A proton enters a uniform magnetic field of 0.2 T with speed v. It describes a semi-circular path of radius 5 cm and comes out of the field. The final speed of the proton will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → v
Magnetic force provides centripetal force:
$qvB = \frac{mv^2}{r}$
Therefore, $r = \frac{mv}{qB}$
Since magnetic force is perpendicular to velocity, it does no work. Hence, kinetic energy and speed remain constant.
Final speed = Initial speed = $v$
$v_{\text{final}} = v$