Moving Charges and Magnetism - Result Question 19
19. A proton moving with a velocity $3 \times 10^{5} m / s$ enters a magnetic field of 0.3 tesla at an angle of $30^{\circ}$ with the field. The radius of curvature of its path will be ( $e / m$ for proton $=10^{8} C / kg$ )
[2000]
(a) $2 cm$
(b) $0.5 cm$
(c) $0.02 cm$
(d) $1.25 cm$
Show Answer
Answer:
Correct Answer: 19. (b)
Solution:
(b) $r=\frac{m v \sin \theta}{B e}=\frac{3 \times 10^{5} \sin 30^{\circ}}{0.3 \times 10^{8}}$
$\frac{3 \times 10^{5} \times \frac{1}{2}}{3 \times 10^{7}}=0.5 \times 10^{-2} m=0.5 cm$.
If the change particle is moving perpendicular to the magnetic field then $q V B=\frac{m v^{2}}{r}$
$\Rightarrow r=\frac{m v}{q B}$
If the charge particle is moving at an, angle (other than $.0^{\circ}, 90^{\circ}, 180^{\circ})$ to the field, then
$ q(v \sin \theta) B=\frac{m(v \sin \theta)^{2}}{r} \Rightarrow r=\frac{m v \sin \theta}{q B} $
