$\begin{array}{1 1} (A) \large\frac{qBd}{2} \\ (B) \large\frac{qBa}{2} \\ (C) \large\frac{qBa^2}{2d} \\ (D) \large\frac{qB}{2} \bigg( \normalsize d + \large\frac{a^2}{d} \bigg)\end{array}$

- The direction of motion of the particle and the direction of the magnetic field are known. Therefore, we can find the direction of the magnetic force acting on a positive and negative charge using Lorent's law.

This question is all about writing radius in an appropriate way

$R^2= y^2 + (r - x)^2 $

$R^2 =\large\frac{ (x^2 + y^2 )}{2x}$ $ = \large\frac{ P}{qB}$

Ans : (d)

The vector product between $v$ and $B$ points upwards in the figure thus indicating that the charge of the particle is positive.

The Force acting on the moving charge $F = qvB$

As a result of the magnetic force, the charged particle will follow a spherical trajectory. The centripetal force $F_c = \large\frac{mv^2}{r}$

Now, the Force acting on the moving charge $F = F_c$, the centripetal force $\rightarrow qvB = \large\frac{mv^2}{r}$

$\Rightarrow r = \large\frac{mv}{qB}$$ = \large\frac{p}{qB}$, where $p$ is the momentum of the particle.

This question is all about figuring out $r$ radius (see figure below) in an appropriate way, $r^2= d^2 + (r - a)^2 \rightarrow r = \large\frac{d^2+a^2}{2d} $ $ = \large\frac{1}{2}$$ \large (\normalsize d + \large\frac{a^2}{d})$

$\Rightarrow p = qBr = \large\frac{qB}{2} \bigg( \normalsize d + \large\frac{a^2}{d} \bigg)$

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