Q: There exists a uniform cylindrically symmetric magnetic field directed along the axis of a cylinder but varying with time as B = k t. If an electron is released from rest in this field at a distance of ‘ r ‘ from the axis of cylinder, its acceleration, just after it is released would be (e and m are the electronic charge and mass respectively)

Solution :$\displaystyle \oint \vec{E}.\vec{dl} = \frac{d \phi}{dt}$

$\displaystyle E (2 \pi r) = \pi r^2 \frac{d B}{dt}$

$\displaystyle E = \frac{r}{2} \frac{d B}{dt}$

$\displaystyle E = \frac{r}{2} \frac{d}{dt}(k t) $

$\displaystyle E = \frac{r}{2} k $

Acceleration $\displaystyle a = \frac{e E}{m} $

$\displaystyle a = \frac{e}{m} \frac{k r}{2} $

$\displaystyle a = \frac{e r k}{2 m} $

Directed along tangent to the circle of radius r whose center lies on the axis of cylinder.