JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A beam of protons with speed \(4 \times 10^{5}\, ms ^{-1}\) enters a uniform magnetic field of \(0.3\, T\) at an angle of \(60^{\circ}\) to the magnetic field. The pitch of the resulting helical path of protons is close to....\(cm\) (Mass of the proton \(=1.67 \times 10^{-27}\, kg\), charge of the proton \(=1.69 \times 10^{-19}\,C\))
- A \(12\)
- B \(4\)
- C \(5\)
- D \(2\)
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
Pitch \(=\frac{2 \pi m }{ qB } v \cos \theta\) Pitch \(=\frac{2(3.14)\left(1.67 \times 10^{-27}\right) \times 4 \times 10^{5} \times \cos 60}{\left(1.69 \times 10^{-19}\right)(0.3)}\) Pitch \(=0.04 m =4 cm\)
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