Answer to Question #313273 in Electricity and Magnetism for Rajat Vats

Question

In a current carrying solenoid, determine the 𝑑𝐵/𝑑𝑥 value at the (i) both ends and (ii) centre of the solenoid.

Answer 1

We know that

(A)

Both end

Magnetic field End of solenoid

$dB=\frac{\mu_0 NIR^2}{2(R^2+x^2)^\frac{3}{2}}$

$\frac{dB}{dx}=-\frac{\mu_0 ni}{2}\int sin\theta d\theta$

$B=-\frac{\mu_0 ni}{2}\int _{\frac{\pi}{2}}^\pi sin\theta d\theta$

$B=\frac{\mu_0 ni}{2}[cos\theta ]_\frac{\pi}{2}^\pi$

$B=\frac{\mu_0 ni}{2}$

(B)

Magnetic field Centre of solenoid

$B_c=-\frac{\mu_0 ni}{2}\int _{0}^\pi sin\theta d\theta$

$B_c=\frac{\mu_0 ni}{2}[cos\theta ]_0^\pi$

$B_c=\mu_0ni$