91Ó°ÊÓ

The radius of the solenoid is a .

The magnetic field inside the solenoid,$B\left(t\right)=B_0\cos \left(\mathrm{Ó\neg }t\right)\hat{z}$

The radius of the circular wire is and resistance is R .

The area of the circular loop is given by,

$$\begin{gathered} \stackrel{\mathbf{â‡\P Ä}}{\mathbf{A}}\mathbf{=}\mathit{Ï€}{\left(\frac{a}{2}\right)}^{\mathbf{2}}\hat{\mathbf{z}} \\ \stackrel{\mathbf{â‡\P Ä}}{\mathbf{A}}\mathbf{=}\frac{\mathbf{Ï€}{\mathbf{a}}^{\mathbf{2}}}{\mathbf{4}}\hat{\mathbf{z}} \end{gathered}$$

The magnetic flux through the loop is given by,

$\mathrm{Ï•}=\stackrel{â‡\P Ä}{B}.\stackrel{â‡\P Ä}{A}$

Substitute $\frac{\mathrm{Ï€}{a}^2}{4}\hat{z}$for$\stackrel{â‡\P Ä}{A}$and$B_0\cos \left(\mathrm{Ó\neg }t\right)\hat{z}$for B into above equation.

$$\begin{gathered} \mathrm{Ï•}=B_0\cos \left(\mathrm{Ó\neg }t\right)\hat{z}\frac{\mathrm{Ï€}{a}^2}{4}\hat{z} \\ \mathrm{Ï•}=B_0\frac{\mathrm{Ï€}{a}^2}{4}\cos \left(\mathrm{Ó\neg }t\right) \end{gathered}$$

The induced emf in any closed loop is equal to the negative of the rate of change of flux in the circuit.

$\mathrm{Î\mu }\left(t\right)=-\frac{d\mathrm{Ï•}}{dt}$

Substitute $B_0\frac{\mathrm{Ï€}{a}^2}{4}\cos \left(\mathrm{Ó\neg }t\right)$for $\mathrm{Ï•}$into above equation.

role="math" localid="1657701318753" $$\begin{gathered} \mathrm{Î\mu }\left(t\right)=-\frac{d}{dt}\left(B_0\frac{\mathrm{Ï€}{a}^2}{4}\cos \left(\mathrm{Ó\neg }t\right)\right) \\ \mathrm{Î\mu }\left(t\right)=-B_0\frac{\mathrm{Ï€}{a}^2}{4}9\left(-\sin \mathrm{Ó\neg }t\right)\mathrm{Ó\neg } \\ \mathrm{Î\mu }\left(t\right)=\frac{B_0\mathrm{Ó\neg }\mathrm{Ï€}{a}^2}{4}\sin \mathrm{Ó\neg }t \end{gathered}$$

According the ohm’s law, the expression for the current is given by,

role="math" localid="1657701356450" $I=\frac{\mathrm{Î\mu }\left(t\right)}{R}$

Substitute role="math" localid="1657701424541" $\frac{B_0\mathrm{Ó\neg }\mathrm{Ï€}{a}^2}{4}\sin \mathrm{Ó\neg }t$for $\mathrm{Î\mu }\left(t\right)$ into above equation.

role="math" localid="1657701523568" $$\begin{gathered} I=\frac{\frac{B_0\mathrm{Ó\neg }\mathrm{Ï€}{a}^2}{4}\sin \mathrm{Ó\neg }t}{R} \\ I=\frac{\frac{B_0\mathrm{Ó\neg }\mathrm{Ï€}{a}^2}{4}\sin \mathrm{Ó\neg }t}{4R} \end{gathered}$$

Hence the current induced in the loop as function of the time is$\frac{B_0\mathrm{Ó\neg }\mathrm{Ï€}{a}^2}{4}\sin \mathrm{Ó\neg }t$ .