Question

A proton of mass $m$ and charge $q$ is moving in a plane with kinetic energy $E$. If there exists a uniform magnetic field $B$, perpendicular to the plane of the motion, the portion will move in a circular path of radius

• A. $\frac{2EM}{qB}$
• B. $\frac{\sqrt{2EM}}{qB}$
• C. $\frac{\sqrt{EM}}{2qB}$
• D. $\sqrt{\frac{2Eq}{mB}}$

Answer 1

Answer: (B)

Given, Kinetic energy $=E$
Mass $=m $
Magnetic field $=B$
Charge $=q$ We know that
$F=q v B \sin \theta$
(motion of a charged particle in a uniform magnetic field) If
$\theta=90^{\circ}$
Then $F=q v B\,\,\,...(i)$
We know that also (centripetal force)
$F=\frac{m v^{2}}{r}\,\,\,...(ii)$
From Eqs. (i) and (ii), we get
$q v B=\frac{m v^{2}}{r}, r=\frac{m v}{q B}$
$\left[\because E=\frac{1}{2} m v^{2}, v=\sqrt{\frac{2 E}{m}}\right]$
$\therefore r=\frac{m \sqrt{\frac{2 E}{m}}}{q B} $
$\Rightarrow r=\frac{\sqrt{2 E m}}{q B}$