Question

A uniform magnetic field of induction $B$ is confined to a cylindrical region of radius $R$. The magnetic field is increasing at a constant ratio of $\dfrac{{dB}}{{dt}}$. An electron of charge $q$, placed at the point $P$ on the periphery of the field experiences an acceleration:

(A) $\dfrac{1}{2}\dfrac{{eR}}{m}\dfrac{{dB}}{{dt}}$ toward left
(B) $\dfrac{1}{2}\dfrac{{eR}}{m}\dfrac{{dB}}{{dt}}$ toward right
(C) $\dfrac{{eR}}{m}\dfrac{{dB}}{{dt}}$ toward left
(D) $\dfrac{1}{2}\dfrac{{eR}}{m}\dfrac{{dB}}{{dt}}$ zero

Answer 1

Here we consider a magnetic field that is uniform and confined to a cylindrical region. The magnetic field is increasing at a constant ratio. If we place an electron at a point on the periphery of the field, it will experience an acceleration. Here we have to find the acceleration experienced by the electron placed at the periphery of the cylindrical region of a uniform magnetic field.

Complete step by step solution:
There will be an electric field induced at the point $P$.
The electric field induced at this point can be written as, $\int {E.dl = \dfrac{{d(AB)}}{{dt}}}$
where $E$ stands for the electric field $\;dl$ stands for the small element of length, $A$ stands for the area of the region, and $B$ stands for the magnetic field.
The above equation can be written as,
$\Rightarrow$ $E.\int {dl = A\dfrac{{dB}}{{dt}}}$
From the figure, we can write
$\int {dl = 2\pi R}$
and $A = \pi {R^2}$
Substituting these values within the above equation, we get
$E\left( {2\pi R} \right) = \left( {\pi {R^2}} \right)\dfrac{{dB}}{{dt}}$
From the above equation, we can write
$E = \dfrac{{\pi {R^2}}}{{2\pi R}}\dfrac{{dB}}{{dt}}$
Cancelling the common terms, we get
$E = \dfrac{R}{2}\dfrac{{dB}}{{dt}}$
Applying the right-hand thumb rule, if the curl of the fingers represents the direction of the induced electric field the thumb points towards the change in the magnetic field.
Hence here the induced electric field will be towards the right.
We know that the force, $F = ma$
From this, we can write
$\Rightarrow$ $a = \dfrac{F}{m}$
We can write the force here as, $F = eE$
where $e$ stands for the charge and $E$ stands for the electric field.
Then the above equation will become
$a = \dfrac{{eE}}{m}$
Substituting $E = \dfrac{R}{2}\dfrac{{dB}}{{dt}}$ in the above equation, we get
$\Rightarrow$ $a = \dfrac{1}{2}\dfrac{{eR}}{m}\dfrac{{dB}}{{dt}}$
The electron will be accelerated towards a direction opposite to the direction of the electric field.
Therefore the acceleration of the electron will be,
$\Rightarrow$ $a = \dfrac{1}{2}\dfrac{{eR}}{m}\dfrac{{dB}}{{dt}}$ toward left
The answer is:

Option (A): $a = \dfrac{1}{2}\dfrac{{eR}}{m}\dfrac{{dB}}{{dt}}$ toward left

Note:
The direction of electric and magnetic fields are found by applying various thumb rules. Whenever the magnetic flux linked with a closed circuit changes an e.m.f is induced in the circuit. This phenomenon is called electromagnetic induction. The e.m.f induced in the circuit is called induced e.m.f.