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JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism

A proton with a kinetic energy of \(2.0\,eV\) moves into a region of uniform magnetic field of magnitude \(\frac{\pi}{2} \times 10^{-3}\,T\). The angle between the direction of magnetic field and velocity of proton is \(60^{\circ}\). The pitch of the helical path taken by the proton is \(..........cm\) (Take, mass of proton \(=1.6 \times 10^{-27}\,kg\) and Charge on proton \(=1.6 \times 10^{-19}\,kg)\)

  1. A \(38\)
  2. B \(41\)
  3. C \(40\)
  4. D \(42\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(40\)

Step-by-step Solution

Detailed explanation

\(B=\frac{\pi}{2} \times 10^{-3}\) \(K E=\frac{1}{2} m V^2\) \(\Rightarrow V=\sqrt{\frac{2 K E}{m}}\) \(=v \cos 60^{\circ} \times \frac{2 \pi m}{e B}\)…
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