In Figure, a wire forms a closed circular loop, of radius R = 2.0mand resistance$\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\Omega }$. The circle is centered on a long straight wire; at time t = 0, the current in the long straight wire is 5.0 Arightward. Thereafter, the current changes according to$\mathit{i}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{0}\mathit{A}\mathbf{-}\left(2.0\frac{A}{s^2}\right){\mathit{t}}^{\mathbf{2}}$. (The straight wire is insulated; so there is no electrical contact between it and the wire of the loop.) What is the magnitude of the current induced in the loop at times t > 0?

1. The radius of the circular loop R = 2.0 m
2. The resistance$4.0\Omega$
3. At time, t = 0, the current in the long straight wire is 5.0 A rightward.
4. The current changes according to$i=5.0A-\left(2.0A/s^2\right)t^2$.
5. The straight wire is insulated.

Applying the right hand rule, find the net flux produced at any time. Substituting this value in eq.30-4, find the emf induced due to change in the magnetic flux. Finally, applying Ohm’s law, find the magnitude of the current induced in the loop.

Right Hand Rule states that if we arrange our thumb, forefinger and middle finger of the right-hand perpendicular to each other, then the thumb points towards the direction of the motion of the conductor relative to the magnetic field, the forefinger points towards the direction of the magnetic field and the middle finger points towards the direction of the induced current.

Faraday's law of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant.

Formulae are as follow:

$$\begin{gathered} \epsilon =-\frac{d{\Phi }_B}{dt} \\ i=\frac{\left|\epsilon \right|}{R} \end{gathered}$$

Where,${\Phi }_B$is magnetic flux, i is current, R is resistance, 𝜀 is emf.

According to Faraday’s law, the emf is induced due to change in the magnetic flux is,

$\epsilon =-\frac{d{\Phi }_B}{dt}.........................(30-4)$

Since,the current in the long straight wire is rightward. Therefore, according to the right-hand rule the field due to the current in the straight wire is out of the page in the upper half of the circle and is into the page in the lower half of the circle.

Hence, the net flux produced at any time is must be zero.

That is,

${\Phi }_B=0$

Substituting in Eq.30-4,

$$\begin{gathered} \epsilon =-\frac{d\left(0\right)}{dt} \\ \epsilon =0 \end{gathered}$$

According to Ohm’s law,

$i=\frac{\left|\epsilon \right|}{R}$

Therefore, the induced current in the circle is,

$$\begin{gathered} i=\frac{0}{R} \\ i=0 \end{gathered}$$

Hence, the induced current in the circle is, i = 0.

Therefore, by using the concept of Right hand rule, Faraday’s law and Ohm’s law, the current induced in the circle can be determined.