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BITSAT2021PhysicsMagnetic Effects of CurrentActual

A proton moving with a velocity 3 10⁵ ~m / 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 =10⁸ C / kg )

Options

  1. A2 ~cm
  2. B0.5 ~cm
  3. C0.02 ~cm
  4. D1.25 ~cm

Correct answer

B. 0.5 ~cm

Step-by-step solution

r= m v B e = 3 10⁵ 30^ 0.3 10⁸ 3 10⁵ 1 2 3 10⁷ =0.5 10⁻²m=0.5 ~cm

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