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Magnetics question

2023 · 30 Jan · Shift 1 · Q40
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  5. /2023 · 30 Jan · Shift 1 · Q40

Magnetics question

2023 · 30 Jan · Shift 1 · Q40

JEE MainPhysicsMagneticsMCQ+4 / −1
A massless square loop, of wire of resistance 10Ω10 \Omega10Ω, supporting a mass of 1 g1 \mathrm{~g}1 g, hangs vertically with one of its sides in a uniform magnetic field of 103G10^{3} \mathrm{G}103G, directed outwards in the shaded region. A dc voltage V\mathrm{V}V is applied to the loop. For what value of V\mathrm{V}V, the magnetic force will exactly balance the weight of the supporting mass of 1 g1 \mathrm{~g}1 g? (If sides of the loop =10 cm, g=10 ms−2=10 \mathrm{~cm}, \mathrm{~g}=10 \mathrm{~ms}^{-2}=10 cm, g=10 ms−2) JEE Main 2023 (Online) 30th January Morning Shift Physics - Magnetic Effect of Current Question 76 English
  1. A
    1 V
  2. B
    110\frac{1}{10}101​ V
  3. C
    10 V
  4. D
    100 V
View written solutionFree

Correct answer: C

  1. Given data

    • Resistance of loop: R=10 ΩR = 10\,\OmegaR=10Ω
    • Supporting mass: m=1 g=10−3 kgm = 1\,\text{g} = 10^{-3}\,\text{kg}m=1g=10−3kg
    • Side of square loop: l=10 cm=0.1 ml = 10\,\text{cm} = 0.1\,\text{m}l=10cm=0.1m
    • Magnetic field: B=103 GB = 10^3\,\text{G}B=103G
    • 1 G=10−4 T1\,\text{G} = 10^{-4}\,\text{T}1G=10−4T, so B=103×10−4=10−1 T=0.1 TB = 10^3 \times 10^{-4} = 10^{-1}\,\text{T} = 0.1\,\text{T}B=103×10−4=10−1T=0.1T
    • Acceleration due to gravity: g=10 m s−2g = 10\,\text{m s}^{-2}g=10m s−2

  2. Weight of the supporting mass

    W=mg=10−3×10=10−2 NW = mg = 10^{-3} \times 10 = 10^{-2}\,\text{N}W=mg=10−3×10=10−2N

  3. Magnetic force on the side inside the field

    Only the horizontal side lying in the magnetic field experiences a vertical magnetic force that can balance the weight.

    Magnetic force on a straight current-carrying wire: F=BIlF = BIlF=BIl

    For equilibrium, BIl=mgBIl = mgBIl=mg

    Substituting values: 0.1×I×0.1=10−20.1 \times I \times 0.1 = 10^{-2}0.1×I×0.1=10−2 0.01I=0.010.01 I = 0.010.01I=0.01 I=1 AI = 1\,\text{A}I=1A

  4. Find the required applied voltage

    Using Ohm’s law: V=IR=1×10=10 VV = IR = 1 \times 10 = 10\,\text{V}V=IR=1×10=10V

  5. Check with options

    • A: 1 V1\,\text{V}1V
    • B: 110 V\frac{1}{10}\,\text{V}101​V
    • C: 10 V10\,\text{V}10V
    • D: 100 V100\,\text{V}100V

    Hence, the correct option is: C: 10 V\boxed{\text{C: }10\,\text{V}}C: 10V​

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