JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
Consider a long thin conducting wire carrying a uniform current I. A particle having mass " M " and charge " \(q\) " is released at a distance " \(a\) " from the wire with a speed \(v_0\) along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance \(x\) from the wire. The value of \(x\) is [ \(\mu_0\) is vacuum permeability]
- A \(a e^{-\frac{4 \pi \mathrm{mv}_{\mathrm{o}}}{\mathrm{q} \mu_{\mathrm{o}} \mathrm{I}}}\)
- B \(a\left[1-\frac{\mathrm{mv}_{\mathrm{o}}}{2 q \mu_{\mathrm{o}} \mathrm{I}}\right]\)
- C \(a\left[1-\frac{\mathrm{mv}}{\mathrm{q} \mu_{\mathrm{o}} \mathrm{I}}\right]\)
- D \(\frac{a}{2}\)
Answer & Solution
Correct Answer
(A) \(a e^{-\frac{4 \pi \mathrm{mv}_{\mathrm{o}}}{\mathrm{q} \mu_{\mathrm{o}} \mathrm{I}}}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & A \rightarrow B \\ & \vec{V}=-v_x \hat{i}+v_y \hat{j} \\ & \vec{B}=\frac{\mu_0 I}{2 \pi r}(-\hat{k}) \\ & \overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}} \times \overrightarrow{\mathrm{B}})=\frac{\mu_0 \mathrm{Iq}}{2 \pi…
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