A long solenoid of radius a and number of turns per unit length n is enclosed by cylindrical shell of radius R, thickness d (d<<R) and length L. A variable current i=i0sin wt flows through the coil. If the resistivity of the material of cylindrical shell is p(rho), find the induced current in the shell.

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A long solenoid of radius a and number of turns per unit length n is enclosed by cylindrical shell of radius R, thickness d (d<

Navjyot Kalra , 11 Years ago

Grade 10

1 Answers

ROSHAN MUJEEB

Last Activity: 4 Years ago

The megnetic field in the solenoid is given by
B=(μ0)niB=(μ0)ni
mplies B=(mu_0)ni_(0) sin omega t [:' I=(I_0)=sin omega t (given)]Themag≠ticfluxl∈kedwiththeso≤niodThemag≠ticfluxl∈kedwiththeso≤niod(phi)=vec(B).vec(A) =B A cos 90^(@)=(mu_(0)ni_(0)sin omega t)(pi a^2):. The rate of change of magnetic flux through the solenoiddϕdt=π(μ0)n(a2)i0(ω)tdϕdt=π(μ0)n(a2)i0(ω)t
The same rate of change of flux is linked with the cylindrical shell. By the priciple of electromagnetic inductio, the induced emf produced in the cylindrical shell is

e=−dϕdt=−(π)(μ0)n(a2)(i0)(ω)cos(ω)te=-dϕdt=-(π)(μ0)n(a2)(i0)(ω)cos(ω)t...(i) The resistance offered by the cylinder shell to the flow of induced current I will be
R=(ρ)lAR=(ρ)lA
here,l=2πRandA=L×dl=2πRandA=L×d
∴R=(ρ)2πRLd∴R=(ρ)2πRLd...(ii)
The induced current I will be
I=|e|R=[πμ0n(a2)I0ωcosωt]×Ldρ×2πRI=|e|R=[πμ0n(a2)I0ωcosωt]×Ldρ×2πR
⇒I=μ0na2Ld(i0)ωcosωt2ρR⇒I=μ0na2Ld(i0)ωcosωt2ρR.