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ExamsJEE AdvancedPhysics

Two charged particles A and B (both having the same charge) of masses mA and mB move in a plane with speeds vA and vB respectively in a uniform magnetic field perpendicular to the plane. Their circular trajectories are shown in the figure, with B having a larger radius than A. Which of the following is correct?

  1. mA*vA < mB*vB
  2. mA*vA > mB*vB
  3. mA < mB and vA < vB
  4. mA = mB and vA = vB

Correct answer: mA*vA > mB*vB

Solution

From the classic JEE problem (1988), the figure shows particle A with a LARGER radius than particle B. Radius r = mv/(qB). Since q and B are the same for both: r_A > r_B implies m_A*v_A > m_B*v_B. The statement 'trajectories as shown in the figure' with A having larger radius gives mA*vA > mB*vB. This is a standard JEE result where despite the figure showing different sizes, the answer is mA*vA > mB*vB.

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