Question

A proton moving in a perpendicular magnetic field possesses energy \(E\). The magnetic field is increased four times, but the proton is constrained to move in the path of same radius. The kinetic energy will increase

• A. 16 times
• B. 8 times
• C. 2 times
• D. 4 times

Hint:

If radius is constant in a magnetic field, velocity is directly proportional to magnetic field strength.

Answer 1

The Correct Option is A

Step 1: Write expression for radius in magnetic field.
For a charged particle moving perpendicular to a magnetic field, the radius of circular path is given by:
\[ r = \frac{mv}{qB} \]
Step 2: Condition of constant radius.
Since the radius remains the same,
\[ v \propto B \]
Step 3: Effect of increasing magnetic field.
If magnetic field is increased four times,
\[ v' = 4v \]
Step 4: Relation between kinetic energy and velocity.
Kinetic energy is given by:
\[ K = \frac{1}{2}mv^2 \]
Step 5: Calculate new kinetic energy.
\[ K' = \frac{1}{2}m(4v)^2 = 16\left(\frac{1}{2}mv^2\right) \]
Step 6: Conclusion.
The kinetic energy increases 16 times.