First slide

Force on a charged particle moving in a magnetic field

Question

A charged particle is projected in a uniform magnetic field making an angle of $30^{o}$ with the field, and radius of its trajectory is found to be 50 cm. When the same particle is projected in the same magnetic field making an angle of $60^{o}$ with the field, what will be the radius of the trajectory?

Moderate

Solution

$r =$ Radius of helical path $= \frac{V \sin θ}{ω} = \frac{m V \sin θ}{Q B}$
$∴ r \propto \sin θ \Rightarrow \frac{r_{2}}{r_{1}} = \frac{\sin θ_{2}}{\sin θ_{1}} \Rightarrow \frac{r_{2}}{50} = \frac{\sin 60^{o}}{\sin 30^{o}}$
$∴ r_{2} = 50 \times \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} c m = 50 \sqrt{3} c m$

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