Question

A tightly-wound solenoid of radius a and length l has n turns per unit length. It carries an electric current i. Consider a length dx of the solenoid at a distance x from one end. This contains n dx turns and may be approximated as a circular current i n dx. (a) Write the magnetic field at the centre of the solenoid due to this circular current. Integrate this expression under proper limits to find the magnetic field at the centre of the solenoid. (b) verify that if l >> a, the field tends to B = µ0ni and if a >> l, the field tends to B=μ0nil2a. Interpret these results.

Answer 1

Explanation:

Step 1:

Given data in the question  

In loop current or circular current = indx

Loop radius has circulating current = r

Distance of the solenoid centre from circular current =$\frac{i}{2}-x$

Step 2:

By circular loop, magnetic field in the centre,

$B=\frac{\mu_{0}}{2} \frac{i r^{2}}{\left(x^{2}+r^{2}\right)^{\frac{3}{2}}}$  

Integration of equation of magnetic field , limit from 0 to 1  

$B=\int d(B) |$

Step 3:

Put the value of B in above integral equation  

we get  

$\begin{aligned}&=\int_{0}^{1} \frac{\mu_{0} a^{2} n i d x}{4 \pi\left(a^{2}+(l-2 x)^{2}\right)^{\frac{3}{2}}}\\&=\int_{0}^{1} \frac{\mu_{0} a^{2} n i d x}{4 \pi a^{3}\left(1+\left(\frac{l-2 x}{a}\right)^{2}\right)^{\frac{3}{2}}}\end{aligned}$

$\begin{aligned}=\frac{\mu_{0} n i}{4 \pi a} \int_{0}^{1} \frac{d x}{\left(1+\left(\frac{l-2 x}{a}\right)^{2}\right)^{\frac{3}{2}}} \\=\frac{\mu_{0} n i}{4 \pi a} \frac{4 \pi a}{\sqrt{1+\left(\frac{2 a}{l}\right)^{2}}}\end{aligned}$

$=\frac{\mu_{0} n i}{\sqrt{1+\left(\frac{2 a}{l}\right)^{2}}}$

Step 4:

(b) If a > > l,

$B=\frac{\mu_{0} n i}{2 a}$