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# A particle having charge $q$ enters a region of uniform magnetic field $B$ directed inwards and is deflected as shown to a distance $d$ above the original line of flight as shown. What is the (magnitude of) momentum of the particle?

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

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• 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.
$R^2= y^2 + (r - x)^2$
$R^2 =\large\frac{ (x^2 + y^2 )}{2x}$ $= \large\frac{ P}{qB}$
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})$
$\Rightarrow p = qBr = \large\frac{qB}{2} \bigg( \normalsize d + \large\frac{a^2}{d} \bigg)$
edited Mar 14, 2014