{\text{ T}} \approx 0.566{\text{ T}}$
Hence, the correct answer is option (D).

Note
Here we do not need to worry about the negative sign which comes in the equation of the Faraday’s law. Here we are only concerned with the magnitude of the emf induced, and not with its polarity. So we can neglect the negative sign in the equation.