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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 proton will move in a circular path of radius
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Verified Answer
The correct answer is:
$\frac{\sqrt{2 E m}}{q B}$
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
$$
F=q v B
$$
We know that also (centripetal force)
$$
F=\frac{m v^{2}}{r}
$$
From Eqs. (i) and (ii), we get
$$
\begin{aligned}
q v B &=\frac{m v^{2}}{r}, r=\frac{m v}{q B}\left[\because E=\frac{1}{2} m v^{2}, v=\sqrt{\frac{2 E}{m}}\right] \\
\therefore \quad r &=\frac{m \sqrt{\frac{2 E}{m}}}{q B} \Rightarrow r=\frac{\sqrt{2 E m}}{q B}
\end{aligned}
$$
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
$$
F=q v B
$$
We know that also (centripetal force)
$$
F=\frac{m v^{2}}{r}
$$
From Eqs. (i) and (ii), we get
$$
\begin{aligned}
q v B &=\frac{m v^{2}}{r}, r=\frac{m v}{q B}\left[\because E=\frac{1}{2} m v^{2}, v=\sqrt{\frac{2 E}{m}}\right] \\
\therefore \quad r &=\frac{m \sqrt{\frac{2 E}{m}}}{q B} \Rightarrow r=\frac{\sqrt{2 E m}}{q B}
\end{aligned}
$$
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