JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
The figure shows a region of length \('l'\) with a uniform magnetic field of \(0.3\, T\) in it and a proton entering the region with velocity \(4 \times 10^{5}\, ms ^{-1}\) making an angle \(60^{\circ}\) with the field. If the proton completes \(10\) revolution by the time it cross the region shown, \(l\) is close to....... \(m\) (mass of proton \(=1.67 \times 10^{-27} \,kg ,\) charge of the proton \(\left.=1.6 \times 10^{-19}\, C \right)\)

- A \(0.11\)
- B \(0.22\)
- C \(0.44\)
- D \(0.88\)
Answer & Solution
Correct Answer
(C) \(0.44\)
Step-by-step Solution
Detailed explanation
\(T =\frac{2 \pi m }{ qB }\) total time \(t=10 T\) Kinematics \(\ell=\frac{ V }{2} t\) \(\ell=\frac{V}{2} 10 \times \frac{2 \pi m}{q B}\) \(=4 \times 10^{5} \times 10 \times \frac{3.14 \times 1.67 \times 1 0^{-27}}{1.6 \times 10^{-19} \times 0.3}\) \(=0.439\)
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