Question

A proton travelling at $23^{\circ}$ w.r.t. the direction of a magnetic field of strength $2.6\, mT$ 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 $\theta=23^{\circ}$,
$B=2.6\, m\, T=2.6 \times 10^{-6}\, T$ and
$F=6.5 \times 10^{-17} N$
We know
$ F=q v B \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} ms ^{-1}$