Question

Find the radius of the circular path described by a proton moving at 15 km/s in a plane perpendicular to a 400- G magnetic field.

Answer 1

When a charge entered perpendicular into a magnetic field, then the magnetic force on the charge rotates the charge in circular path. The magnetic force on the charge is equal to the centripetal force.

$$ \begin{aligned} F_{B} &=F_{c} \\ B q v \sin 90^{\circ} &=\frac{m v^{2}}{r} \\ r &=\frac{m v}{B q}=\frac{\left(1.67 \times 10^{-27} \mathrm{~kg}\right)\left(15 \times 10^{3} \mathrm{~m} / \mathrm{s}\right)}{\left(400 \mathrm{G} \frac{1 \mathrm{~T}}{10^{4} \mathrm{G}}\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)} \\ &=3.9 \times 10^{-3} \mathrm{~m}=3.9 \mathrm{~mm} \end{aligned} $$