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  • In the middle of a long solenoid, there is a coaxial ring of square cross-section, made of conductivity material of resistivity . The thickness of the r

In the middle of a long solenoid, there is a coaxial ring of square cross-section, made of conductivity material of resistivity \rho. The thickness of the ring is equal to h, and its inside and outside radii are equal to 'a' and 'b' respectively. What is the total current induced where the magnetic field varies with time as B=\beta t ?

Option: 1

\mathrm{\frac{h \beta\left(b^2-a^2\right)}{4 \rho}}


Option: 2

\mathrm{\frac{h \beta\left(b^2-a^2\right)}{8 \rho}}


Option: 3

\mathrm{\frac{h \beta\left(b^2-a^2\right)}{16 \rho}}


Option: 4

\mathrm{\frac{h \beta\left(b^2-a^2\right)}{2 \rho}}


Answers (1)

best_answer

According to the problem, the e.m.f. induced in an elementary ring of radius r and width dr is \mathrm{\pi r^2 \beta}. The conductance of this ring is \mathrm{d\left(\frac{1}{R}\right)=\left(\frac{h d r}{\rho 2 \pi r}\right)}

thus, the current induced is \mathrm{\mathrm{d}I=h r d r \beta / 2 \rho}

 Upon integration, we get the total current

\mathrm{I=\int_a^b \frac{h r d r \beta}{2 \rho}=\frac{h \beta\left(b^2-a^2\right)}{4 \rho}}

Posted by

Ajit Kumar Dubey

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