A positively charged particle having charge q and mass m enters in isosceles right angle triangular region having uniform magnetic field B as shown in gravity free space. The maximum value of v so that particle must comes out through the region from side PQ is

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(2 − 1) B q L 2 m

B q L 22 m

$\frac{(\sqrt{2} - 1) B q L}{m}$

B q L 2 m

answer is B.

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Detailed Solution

For v to be maximum the line $x + y = L$ must be tangent to the circle
$x 2 + (y − R) 2 = R 2$

Hence $R = (\sqrt{2} - 1) L$

$V_{\max} = (\sqrt{2} - 1) \frac{LBq}{m}$

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