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A proton is moving with a uniform velocity of $10^{6} \mathrm{ms}^{-1}$ along the $Y$ -axis, under the joint action of a magnetic field along $Z$ -axis and an electric field of magnitude $2 \times 10^{4} \mathrm{Vm}^{-1}$ along the negative $X$ -axis. If the electric field is switched off, the proton starts moving in a circle. The radius of the circle is nearly given: $\frac{e}{m}$ ratio for proton $=10^{4} \mathrm{Ckg}^{-1}$
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Verified Answer
The correct answer is:
$0.5 \mathrm{m}$
Velocity of proton $=10^{6} \mathrm{m} / \mathrm{s}$ along $y$ -direction
$$
\text { Electric field }=2 \times 10^{4} \mathrm{V} / \mathrm{m}
$$
$\frac{e}{=}$ for proton $=10^{-8} \mathrm{C} / \mathrm{kg}$
it
The magnetic field,
$$
\begin{aligned}
B &=\frac{E}{v} \quad\left(\because q, v B=q_{\rho} E\right) \\
&=\frac{2 \times 10^{4}}{10^{6}}=2 \times 10^{-2} \mathrm{T}
\end{aligned}
$$
The radius of circular path
$$
\begin{aligned}
\gamma &=\frac{m v}{q_{p} \cdot B}=\frac{10^{-6} \times 10^{6}}{2 \times 10^{-2}} \\
&=\frac{1}{2}=0 \cdot 5 \mathrm{m}
\end{aligned}
$$
$$
\text { Electric field }=2 \times 10^{4} \mathrm{V} / \mathrm{m}
$$
$\frac{e}{=}$ for proton $=10^{-8} \mathrm{C} / \mathrm{kg}$
it
The magnetic field,
$$
\begin{aligned}
B &=\frac{E}{v} \quad\left(\because q, v B=q_{\rho} E\right) \\
&=\frac{2 \times 10^{4}}{10^{6}}=2 \times 10^{-2} \mathrm{T}
\end{aligned}
$$
The radius of circular path
$$
\begin{aligned}
\gamma &=\frac{m v}{q_{p} \cdot B}=\frac{10^{-6} \times 10^{6}}{2 \times 10^{-2}} \\
&=\frac{1}{2}=0 \cdot 5 \mathrm{m}
\end{aligned}
$$
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