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Q.

Each of the lettered points at the corners of the cube in figure represents a positive charge $q$ moving with a velocity of magnitude $v$ in the direction indicated. The region in the figure is in a uniform magnetic field $B$ , parallel to the $x$ -axis and directed toward the right. Find the magnitude and direction of the force on each charge.
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a

$F_{a} = q v B \hat{k}, F_{b} = + q v B \hat{j}, F_{c} = 0, F_{d} = \frac{- q v B}{\sqrt{2}} \hat{j}$ $F_{e} = \frac{q v B}{\sqrt{2}} (\hat{j} - \hat{k})$

b

$F_{a} = - q v B \hat{k}, F_{b} = - q v B \hat{j}, F_{c} = 0, F_{d} = \frac{- q v B}{\sqrt{2}} \hat{j}$ $F_{e} = \frac{q v B}{\sqrt{2}} (- \hat{j} + \hat{k})$

c

$F_{a} = - q v B \hat{k}, F_{b} = + q v B \hat{j}, F_{c} = 0, F_{d} = \frac{- q v B}{\sqrt{3}} \hat{j}$ $F_{e} = \frac{q v B}{\sqrt{3}} (- \hat{j} - \hat{k})$

d

$F_{a} = - q v B \hat{k}, F_{b} = + q v B \hat{j}, F_{c} = 0, F_{d} = \frac{- q v B}{\sqrt{2}} \hat{j}$ $F_{e} = \frac{q v B}{\sqrt{2}} (- \hat{j} - \hat{k})$

answer is A.

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Detailed Solution

Apply $F = q (v \times B)$

For example, let us apply for charge particle as follow,

$F_{e} = q [(\frac{v}{\sqrt{2}} \hat{j} - \frac{v}{\sqrt{2}} \hat{k}) \times (B \hat{i})] = \frac{q v B}{\sqrt{2}} (- \hat{j} - \hat{k})$

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