Question

A small point mass with <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $m=1\text{kg}$ and charge <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $q=1\text{C}$ enters perpendicularly into a triangular region of uniform magnetic field strength <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $B=2\text{T}$ as shown in figure.
The maximum velocity (in $\text{m/s}$) of mass such that it completes a semicircle in magnetic field region is (Given your answer as an integer)

Answer 1

If the particle completes semicircle it must emerge out from diametrically opposite point.


The charge touches the <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\text{BC}$ at point $\text{P}$ , so $\text{BC}$ is a tangent and $\text{OP}\perp$ $\text{BC}$ will be the radius of semicircle.

In $\Delta ABC$

$\sin \theta =\frac{AC}{BC}\text{(or)}$

$\sin \theta =\frac{4}{5}$

$\theta ={53}^{\circ }$

In $\Delta OPC$

$\angle COP={90}^{\circ }-\angle OCP={90}^{\circ }-{37}^{\circ }={53}^{\circ }$

$\cos {53}^{\circ }=\frac{OP}{OC}$

$\frac{3}{5}=\frac{R}{4-R}$

$\Rightarrow 8R=12$

$\therefore R=\frac{12}{8}=\frac{3}{2}\text{m}$

Thus ${}^{\prime }{R}^{\prime }$ is radius of semicircle, Considering the tangent $\text{BC}$ for upper limit.

$R_{max}=\frac{m{v}_{max}}{qB}$

$v_{max}=\frac{qB{R}_{max}}{m}$

$v_{max}=\frac{\left(1\right)\left(2\right)\times \left(\frac{3}{2}\right)}{1}$

$\therefore v_{max}=3\text{m}/\text{s}$

Accepted answer : $3$

Why this question ? It challenges you uniquely to apply the concept of symmetrical motion of charge particle in magnetic field region along with sound observation of geometry involved.