A proton is moving with a uniform velocity of $\$ 0 \mathrm{~ns}^{-1}$ along the $Y$-axis, under the joint action of a magnetic field aloig $Z$-axis and an electric field of magnitude $2 \times 10 \mathrm{Vm}^{-1}$ along the negativeX-axis. If the electric field is switched off, the proton starts moving in a circle. The radius of the circle is nearly $\left(\right.$ given $\frac{e}{m}$ ratio for proton $\left.=10 \mathrm{Ckg}^{-1}\right)$

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A proton is moving with a uniform velocity of $\$ 0 \mathrm{~ns}^{-1}$ along the $Y$-axis, under the joint action of a magnetic field aloig $Z$-axis and an electric field of magnitude $2 \times 10 \mathrm{Vm}^{-1}$ along the negativeX-axis. If the electric field is switched off, the proton starts moving in a circle. The radius of the circle is nearly $\left(\right.$ given $\frac{e}{m}$ ratio for proton $\left.=10 \mathrm{Ckg}^{-1}\right)$

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kritika · Answer 1 · 2021-12-14T08:41:50+0000

Ans: (A)
Initially $F_{E}=F_{m}$ $\Rightarrow \mathrm{B}=\mathrm{E} / \mathrm{V}=(2 / 100) \mathrm{T}$
Now when $E$ is switched off,
$\mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{10^{6} \times 100}{10^{8} \times 2}=\frac{1}{2}=0.5 \mathrm{~m}$

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