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  • Consider a circular current-carrying loop of radius R in the x-y plane with centre at origin.

Consider a circular current-carrying loop of radius R in the x-y plane with centre at origin. Consider the line integral

\tau (L)=\left | \int_{-L}^{L} B.dl\right | taken along z-axis

 

(a) Show that \tau (L) monotonically increases with L.

b) Use an appropriate Amperian loop to show that \tau (\infty )=\mu_{0}I , where I is the current in the wire.

(c) Verify directly the above result.

(d) Suppose we replace the circular coil by a square coil of sides R carrying the same current I. What can you say about \tau (L )and \tau (\infty )

Answers (1)

(a) In a circular current-carrying loop in x-y plane, the magnetic field acts along z-axis as shown in below.

\tau (L)=\left | \int_{-L}^{L} B.dl\right |=\int_{-L}^{+L}Bdl \cos 0^{\circ}=\int_{-L}^{+L}Bdl= 2BL

\therefore \tau (L) is monotonically increasing function of L

(b) Now consider an Amperean loop around the circular coil of such a large radius that L\rightarrow \infty. Since this loop encloses a current I, now using the Amperan's law

\tau (\infty )=\oint_{-\infty }^{+\infty }\overrightarrow{B}.\overrightarrow{dl}=\mu _{0}I

(c) The magnetic field at the axis (z-axis) of the circular coil is given by 

B){z}=\frac{\mu _{0}IR^{2}}{2(z^{2}+R^{2})^{\frac{3}{2}}}

Now intergarting

\int_{-\infty }^{+\infty }B_{z}=\int_{-\infty }^{+\infty }\frac{\mu _{0}IR^{2}}{2(z^{2}+R^{2})^{\frac{3}{2}}}dz

Let z = R tan \theta \ so \ that \ dz = r\sec ^{2}\theta d\theta

and

 (z^{2}+R^{2})^{\frac{3}{2}}=(R^{2}\tan ^{2}\theta +R^{2})^{\frac{3}{2}}\\ = R^{3}\sec^{3}\theta(as 1 +\tan ^{2}\theta \sec^{2}\theta)

Thus,

\int_{-\infty }^{+\infty }B_{z}=\frac{\mu _{0}I}{2}\int_{-\frac{\pi}{2} }^{+\frac{\pi} {2}}\frac{R^{2}(R\sec ^{2}\theta d \theta)}{R^{3}\sec ^{3}\theta }

=\frac{\mu _{0}I}{2}\int_{-\frac{\pi}{2} }^{+\frac{\pi} {2}}\cos \theta d \theta =\mu _{0}I

(d) As we know (B_{z})_{square}<(B_{z})_{circular coil}

For the same current and side of the square equal to radius of the coil 

\tau(\infty )_{square}<\tau(\infty )_{circular coil}

By using the same argument as we done in the case (b), it can be shown that 

\tau(\infty )_{square}=\tau(\infty )_{circular coil}

 

 

Posted by

infoexpert24

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