Question

A charged particle gains a speed of $10^6$ m/s, when accelerated from rest through a potential difference $10$ kV. It enters a region of magnetic field of $0.4$ T such that it enters the magnetic field perpendicularly. The radius of circular path described by it is :

• A. 5 cm
• B. 5 mm
• C. 0.5 m
• D. 50 cm

Answer 1

Answer: (A)

5 cm

Answer 2

The kinetic energy gained by the charged particle is equal to the work done by the electric field: $q\Delta V = \frac{1}{2}mv^2$ This gives the charge-to-mass ratio: $\frac{q}{m} = \frac{v^2}{2\Delta V}$

When the particle enters the magnetic field perpendicularly, the magnetic force provides the centripetal force for circular motion: $qvB = \frac{mv^2}{r}$ Rearranging for the radius $r$: $r = \frac{mv}{qB}$

Substitute the expression for $\frac{q}{m}$ into the radius equation: $r = \frac{v}{B} \left(\frac{m}{q}\right) = \frac{v}{B} \left(\frac{2\Delta V}{v^2}\right) = \frac{2\Delta V}{vB}$

Given values: Potential difference, $\Delta V = 10 \text{ kV} = 10 \times 10^3 \text{ V}$ Speed, $v = 10^6 \text{ m/s}$ Magnetic field, $B = 0.4 \text{ T}$

Plugging in the values: $r = \frac{2 \times (10 \times 10^3 \text{ V})}{(10^6 \text{ m/s}) \times (0.4 \text{ T})} = \frac{2 \times 10^4}{0.4 \times 10^6} = \frac{2}{0.4} \times 10^{4-6} = 5 \times 10^{-2} \text{ m}$ $r = 0.05 \text{ m} = 5 \text{ cm}$