Question #313273

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


1
Expert's answer
2022-03-21T12:47:25-0400

We know that

(A)



Both end


Magnetic field End of solenoid

dB=μ0NIR22(R2+x2)32dB=\frac{\mu_0 NIR^2}{2(R^2+x^2)^\frac{3}{2}}

dBdx=μ0ni2sinθdθ\frac{dB}{dx}=-\frac{\mu_0 ni}{2}\int sin\theta d\theta

B=μ0ni2π2πsinθdθB=-\frac{\mu_0 ni}{2}\int _{\frac{\pi}{2}}^\pi sin\theta d\theta

B=μ0ni2[cosθ]π2πB=\frac{\mu_0 ni}{2}[cos\theta ]_\frac{\pi}{2}^\pi

B=μ0ni2B=\frac{\mu_0 ni}{2}

(B)

Magnetic field Centre of solenoid

Bc=μ0ni20πsinθdθB_c=-\frac{\mu_0 ni}{2}\int _{0}^\pi sin\theta d\theta

Bc=μ0ni2[cosθ]0πB_c=\frac{\mu_0 ni}{2}[cos\theta ]_0^\pi

Bc=μ0niB_c=\mu_0ni


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