Asked in MDCAT Test Series 26 — Electromagnetic InductionModerate

One proton beam enters a magnetic field of 10-4T normally, a specific charge of 1011C/kg, and a velocity of 107m/s. What is the radius of the circle described by it?

Correct answer: B. 1 m

  • A. 0.1 cm
  • B. 1 m
  • C. 10 cm
  • D. None of these

Explanation

The radius of the circle described by a charged particle moving perpendicular to a magnetic field is given by the formula:r = mv/(qB), where:m is the mass of the particle, q is the charge, v is the velocity, and B is the magnetic field strength.For a proton, the specific charge q/m is given as 1011 C/kg. Using the provided velocity v = 107 m/s and magnetic field strength B = 10-4 T:r = v/(q/m × B) = 107 / (1011 × 10-4) = 1 mThus, the correct answer is Option B: 1 m. Options A and C result from calculation errors or incorrect unit conversions. Option D is incorrect because a correct answer is provided in the options.

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About Electromagnetic Induction

Changing magnetic flux induces an emf according to Faraday's law, while Lenz's law gives the direction that opposes the change producing it. Transformers apply induction between coils to step alternating voltage up or down, with turns ratio, current transformation and energy losses determining practical operation.

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