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

Home » Electricity and Magnetism » Moving Charges and Magnetism » Question

Moving Charges and Magnetism

  1. 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°
Be0.3 × 108

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



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