Question

A proton enters a magnetic field of flux density $1.5Wb/m^2$ with a speed of $2\times {10}^7$ m/s at angle of ${30}^o$ with the field. The force on a proton will be

• A. $0.44\times {10}^{-12}N$
• B. $2.4\times {10}^{-12}N$
• C. $24\times {10}^{-12}N$
• D. $0.024\times {10}^{-12}N$

Answer 1

The correct option is B $2.4\times {10}^{-12}N$

Angle between velocity of proton and magnetic field $\theta ={30}^o$

Velocity of proton $v=2\times {20}^{7}m/s$

Magnetic field $B=1.5Wb/m^2$

Magnetic force acting on the proton $F=qvB\sin \theta$

$\therefore$ $F=(1.6\times {10}^{-19})(2\times {10}^7)\left(1.5\right)\times 0.5$

$⟹$ $F=2.4\times {10}^{-12}N$