In a current carrying solenoid, determine the 𝑑𝐵/𝑑𝑥 value at the (i) both ends and (ii) centre of the solenoid.
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}}dB=2(R2+x2)23μ0NIR2
dBdx=−μ0ni2∫sinθdθ\frac{dB}{dx}=-\frac{\mu_0 ni}{2}\int sin\theta d\thetadxdB=−2μ0ni∫sinθdθ
B=−μ0ni2∫π2πsinθdθB=-\frac{\mu_0 ni}{2}\int _{\frac{\pi}{2}}^\pi sin\theta d\thetaB=−2μ0ni∫2ππsinθdθ
B=μ0ni2[cosθ]π2πB=\frac{\mu_0 ni}{2}[cos\theta ]_\frac{\pi}{2}^\piB=2μ0ni[cosθ]2ππ
B=μ0ni2B=\frac{\mu_0 ni}{2}B=2μ0ni
(B)
Magnetic field Centre of solenoid
Bc=−μ0ni2∫0πsinθdθB_c=-\frac{\mu_0 ni}{2}\int _{0}^\pi sin\theta d\thetaBc=−2μ0ni∫0πsinθdθ
Bc=μ0ni2[cosθ]0πB_c=\frac{\mu_0 ni}{2}[cos\theta ]_0^\piBc=2μ0ni[cosθ]0π
Bc=μ0niB_c=\mu_0niBc=μ0ni
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