JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A proton and a deutron ( \(\mathrm{q}=+\mathrm{e}, m=2.0 \mathrm{u})\) having same kinetic energies enter a region of uniform magnetic field \(\vec{B}\), moving perpendicular to \(\vec{B}\). The ratio of the radius \(r_d\) of deutron path to the radius \(r_p\) of the proton path is _______.
- A \(1: 1\)
- B \(1: \sqrt{2}\)
- C \(\sqrt{2}: 1\)
- D \(1: 2\)
Answer & Solution
Correct Answer
(C) \(\sqrt{2}: 1\)
Step-by-step Solution
Detailed explanation
In uniform magnetic field, \(\mathrm{R}=\frac{\mathrm{m} v}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{~m}(\mathrm{~K} \cdot \mathrm{E})}}{\mathrm{qB}}\) Since same \(K.E\) \(\mathrm{R} \propto \frac{\sqrt{\mathrm{m}}}{\mathrm{q}}\)…
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