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} $