WBJEE · Physics · Magnetic Effects of Current
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 \(=10^{4} \mathrm{Ckg}^{-1}\)
- A \(0.5 \mathrm{m}\)
- B \(0.2 \mathrm{m}\)
- C \(0.1 \mathrm{m}\)
- D \(0.05 \mathrm{m}\)
Answer & Solution
Correct Answer
(A) \(0.5 \mathrm{m}\)
Step-by-step Solution
Detailed explanation
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,…
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