SATHEE: Moving Charges and Magnetism

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 $= 10^{8} C / k g$ )

[2000]

(a) $2 c m$

(b) $0.5 c m$

(d) $1.25 c m$

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Answer:

Correct Answer: 19. (b)

Solution:

(b) $r = \frac{m v \sin θ}{B e} = \frac{3 \times 10^{5} \sin 30^{\circ}}{0.3 \times 10^{8}}$

$\frac{3 \times 10^{5} \times \frac{1}{2}}{3 \times 10^{7}} = 0.5 \times 10^{- 2} m = 0.5 c m$ .

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

$q (v \sin θ) B = \frac{m (v \sin θ)^{2}}{r} \Rightarrow r = \frac{m v \sin θ}{q B}$

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Moving Charges and Magnetism - Result Question 19

19. A proton moving with a velocity 3×105m/s enters a magnetic field of 0.3 tesla at an angle of 30∘ with the field. The radius of curvature of its path will be ( e/m for proton =108C/kg )

Answer:

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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 $= 10^{8} C / k g$ )