Question

A circular loop of radius $r$ is made of a wire of circular cross section of diameter $a$. When current $I$ flows through the loop the magnetic flux linked with the loop due to self induced magnetic field is given by, $\varphi ={\mu }_{o}r[ln\left(\frac{16r}{a}\right)-\frac{7}{4}]I.$ The resistivity of the material of the wire is $\rho$ and $r>>a$. Switch $S$ is closed at time $t=0$ so as to connect the loop to a cell of emf $V$.
Choose the correct statements

• A. Self inductance of the loop is, $L={\mu }_{o}r[ln\left(\frac{16r}{a}\right)-\frac{7}{4}]$
• B. Resistance of the loop is, $R=\frac{8\rho r}{a^2}$
• C. Current in the loop is, $i=\frac{V{a}^2}{8\rho r}[1-e^{\frac{-t}{\tau }}]$ where $\tau =\frac{L}{R}=\frac{{\mu }_0{a}^2}{8\rho }[ln\left(\frac{16r}{a}\right)-\frac{7}{4}]$
• D. Current in the loop is, $i=\frac{V{a}^2}{8\rho r}[1-e^{\frac{-2t}{\tau }}]$
  where $\tau =\frac{L}{R}=\frac{{\mu }_o{a}^2}{\rho }[ln\left(\frac{16r}{a}\right)-\frac{7}{4}]$
  

Answer 1

Answer: (A), (B), (C)

The correct options are
A Self inductance of the loop is, $L={\mu }_{o}r[ln\left(\frac{16r}{a}\right)-\frac{7}{4}]$
B Resistance of the loop is, $R=\frac{8\rho r}{a^2}$
C Current in the loop is, $i=\frac{V{a}^2}{8\rho r}[1-e^{\frac{-t}{\tau }}]$ where $\tau =\frac{L}{R}=\frac{{\mu }_0{a}^2}{8\rho }[ln\left(\frac{16r}{a}\right)-\frac{7}{4}]$

Self inductance of the loop is,
$L=\frac{\varphi }{I}={\mu }_{0}r[ln\left(\frac{16r}{a}\right)-\frac{7}{4}]$

Resistance of the loop is, $R=\frac{\rho l}{A}=\frac{\rho \left(2\pi r\right)}{\pi {\left(\frac{a}{2}\right)}^2}=\frac{8\rho r}{a^2}$

The circuit is a simple L-R circuit, hence, Current, $i=\frac{V}{R}[1-e^{\frac{-t}{\tau }}]=\frac{V{a}^2}{8\rho r}[1-e^{\frac{-t}{\tau }}]$
where, $\tau =\frac{L}{R}=\frac{{\mu }_0{a}^2}{8\rho }[ln\left(\frac{16r}{a}\right)-\frac{7}{4}]$