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 \(=1.6 \times 10^{-19}\,kg)\)
- A \(38\)
- B \(41\)
- C \(40\)
- D \(42\)
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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