An electricity and magnetism problem by Akshat Sharda

Electricity and Magnetism Level 3

Consider a infinitely long wire along z z axis carrying current I I . If it is given that A ( a , 0 ) A(a,0) & B ( a , b ) B(a,b) and path A B AB is a straight line, then find the value of ∫ A B B ⃗ . d l ⃗ \int_{A}^{B} \vec{B} . d\vec{l} .

Details

∫ A B B ⃗ . d l ⃗ \int_{A}^{B} \vec{B} . d\vec{l} denotes ∫ B ⃗ . d l ⃗ \int \vec{B}.d\vec{l} on path A B AB from A A to B B .

None of these. μ o I 2 π tan ⁡ − 1 ( a b ) \frac{\mu_{o}I}{2\pi} \tan^{-1}\left(\frac{a}{b}\right) 2 μ o I π tan ⁡ − 1 ( a b ) \frac{2\mu_{o}I}{\pi} \tan^{-1}\left(\frac{a}{b}\right) μ o I 2 π tan ⁡ − 1 ( b a ) \frac{\mu_{o}I}{2\pi} \tan^{-1}\left(\frac{b}{a}\right) μ o I π tan ⁡ − 1 ( b a ) \frac{\mu_{o}I}{\pi} \tan^{-1}\left(\frac{b}{a}\right)

Akshat Sharda by Akshat Sharda , 21, India

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1 solution

B ˉ = μ 0 I 2 π r e θ \bar{B} = \frac{\mu_0 I}{2\pi r} e_\theta ; d l ˉ = d y e y d\bar{l} = dy e_y

Converting B ˉ \bar{B} to cartesian coordinates with r = x 2 + y 2 r = \sqrt{x^2 + y^2} and e θ = − y e x + x e y x 2 + y 2 e_\theta = \frac{-ye_x+xe_y}{\sqrt{x^2+y^2}}

Given integral becomes I = ∫ A B B ˉ ⋅ d l ˉ = ∫ 0 b μ 0 I 2 π a 2 + y 2 a a 2 + y 2 d y = μ 0 I 2 π t a n − 1 ( b a ) I = \int_A^B \bar{B}\cdot d\bar{l} = \int_0^b\frac{\mu_0 I}{2\pi \sqrt{a^2+y^2}} \frac{a}{\sqrt{a^2+y^2}} dy = \frac{\mu_0 I}{2\pi} tan^{-1}\left(\frac{b}{a}\right)

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