Question

A proton is moving perpendicular to a uniform magnetic field of $2.5$ tesla with $2MeV$ kinetic energy. The force on proton is ________N. (Mass of proton $==1.6\times {10}^{-27}kg$. Charge of proton $=1.6\times {10}^{-19}C$

• A. $8\times {10}^{-12}$
• B. $8\times {10}^{-11}$
• C. $3\times {10}^{-11}$
• D. $3\times {10}^{-10}$

Answer 1

The correct option is A $8\times {10}^{-12}$

Force on moving charge in the magnetic field

$F=qvB\sin \theta$
but $\theta =90$

$F=qvB$

Hence velocity $v=\sqrt{\frac{2E}{m}}$ ($E$ is kinetic energy of proton)]

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

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

putting the values we get
$F=1.6\times {10}^{-19}\times 2\times {10}^7\times 2.5$

$=8\times {10}^{-12}N$