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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)
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                        - 2 cm
- 0.5 cm
- 0.02 cm
- 1.25 cm
 
Correct Option: B
| r = | = | |||
| Be | 0.3 × 108 | 
| = 0.5 × 10-2 m = 0.5 cm. | ||
| 3 × 107 | 
 
	