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Magnetics question

2023 · 13 Apr · Shift 2 · Q65
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  5. /2023 · 13 Apr · Shift 2 · Q65

Magnetics question

2023 · 13 Apr · Shift 2 · Q65

JEE MainPhysicsMagneticsNumerical+4 / −1
A straight wire AB\mathrm{AB}AB of mass 40 g40 \mathrm{~g}40 g and length 50 cm50 \mathrm{~cm}50 cm is suspended by a pair of flexible leads in uniform magnetic field of magnitude 0.40 T0.40 \mathrm{~T}0.40 T as shown in the figure. The magnitude of the current required in the wire to remove the tension in the supporting leads is ‾\underline{\hspace{2cm}}​ A. (\left(\right.( Take g=10 ms−2g=10 \mathrm{~ms}^{-2}g=10 ms−2 ). JEE Main 2023 (Online) 13th April Evening Shift Physics - Magnetic Effect of Current Question 62 English
Numerical answer
View written solutionFree

Correct answer: 2

  1. Given data

    • Mass of wire: m=40 g=0.04 kgm = 40\text{ g} = 0.04\text{ kg}m=40 g=0.04 kg
    • Length of wire: L=50 cm=0.50 mL = 50\text{ cm} = 0.50\text{ m}L=50 cm=0.50 m
    • Magnetic field: B=0.40 TB = 0.40\text{ T}B=0.40 T
    • Acceleration due to gravity: g=10 m s−2g = 10\text{ m s}^{-2}g=10 m s−2

  2. Condition for zero tension

    To remove the tension in the supporting leads, the magnetic force on the wire must balance its weight.

    So, FB=mgF_B = mgFB​=mg

  3. Magnetic force on a current-carrying straight wire

    The force on a wire carrying current in a magnetic field is FB=BILsin⁡θF_B = BIL\sin\thetaFB​=BILsinθ

    From the figure/setup, the wire is perpendicular to the magnetic field, so sin⁡θ=1\sin\theta = 1sinθ=1

    Hence, BIL=mgBIL = mgBIL=mg

  4. Substitute the values

    0.40×I×0.50=0.04×100.40 \times I \times 0.50 = 0.04 \times 100.40×I×0.50=0.04×10

    0.20I=0.400.20I = 0.400.20I=0.40

    I=0.400.20=2I = \frac{0.40}{0.20} = 2I=0.200.40​=2

  5. Final answer

    The required current is 2 A\boxed{2\text{ A}}2 A​

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