Question

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

Answer 1

Answer:

b. μ0I/4π∫ dx b/(b2 + x2)³/² j

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 = $\frac{\mu _0idx}{4\pi(x^2+b^2)^2 }$

component of magnetic field along y- axis at point P

$\frac{\mu _0idx}{4\pi(x^2+b^2)^2 }cos\theta$

where θ is angle between y - axis and dE .

component of magnetic field along y- axis at point P

$\frac{\mu _0idx}{4\pi(x^2+b^2)^2 }\times \frac{b}{\sqrt{x^2+b^2} }$

$\frac{\mu _0ibdx}{4\pi(x^2+b^2)^\frac{3}{2} }$

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

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