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Magnetics question

2023 · 1 Feb · Shift 1 · Q48
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  5. /2023 · 1 Feb · Shift 1 · Q48

Magnetics question

2023 · 1 Feb · Shift 1 · Q48

JEE MainPhysicsMagneticsMCQ+4 / −1
Find the magnetic field at the point P\mathrm{P}P in figure. The curved portion is a semicircle connected to two long straight wires. JEE Main 2023 (Online) 1st February Morning Shift Physics - Magnetic Effect of Current Question 80 English
  1. A
    μ0i2r(12+12π)\frac{\mu_{0} i}{2 r}\left(\frac{1}{2}+\frac{1}{2 \pi}\right)2rμ0​i​(21​+2π1​)
  2. B
    μoi2r(1+2π)\frac{\mu_{\mathrm{o}} i}{2 r}\left(1+\frac{2}{\pi}\right)2rμo​i​(1+π2​)
  3. C
    μ0i˙2r(12+1π)\frac{\mu_{0} \dot{i}}{2 r}\left(\frac{1}{2}+\frac{1}{\pi}\right)2rμ0​i˙​(21​+π1​)
  4. D
    μ0i2r(1+1π)\frac{\mu_{0} i}{2 r}\left(1+\frac{1}{\pi}\right)2rμ0​i​(1+π1​)
View written solutionFree

Correct answer: C

  1. Interpret the figure

    The conductor consists of:

    • a semicircular arc of radius rrr,
    • connected to two long straight wires, one on each side.

    We need the magnetic field at the center point PPP of the semicircle.

  2. Magnetic field due to the semicircular arc

    For a circular arc of angle θ\thetaθ (in radians), magnetic field at the center is

    Barc=μ0iθ4πr.B_{\text{arc}}=\frac{\mu_0 i\theta}{4\pi r}.Barc​=4πrμ0​iθ​.

    For a semicircle, θ=π\theta=\piθ=π. Hence

    Bsemi=μ0iπ4πr=μ0i4r.B_{\text{semi}}=\frac{\mu_0 i\pi}{4\pi r}=\frac{\mu_0 i}{4r}.Bsemi​=4πrμ0​iπ​=4rμ0​i​.

  3. Magnetic field due to each long straight wire

    Each straight portion is effectively a semi-infinite straight wire at perpendicular distance rrr from point PPP.

    Magnetic field due to a semi-infinite straight wire at distance rrr is

    Bsemi-inf=μ0i4πr.B_{\text{semi-inf}}=\frac{\mu_0 i}{4\pi r}.Bsemi-inf​=4πrμ0​i​.

    Since there are two such straight wires,

    Bstraight total=2×μ0i4πr=μ0i2πr.B_{\text{straight total}}=2\times \frac{\mu_0 i}{4\pi r}=\frac{\mu_0 i}{2\pi r}.Bstraight total​=2×4πrμ0​i​=2πrμ0​i​.

  4. Direction of fields

    By the right-hand rule, the magnetic fields due to the semicircle and both straight wires at PPP are in the same direction, so they add.

  5. Net magnetic field

    Therefore,

    B=Bsemi+Bstraight total=μ0i4r+μ0i2πr.B=B_{\text{semi}}+B_{\text{straight total}} =\frac{\mu_0 i}{4r}+\frac{\mu_0 i}{2\pi r}.B=Bsemi​+Bstraight total​=4rμ0​i​+2πrμ0​i​.

    Taking common factor μ0i2r\frac{\mu_0 i}{2r}2rμ0​i​,

    B=μ0i2r(12+1π).B=\frac{\mu_0 i}{2r}\left(\frac{1}{2}+\frac{1}{\pi}\right).B=2rμ0​i​(21​+π1​).

  6. Compare with options

    This matches:

    C: μ0i2r(12+1π)\boxed{\text{C: }\frac{\mu_0 i}{2r}\left(\frac{1}{2}+\frac{1}{\pi}\right)}C: 2rμ0​i​(21​+π1​)​

  7. Comparison with stored answer

    Stored correct answer is A:

    μ0i2r(12+12π)\frac{\mu_0 i}{2r}\left(\frac{1}{2}+\frac{1}{2\pi}\right)2rμ0​i​(21​+2π1​)

    But this would correspond to adding only one semi-infinite straight-wire contribution instead of two. Since the conductor has two long straight wires, both contribute equally and in the same direction.

    Therefore, the stored answer appears incorrect.

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