A proton moves with speed \(v\) in a uniform magnetic field. Its velocity makes an angle of \(30^\circ\) with the field, and the magnetic force on it is \(F\). The proton now moves with speed \(v/2\) at right angles to the same field. The magnetic force on it becomes
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(F\)
The magnetic force on a moving charge has magnitude \( qvB\sin\theta \), where \( \theta \) is the angle between the velocity and the field. Here both the speed and the angle change.
In the first case \( \theta=30^\circ \).
\( \begin{aligned} F &= qvB\sin30^\circ \\ &= \frac{qvB}{2} \end{aligned} \)
In the second case the speed is \( v/2 \) and \( \theta=90^\circ \).
\( \begin{aligned} F' &= q\left(\frac{v}{2}\right)B\sin90^\circ \\ &= \frac{qvB}{2} \\ &= F \end{aligned} \)
Halving the speed halves the force, and going from \( \sin30^\circ \) to \( \sin90^\circ \) doubles it. Hence, the force stays \( F \).
NCERT (2026-27):
Magnetic Field, Lorentz Force – “… F = q [ v × B] =q v B sin θ n̂ , where θ is the angle between v and B …”