Question

A particle with velocity $2\times {10}^{4}m/s$ enters the uniform magnetic field perpendicularly. Calculate the magnitude of the magnetic force on this particle if charge of particle is -0.04 C and magnetic field of strength is B= 0.5 T.

• A. 4 N
• B. 8 N
• C. 40 N
• D. 80 N
• E. 400 N

Answer 1

The correct option is E 400 N

Magnetic force $F=q(v\times B)$

As $v\perp B$ $⟹|F|=|q|vB$

$\therefore$ $|F|=0.04\times (2\times {10}^4)\times 0.5=400$ $N$