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 rate of $dB/dtT\text{ }{{s}^{-1}}$. An electron 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. Zero

Answer 1

Hint: The increment of magnetic field induces an emf to the electron placed in the field. This induced emf is due to change in flux. The magnetic flux is a product of magnetic field and area of cross-section. We can assign this to the induced emf equation to find the acceleration.

Complete step by step answer:
An emf is induced at the point $P$ due to the presence of charge and magnetic field.
Induced emf can be written as the negative rate of change in flux.
$V=-\dfrac{d\phi }{dt}$…………………(1)
Magnetic flux can be written as the product of magnetic field and area.
$\phi =\overrightarrow{B}\cdot \overrightarrow{A}$
Here the magnetic field and area vector will be in the same direction.
Therefore, $\phi =BA$……………………(2)
We can assign this into the equation 1
Therefore, the induced emf will be,
$V=-\dfrac{d(BA)}{dt}$
Since Area is not changing with time, we can take it out from the differential operator.
$V=-A\dfrac{dB}{dt}$…………………………..(3)
Electric field is the negative of potential gradient.
$\overrightarrow{E}=\dfrac{-dV}{dx}$
We can rewrite this equation as,
$-\overrightarrow{E}\cdot d\vec{x}=dV$
$-\int{\overrightarrow{E}\cdot d\vec{x}}=V$
We can put $V=-A\dfrac{dB}{dt}$in this equation.
$-\int{\overrightarrow{E}\cdot d\vec{x}}=-A\dfrac{dB}{dt}$……………………(4)
Radius of the circle is $R$, then
$\int{dx=2\pi R}$, since we are dealing with the circle.
The direction of the electric field can be found by using right hand rule. According to this, the right thumb points towards the direction opposite to the change in magnetic field and the other fingers will be in the direction of the induced electric field.
Then the equation 4 will be,
$E.2\pi R=A\dfrac{dB}{dt}$
The area of the circle has to include in this equation instead of $A$
$E.2\pi R=\pi {{R}^{2}}\dfrac{dB}{dt}$
So, we can write the electric field as,
$E=\dfrac{R}{2}\dfrac{dB}{dt}$……………………….(5)
To find the acceleration, first we have to find the force from the electric field.
$F=qE$
Here, the electron is the charge.
$F=\dfrac{eR}{2}\dfrac{dB}{dt}$
$F=ma$
$ma=\dfrac{eR}{2}\dfrac{dB}{dt}$
Then, the acceleration can be written as,
$a=\dfrac{eR}{2m}\dfrac{dB}{dt}$
The current is moving towards the right direction of the circle. So, the electron flow will be in the opposite direction. The induced electric field is towards the right direction. Hence the acceleration will be in the left direction.

So, the correct option is A.

Note: We are using right hand rule to find the direction of the electric field. The acceleration will be in the opposite direction of the electric field. Since we are dealing with electrons. Electrons will flow towards the opposite direction of current.