A proton moving with a velocity 3 × 105 m/s enters a magnetic

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

1. 2 cm
2. 0.5 cm
3. 0.02 cm
4. 1.25 cm

Correct Option: B

r =	mv sinθ	=	3 × 10⁵ sin 30°
	Be		0.3 × 10⁸

	3 × 10⁵ × (1/2)	= 0.5 × 10^-2 m = 0.5 cm.
	3 × 10⁷