Question

A particle having charge $q$ enters a region of uniform magnetic field $\overrightarrow{B}$ (directed perpendicularly inwards to the plane of paper), and is deflected a distance $y$ after travelling a distance $x$ in horizontal direction as shown in figure. Then the magnitude of momentum of the particle is :

• A. $\frac{qB{y}^2}{2x}$
• B. $\frac{qBy}{x}$
• C. $\frac{qB}{2}(\frac{y^2}{x}+x)$
• D. $\frac{qB}{2}(\frac{x^2}{y}+y)$

Answer 1

The correct option is D $\frac{qB}{2}(\frac{x^2}{y}+y)$

The given path followed by the particle is a part of the circle of radius $R$.


And we know that $R=\frac{mv}{qB}$

$R=\frac{p}{qB}$ $(p=$ momentum of particle)

From geometry:

$AB=\sqrt{x^2+y^2}$

$AC=BC=\frac{\sqrt{x^2+y^2}}{2}$

Thus, $AB=2BC=2R\sin \theta$

$\Rightarrow \sqrt{x^2+y^2}=2R\left(\frac{y}{\sqrt{x^2+y^2}}\right)$

$\frac{x^2+y^2}{y}=2R$

$\frac{x^2+y^2}{y}=\frac{2p}{qB}(\because R=p/qB)$

$\therefore p=\frac{qB}{2}(\frac{x^2}{y}+y)$

Hence, option $\left(d\right)$ is the right choice.

Why this question ? Tip: The path followed by a moving charge particle in a uniform magnetic field $(\overrightarrow{V}\perp \overrightarrow{B})$ will be circular. Tip: The parameters $y\&x$ can be related by using geometry for the circular arc.