Question

Find the distance travelled by the second particle during that interval mentioned in the above problem

• A. $\sqrt{\frac{Vm}{q}}\frac{2\pi }{B}cos\theta$
• B. $\sqrt{\frac{2Vm}{3q}}\frac{2\pi }{B}cos\theta$
• C. $\sqrt{\frac{2Vm}{q}}\frac{2\pi }{B}cos\theta$
• D. $\frac{2}{3}\sqrt{\frac{Vm}{q}}\frac{\pi }{m}cos\theta$

Answer 1

Answer: (C)

$\sqrt{\frac{2Vm}{q}}\frac{2\pi }{B}cos\theta$

Velocity of the first particle is due to acceleration by a potential V.

$\frac{1}{2}m{v}^2=qV$

$\Rightarrow v=\sqrt{\frac{2qV}{m}}$

Component of velocity parallel to magnetic field B is $v_{\parallel }=v\cos \theta$

Velocity of the second particle needs to be equal to the $v_{\parallel }$ of the first particle.

Distance travelled by the 2nd particle when both particles meet is

$d=\sqrt{\frac{2qV}{m}}\cos \theta \times T=\sqrt{\frac{2qV}{m}}\cos \theta \times \frac{2\pi m}{qB}=\sqrt{\frac{2Vm}{q}}\frac{2\pi \cos \theta }{B}$