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Magnetism question

2024 · Shift 1 · Q36
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  5. /2024 · Shift 1 · Q36

Magnetism question

2024 · Shift 1 · Q36

JEE AdvancedPhysicsMagnetismMCQ+3 / −1
An infinitely long wire, located on the zzz-axis, carries a current III along the +z+z+z-direction and produces the magnetic field B⃗\vec{B}B. The magnitude of the line integral ∫B⃗⋅dl→\int \vec{B} \cdot \overrightarrow{d l}∫B⋅dl along a straight line from the point (−3a,a,0)(-\sqrt{3} a, a, 0)(−3​a,a,0) to (a,a,0)(a, a, 0)(a,a,0) is given by [μ0\mu_0μ0​ is the magnetic permeability of free space.]
  1. A
    7μ0I24{{7{\mu _0}I} \over {24}}247μ0​I​
  2. B
    7μ0I12{{7{\mu _0}I} \over {12}}127μ0​I​
  3. C
    μ0I8{{{\mu _0}I} \over {8}}8μ0​I​
  4. D
    μ0I6{{{\mu _0}I} \over {6}}6μ0​I​
View written solutionFree

Correct answer: A

  1. Magnetic field due to an infinite straight wire
    For a wire along the zzz-axis carrying current III in the +z+z+z direction, the magnetic field at a point (x,y,0)(x,y,0)(x,y,0) is
B⃗=μ0I2πr ϕ^\vec B=\frac{\mu_0 I}{2\pi r}\,\hat\phiB=2πrμ0​I​ϕ^​

where r=x2+y2r=\sqrt{x^2+y^2}r=x2+y2​. In Cartesian form,

ϕ^=(−yr)i^+(xr)j^\hat\phi=\left(-\frac{y}{r}\right)\hat i+\left(\frac{x}{r}\right)\hat jϕ^​=(−ry​)i^+(rx​)j^​

so

B⃗=μ0I2π(x2+y2) (−yi^+xj^).\vec B=\frac{\mu_0 I}{2\pi (x^2+y^2)}\,(-y\hat i+x\hat j).B=2π(x2+y2)μ0​I​(−yi^+xj^​).
  1. Path of integration
    The straight line is from (−3a,a,0)(-\sqrt{3}a,a,0)(−3​a,a,0) to (a,a,0)(a,a,0)(a,a,0).
    Since y=ay=ay=a is constant, this is a horizontal line segment with
y=a,x:−3a→a.y=a,\qquad x:-\sqrt{3}a\to a.y=a,x:−3​a→a.

Hence,

overrightarrowdl=dx i^.overrightarrow{dl}=dx\,\hat i.overrightarrowdl=dxi^.
  1. Compute B⃗⋅dl⃗\vec B\cdot d\vec lB⋅dl
    Substitute y=ay=ay=a into B⃗\vec BB:
B⃗=μ0I2π(x2+a2)(−ai^+xj^).\vec B=\frac{\mu_0 I}{2\pi (x^2+a^2)}(-a\hat i+x\hat j).B=2π(x2+a2)μ0​I​(−ai^+xj^​).

Therefore,

B⃗⋅dl⃗=μ0I2π(x2+a2)(−ai^+xj^)⋅(dx i^)\vec B\cdot d\vec l=\frac{\mu_0 I}{2\pi (x^2+a^2)}(-a\hat i+x\hat j)\cdot (dx\,\hat i)B⋅dl=2π(x2+a2)μ0​I​(−ai^+xj^​)⋅(dxi^) B⃗⋅dl⃗=−μ0I2πa dxx2+a2.\vec B\cdot d\vec l=-\frac{\mu_0 I}{2\pi}\frac{a\,dx}{x^2+a^2}.B⋅dl=−2πμ0​I​x2+a2adx​.
  1. Integrate along the path
∫B⃗⋅dl⃗=−μ0I2π∫−3aaa dxx2+a2.\int \vec B\cdot d\vec l=-\frac{\mu_0 I}{2\pi}\int_{-\sqrt{3}a}^{a}\frac{a\,dx}{x^2+a^2}.∫B⋅dl=−2πμ0​I​∫−3​aa​x2+a2adx​.

Use

∫dxx2+a2=1atan⁡−1(xa).\int \frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right).∫x2+a2dx​=a1​tan−1(ax​).

So,

∫−3aaa dxx2+a2=[tan⁡−1(xa)]−3aa.\int_{-\sqrt{3}a}^{a}\frac{a\,dx}{x^2+a^2} =\left[\tan^{-1}\left(\frac{x}{a}\right)\right]_{-\sqrt{3}a}^{a}.∫−3​aa​x2+a2adx​=[tan−1(ax​)]−3​aa​.

Now,

tan⁡−1(1)=π4,tan⁡−1(−3)=−π3.\tan^{-1}(1)=\frac{\pi}{4},\qquad \tan^{-1}(-\sqrt{3})=-\frac{\pi}{3}.tan−1(1)=4π​,tan−1(−3​)=−3π​.

Thus,

[tan⁡−1(xa)]−3aa=π4−(−π3)=7π12.\left[\tan^{-1}\left(\frac{x}{a}\right)\right]_{-\sqrt{3}a}^{a} =\frac{\pi}{4}-\left(-\frac{\pi}{3}\right)=\frac{7\pi}{12}.[tan−1(ax​)]−3​aa​=4π​−(−3π​)=127π​.

Hence,

∫B⃗⋅dl⃗=−μ0I2π⋅7π12=−7μ0I24.\int \vec B\cdot d\vec l=-\frac{\mu_0 I}{2\pi}\cdot \frac{7\pi}{12} =-\frac{7\mu_0 I}{24}.∫B⋅dl=−2πμ0​I​⋅127π​=−247μ0​I​.
  1. Magnitude of the line integral
    The question asks for the magnitude, so
∣∫B⃗⋅dl⃗∣=7μ0I24.\left|\int \vec B\cdot d\vec l\right|=\frac{7\mu_0 I}{24}.​∫B⋅dl​=247μ0​I​.
  1. Option matching
    This matches Option A:
7μ0I24\boxed{\frac{7\mu_0 I}{24}}247μ0​I​​
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