Question

A proton enters a magnetic field of $1.5weber/m^2$ with a velocity of $2\times {10}^{7}m/sec$ at an angle of ${30}^o$ with the field. The force on the proton will be:

• A. $2.4\times {10}^{-12}N$
• B. $0.24\times {10}^{-12}N$
• C. $24\times {10}^{-12}N$
• D. $0.024\times {10}^{-12}N$

Answer 1

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

Given that,

Charge on proton$q=1.602\times {10}^{-19}C$

Velocity $v=2\times {10}^{7}m/s$

Angle $\theta ={30}^0$

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

We know that, the magnitude of the magnetic force on a charge particle

$F=qvB\sin \theta$

$F=1.6\times {10}^{-19}\times 2\times {10}^7\times 1.5\times \sin {30}^0$

$F=4.8\times {10}^{-12}\times \frac{1}{2}$

$F=2.4\times {10}^{-12}N$

Hence, the force on the proton is $2.4\times {10}^{-12}N$