Question

A proton travelling at $23^{\circ }$ w.r.t. the direction of a magnetic field of strength $2.6mT$ experiences a magnetic force of $6.5\times 10^{-17}N$ what is the speed of the proton?

• A. $2\times 10^{-5}m/s$
• B. $4\times 10^{5}m/s$
• C. $5\times 10^{-5}m/s$
• D. $6\times 10^{-5}m/s$

Answer 1

Answer: (B)

Given $\theta =23^{\circ }$,
$B=2.6mT=2.6\times 10^{-6}T$ and
$F=6.5\times 10^{-17}N$
We know
$F=qvB\sin \theta$
$6.5\times 10^{-17}=1.6\times 10^{-19}\times v\times 2.6\times 10^{-6}\times \sin 23^{\circ }$
$v=\frac{6.5\times 10^{-17}}{2.6\times 10^{-6}\times 1.6\times 10^{-19}\times 0.39}$
$v=4\times 10^{5}m{s}^{-1}$