A proton is moving in a circular path in a magnetic field perpendicular to its plane. It possesses kinetic energy E. If the magnetic field increases to four times and the proton is constrained to move along the same circle then its kinetic energy will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 16 E
For a charge moving perpendicular to a uniform magnetic field, the magnetic force provides the centripetal force:
\( \displaystyle qvB = \frac{mv^{2}}{r} \;\Rightarrow\; r = \frac{mv}{qB} = \frac{p}{qB} \)
So for a fixed radius the momentum is proportional to the field: \( p = qBr \). The kinetic energy in terms of momentum is
\( \displaystyle E = \frac{p^{2}}{2m} = \frac{q^{2}B^{2}r^{2}}{2m} \;\Rightarrow\; E \propto B^{2}\ \text{(same } r\text{)} \)
\( \displaystyle \frac{E'}{E} = \left(\frac{4B}{B}\right)^{2} = 16 \;\Rightarrow\; E' = 16E \)