1. Motion of a charged particle in a uniform magnetic field

A proton enters a uniform magnetic field $B=0.3\,\text{T}$ with speed $v=4\times 10^5\,\text{m s}^{-1}$ and makes angle $60^\circ$ with the field.

Its velocity has two components:

• parallel to field: $v_\parallel = v\cos 60^\circ = 4\times 10^5 \times \frac12 = 2\times 10^5\,\text{m s}^{-1}$
• perpendicular to field: $v_\perp = v\sin 60^\circ = 4\times 10^5\times \frac{\sqrt3}{2}$

Because of these components, the proton follows a helical path.

2. Time period of revolution in magnetic field

The time period of circular motion due to the perpendicular component is $T=\frac{2\pi m}{qB}$

Substitute the values: $T=\frac{2\pi(1.67\times 10^{-27})}{(1.6\times 10^{-19})(0.3)}$

First calculate denominator: $qB = 1.6\times 10^{-19}\times 0.3 = 4.8\times 10^{-20}$

So, $T=2\pi\times \frac{1.67\times 10^{-27}}{4.8\times 10^{-20}}$ $T=2\pi\times 0.3479\times 10^{-7}$ $T\approx 2.186\times 10^{-7}\,\text{s}$

3. Time for 10 revolutions

If the proton completes 10 revolutions, total time spent in the field is $t=10T=10\times 2.186\times 10^{-7}$ $t=2.186\times 10^{-6}\,\text{s}$

4. Distance travelled along the field direction

The proton crosses the region along the magnetic field direction with constant velocity $v_\parallel$. Thus the length of the region is $l=v_\parallel t$

So, $l=(2\times 10^5)(2.186\times 10^{-6})$ $l=4.372\times 10^{-1}\,\text{m}$ $l\approx 0.44\,\text{m}$

5. Option check

• A: $0.22\,\text{m}$ ❌
• B: $0.11\,\text{m}$ ❌
• C: $0.88\,\text{m}$ ❌
• D: $0.44\,\text{m}$ ✅

Hence, the correct answer is Option D.