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Consider a flat, circular current loop of radius R carrying a current I. Choose the x axis to be along the axis of the loop, with the origin at the loop’s center. Plot a graph of the ratio of the magnitude of the magnetic field at coordinate x to that at the origin for x = 0 to x = 5R. It may be helpful to use a programmable calculator or a computer to solve this problem.

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Answered 2 years ago
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Each segment on the loop applies a magnetic field, so the total magnetic field exerted by NN loops at any point xx is given by

Bx=μo2IR2(x2+R2)3/2\begin{equation} B_{x} = \frac{\mu_{o}}{2} \frac{ I R^{2}}{\left(x^{2}+R^{2}\right)^{3 / 2}} \end{equation}

At the origin x=0x=0, we get the magnetic field by

B0,0=μo2IR2(02+R2)3/2=μoI2RB_{0,0} = \frac{\mu_{o}}{2} \frac{I R^{2}}{\left(0^{2}+R^{2}\right)^{3 / 2}} = \frac{\mu_{o} I}{2R}

Now, we can get the fraction B0,0/BxB_{0,0}/B_{x} by

BxB0,0=(μoIR2)/(x2+R2)3/2(μoI/2R)=R3(x2+R2)3/2\begin{equation} \frac{B_{x} }{B_{0,0}} = \frac{(\mu_o IR^2)/( x^2+R^2)^{3/2} }{(\mu_{o} I/2R)}= \frac{R^3}{( x^2+R^2)^{3/2} } \end{equation}

Now, we find the value of B0,0Bx\frac{B_{0,0} }{B_{x}} at x=0,R,2R,3R,4R,5Rx = 0, R, 2R, 3R,4R , 5R

B0,0B0=1B0,0BR=0.35B0,0B2R=0.09B0,0B3R=0.03B0,0B4R=0.01B0,0B5R=0.007\begin{gather*} \frac{B_{0,0} }{B_{0}} = 1 \\ \frac{B_{0,0} }{B_{R}} = 0.35 \\ \frac{B_{0,0} }{B_{2R}} = 0.09\\ \frac{B_{0,0} }{B_{3R}} = 0.03\\ \frac{B_{0,0} }{B_{4R}} = 0.01\\ \frac{B_{0,0} }{B_{5R}} = 0.007\\ \end{gather*}

Now, plot a graph between these values versus the values of xx to get the next graph

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