Question

If a proton of kinetic energy 8.35 MeV enters a uniform magnetic field of 10 T at right angles to the direction of the field, then the force acting on the proton is

• A. $48 \times 10^{-12}$ N
• B. $16 \times 10^{-12}$ N
• C. $64 \times 10^{-12}$ N
• D. $32 \times 10^{-12}$ N

Hint:

For charged particles moving in a magnetic field, always remember to convert kinetic energy to velocity using the relationship $K.E = \frac{1}{2}mv^2$.

Answer 1

The Correct Option is A

The force on a charged particle moving in a magnetic field is given by: $$ F = qvB $$ where $q$ is the charge, $v$ is the velocity, and $B$ is the magnetic field. First, we convert the kinetic energy to velocity, then calculate the force. Substituting the given values, we get the force acting on the proton to be $48 \times 10^{-12}$ N.