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Magnetics question

2004 · Shift 0 · Q128
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Magnetics question

2004 · Shift 0 · Q128

JEE MainPhysicsMagneticsMCQ+4 / −1
A current iii ampere flows along an infinitely long straight thin walled tube, then the magnetic induction at any point inside the tube is
  1. A
    μ04π,2ir{{{\mu _0}} \over {4\pi }},{{2i} \over r}4πμ0​​,r2i​ tesla
  2. B
    zero
  3. C
    infinite
  4. D
    2ir{{2i} \over r}r2i​ tesla
View written solutionFree

Correct answer: B

  1. Interpret the geometry

    The conductor is an infinitely long straight thin-walled tube carrying current iii along its length.

    This means the current is distributed symmetrically over the cylindrical surface.

  2. Use Ampere’s circuital law

    For a circular Amperian loop of radius rrr taken inside the tube, symmetry implies that if a magnetic field exists, it would be tangential and have constant magnitude on the loop.

    Ampere’s law gives ∮B⃗⋅dl⃗=μ0Ienc.\oint \vec B\cdot d\vec l = \mu_0 I_{\text{enc}}.∮B⋅dl=μ0​Ienc​.

  3. Find enclosed current for a loop inside the tube

    Since the current flows only on the thin wall of the tube, an Amperian circle drawn entirely inside the hollow region encloses Ienc=0.I_{\text{enc}}=0.Ienc​=0.

    Therefore, ∮B⃗⋅dl⃗=0.\oint \vec B\cdot d\vec l = 0.∮B⋅dl=0.

  4. Apply symmetry

    On the circular loop, B(2πr)=0.B(2\pi r)=0.B(2πr)=0. Hence, B=0.B=0.B=0.

  5. Conclusion

    The magnetic induction at any point inside the tube is 0.\boxed{0}.0​.

  6. Check options

    • A: μ04π⋅2ir\dfrac{\mu_0}{4\pi}\cdot \dfrac{2i}{r}4πμ0​​⋅r2i​ tesla — incorrect
    • B: zero — correct
    • C: infinite — incorrect
    • D: 2ir\dfrac{2i}{r}r2i​ tesla — incorrect

Therefore, the correct option is B.

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