Question

A positive charge q is projected in magnetic field of width $\frac{mv}{\sqrt{2}qB}$ with velocity v as shown in figure. Then time taken by charged particle to emerge from the magnetic field is :

• A. $\frac{m}{\sqrt{2}qB}$
• B. $\frac{\pi m}{4qB}$
• C. $\frac{\pi m}{2qB}$
• D. $\frac{\pi m}{\sqrt{2}qB}$

Answer 1

The correct option is B $\frac{\pi m}{4qB}$

$sin\theta =\frac{x}{R}=\frac{\left(\frac{mv}{\sqrt{2}qB}\right)}{\left(\frac{mv}{qB}\right)}\Rightarrow sin\theta =\frac{1}{\sqrt{2}}\Rightarrow \theta =\frac{\pi }{4}\therefore Timet=\frac{\theta }{\omega }t=\frac{\pi }{4}\frac{m}{qB}$