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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 is p. The radius of the helix will be |
$\frac{p}{\pi}$ $\frac{p}{2\pi}$ $\sqrt{2}p$ $\frac{p}{2\sqrt{\pi}}$ |
$\frac{p}{2\pi}$ |
Pitch P = $\frac{2\pi mv sin\theta}{qB} = 2\pi R$ Since R = $\frac{mvcos\theta}{qB}$ $\Rightarrow R = \frac{P}{2\pi}$ |
