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(D) The radius of a charged particle moving in a uniform magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$,where $K$ is the kinet

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$10$

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(D) The radius of a charged particle moving in a uniform magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$,where $K$ is the kinetic energy.Since the path $(r)$ and the magnetic field $(B)$ are the same for both particles,we have $\frac{\sqrt{2m_{\alpha}K_{\alpha}}}{q_{\alpha}B} = \frac{\sqrt{2m_{p}K_{p}}}{q_{p}B}$.Squaring both sides and rearranging,we get $\frac{2m_{\alpha}K_{\alpha}}{q_{\alpha}^2} = \frac{2m_{p}K_{p}}{q_{p}^2}$,which implies $K_{p} = K_{\alpha} \cdot \left(\frac{m_{\alpha}}{m_{p}}\right) \cdot \left(\frac{q_{p}}{q_{\alpha}}\right)^2$.Given $m_{\alpha} = 4m_{p}$ and $q_{\alpha} = 2q_{p}$,we substitute these values:$K_{p} = 10 \ eV \cdot \left(\frac{4m_{p}}{m_{p}}\right) \cdot \left(\frac{q_{p}}{2q_{p}}\right)^2 = 10 \ eV \cdot 4 \cdot \frac{1}{4} = 10 \ eV$.

List-$I$	List-$II$
$a$. Fleming's left-hand rule	$e$. Direction of induced current
$b$. Fleming's right-hand rule	$f$. South pole
$c$. Clockwise current	$g$. North pole
$d$. Anticlockwise current	$h$. Direction of force

An $\alpha$ particle of energy $10 \ eV$ is moving in a circular path in a uniform magnetic field. The energy of a proton moving in the same path and the same magnetic field will be [mass of $\alpha$ particle $= 4 \times$ mass of proton]. (in $eV$)

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The figure shows a circular coil carrying current $i$ kept very close but not touching at a point $A$ on a straight conductor carrying the same current $i$. The magnitude of magnetic induction at the centre $O$ of the circular coil will be:

Match the following:
List-$I$ List-$II$
$a$. Fleming's left-hand rule $e$. Direction of induced current
$b$. Fleming's right-hand rule $f$. South pole
$c$. Clockwise current $g$. North pole
$d$. Anticlockwise current $h$. Direction of force

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