Question

A proton moves with a speed of $5.0\times {10}^6$m/s along the x-axis. It enters a region where there is a magnetic field of magnitude $2.0$. Tesla directed at an angle of ${30}^o$ to the x-axis and lying in the xy-plane. The magnitude of the magnetic force on the proton is?

• A. $0.8\times {10}^{-13}$N
• B. $1.6\times {10}^{-13}$N
• C. $8.0\times {10}^{-13}$N
• D. $8.01\times {10}^{-13}$N
• E. $16\times {10}^{-13}$N

Answer 1

The correct option is C $8.0\times {10}^{-13}$N

Given, The speed of proton,

$v=5.0\times {10}^{6}m/s$

Magnetic field, $B=2.0$ Tesla

Angle $\theta ={30}^o$

Charge on proton, $q=1.6\times {10}^{-19}C$

When the proton enters in magnitude field, it experience Lorentz force

$F=q(v\times B)$

$|F|=qvB\sin \theta$

$F=1.6\times {10}^{-19}\times 5\times {10}^6\times 2\times \sin {30}^o$

$=8.0\times {10}^{-13}N$.