Question

A charged particle enters a uniform magnetic field with velocity vector at an angle of 45° with the magnetic field . The pitch of the helical path followed by the particle is p . The radius of the helix will be (1) p/√2π. (2) √2π (3) p/2π (4) √2p/π

Answer 1

Answer on Question #75000, Physics / Electromagnetism

Question. A charged particle enters a uniform magnetic field with velocity vector at an angle of $45{}^{\circ}$ with the magnetic field. The pitch of the helical path followed by the particle is $p$. The radius of the helix will be (1) $p / \sqrt{2}\pi$; (2) $\sqrt{2}\pi$; (3) $p / 2\pi$; (4) $\sqrt{2p} / \pi$.

Solution.

According to the second Newton's law

$$F = m a$$

So,

$$q v B \sin \alpha = m \frac {(v \sin \alpha) ^ {2}}{R} \rightarrow R = \frac {m v \sin 4 5 {}^ {\circ}}{q B} = \frac {m v}{\sqrt {2} q B}$$

The pitch of the helical path

$$p = v \cos \alpha \cdot T = v \cos \alpha \cdot \frac {2 \pi R}{v \sin \alpha} = \cos 4 5 {}^ {\circ} \cdot \frac {2 \pi R}{\sin 4 5 {}^ {\circ}} = 2 \pi R \rightarrow$$ $$R = \frac {p}{2 \pi}$$

Answer. (3) $R = p / 2\pi$

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