AP EAMCET · PHYSICS · Magnetic Effects of Current
A proton and an \(\alpha\)-particle are simultaneously projected in opposite direction into a region of uniform magnetic field of \(2 \mathrm{mT}\) perpendicular to the direction of the field. After some time it is found that the velocity of proton has changed in direction by \(90^{\circ}\). Then at this time, the angle between the velocity vectors of proton and \(\alpha\)-particle is
- A \(60^{\circ}\)
- B \(90^{\circ}\)
- C \(45^{\circ}\)
- D \(180^{\circ}\)
Answer & Solution
Correct Answer
(C) \(45^{\circ}\)
Step-by-step Solution
Detailed explanation
Key Idea In a circular motion, a body changes its direction by \(90^{\circ}\) in one-fourth of its time period. Given, magnetic field, \(B=2 \mathrm{mT}\) Let \(T_p\) be the time-period of revolution of proton in \(T_p=\frac{2 \pi m_p}{e B}\) ...(i)…
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