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Q. A particle moving with velocity $v$ having specific charge $\left(\right.q/m\left.\right)$ enters a region of magnetic field $B$ having width $d=\frac{3 mv}{5 qB}$ at angle $53^\circ $ to the boundary of magnetic field. Find the angle $\theta $ in the diagram.
Question

NTA AbhyasNTA Abhyas 2022

Solution:

$R=\frac{mv}{qB}$
$d=Rcos53°$
Solution
$\Rightarrow \theta =90°$