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A segment of wire of length L is along the x axis centered at x=0. Which of the following is a correct integral expression for the magnetic field at point P (centered on the wire segment at y=b) due to the current I flowing left to right in the segment of length L? In all answers below the limits of integration are from -L/2 to L/2.

a. μ0I/4π∫ dx b/(b2 + x2)3/2 kb. μ0I/4π∫ dx b/(b2 + x2)3/2 j c. μ0I/4π∫ dx /(b2 + x2) kd. -μ0I/4π∫ dx /(b2 + x2)1/2 ke. none of 1-5

User Jack Wild
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1 Answer

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Answer:

b. μ0I/4π∫ dx b/(b2 + x2)³/² j

Step-by-step explanation:

The wire of length L centered at origin ( x =0 and y=0 ) carries current of I . We have to find magnetic field at point ( x = 0 , y = b ) .

First of all we shall consider magnetic field due to current element idx which is at x distance away from origin . magnetic field

dB =
(\mu _0idx)/(4\pi(x^2+b^2)^2 )

component of magnetic field along y- axis at point P


(\mu _0idx)/(4\pi(x^2+b^2)^2 )cos\theta

where θ is angle between y - axis and dE .

component of magnetic field along y- axis at point P


(\mu _0idx)/(4\pi(x^2+b^2)^2 )* (b)/(√(x^2+b^2) )


(\mu _0ibdx)/(4\pi(x^2+b^2)^(3)/(2) )

The same magnetic field will also exist due to current element dx at x distance away on negative x - axis

The perpendicular component will cancel out .

This is magnetic field dE due to small current element

Magnetic field due to whole wire


\int\limits^(L)/(2) _(-L )/( 2 ) }(\mu _0ibdx)/(4\pi(x^2+b^2)^(3)/(2) ) \, dx

User Dreamzboy
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