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JEE Main 2023 (13 Apr, Shift 2) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from JEE Main 2023 (13 Apr, Shift 2)

Q26JEE Main 2023 · 13 APR · SHIFT 2Easy

Force on Current-Carrying Conductors

A straight wire ABAB of mass 40 g40\text{ g} and length 50 cm50\text{ cm} is suspended by a pair of flexible leads in uniform magnetic field of magnitude 0.40 T0.40\text{ T} as shown in the figure. The magnitude of the current required in the wire to remove the tension in the supporting leads is ___________ A.

(( Take g=10 m s−2g = 10\text{ m s}^{-2} ).

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 2

  1. Tension becomes zero when the upward magnetic force equals the weight: BIL=mgBIL = mg.
  2. I=mgBL=0.04×100.40×0.50=2 AI = \dfrac{mg}{BL} = \dfrac{0.04 \times 10}{0.40 \times 0.50} = 2\text{ A}
  3. So the current is 2 A.
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