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A small circular loop of conducting wire has radius a and carries current I. It is placed in a uniform magnetic field B perpendicular to its plane such that when rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period T. If the mass of the loop is m then

Asked in JEE Main 9th Jan 2nd Shift 2020 · Torque and potential energy

• (1) T = √(2m)/(IB)
• (2) T = √(π m)/(IB)
• (3) T = √(π m)/(2IB)
• (4) T = √(2π m)/(IB)

Step-by-step solution

Why the other options are wrong

• (1) Dropped the π factors from both the period formula and the moment.
• (2) Used moment of inertia ma² (about the axis) instead of ma²/2 about a diameter.
• (3) Used moment of inertia ma²/4 (disc about diameter) instead of ring value ma²/2.

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