Question

A proton of energy 200 MeV enters the magnetic field of 5 T. If direction of field is from south to north and motion is upward, the force acting on it will be

• A. Zero
• B.
• C.
• D.

Answer 1

Answer: (B)

Answer 2

$F = q v B$ also Kinetic energy

$K = \frac { 1 } { 2 } m v ^ { 2 }$ $\Rightarrow v = \sqrt { \frac { 2 K } { m } }$

∴ $F = q \sqrt { \frac { 2 K } { m } } B$

$= 1.6 \times 10 ^ { - 19 } \sqrt { \frac { 2 \times 200 \times 10 ^ { 6 } \times 1.6 \times 10 ^ { - 19 } } { 1.67 \times 10 ^ { - 27 } } } \times 5$