91影视

Length a =12 cm

Length b =16 cm

Current$i=4.5t^2-10t$

Time t=3 s

First find the net flux through the lower part of loop and substituting it in Faraday鈥檚 law, calculate the emf induced in the loop. Secondly, using Lenz鈥檚 law, find the direction of the induced current.

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

Lenz's law states that the current induced in a circuit due to a change in a magnetic field is directed to oppose the change in flux and to exert a mechanical force that opposes the motion.

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, and the forefinger points towards the direction of the magnetic field and the middle finger points towards the direction of the induced current.

Formulae are as follows:

$$\begin{gathered} 蠒=鈭�\stackrel{鈫�}{B}.d\stackrel{鈫�}{A} \\ 蔚=-\frac{d蠒}{dt} \end{gathered}$$

Where,$桅$is magnetic flux, B is magnetic field, A is area,饾渶 is emf.

Emf in the square loop at, t =3 s

Since, the current is towards the right, then by using the right-hand rule, conclude that the magnetic field through the part of the rectangle above the wire is out of the page and that through the part below the fire is into the page.

The height of the upper part is b-a. The field below the wire up to the distanceb-a is the same as that above the wire but opposite. Hence, both the fields cancel each other and we get the non-zero value of the field only in the region fromb-a toa

Magnetic flux is given by,

$蠒=鈭�BdA$

The magnetic field outside, due to the long straight wire is,

$B=\frac{{渭}_{0}i}{2蟿蟿r}$

The area of the small strip is,$dA=bdr$,

$$\begin{gathered} 蠒=\frac{{渭}_{0}ib}{2蟿蟿}{鈭�}_{b-a}^a\frac{1}{r}dr \\ 蠒=\frac{{渭}_{0}ib}{2蟿蟿}{\left[Inr\right]}_{b-a}^a \\ 蠒=\frac{{渭}_{0}ib}{2蟿蟿}\left[In\frac{a}{b-a}\right] \end{gathered}$$

Using faradays law,

$蔚=-\frac{d蠒}{dt}$

$蔚=-\frac{d}{dt}\left(\frac{{渭}_{0}ib}{2蟿蟿}In\frac{a}{b-a}\right)$

$蔚=-\frac{{渭}_{0}ib}{2蟿蟿}In\left(\frac{a}{b-a}\right)\frac{di}{dt}$

Since,localid="1661854401943" $i=4.5t^2-10t,$

Therefore,

$$\begin{gathered} \frac{di}{dt}=\frac{d}{dt}\left(4.5t^2-10t\right) \\ \frac{di}{dt}=9t-10 \end{gathered}$$

Substituting it in the equation for emf,

$蔚=-\frac{{渭}_{0}b\left(9t-10\right)}{2蟿蟿}In\left(\frac{a}{b-a}\right)$

Substituting the given values,$\mathrm{a}=0.120\mathrm{m},\mathrm{b}=0.160\mathrm{m},\mathrm{t}=3\mathrm{s}$

$$\begin{gathered} \left|蔚\right|=\frac{\left(4蟿蟿脳{10}^{-7}\right)\left(0.16m\right)\left(9脳3-10\right)}{2蟿蟿}In\left(\frac{0.12}{0.16-0.12}\right) \\ \left|蔚\right|=5.98脳{10}^{-7}V \end{gathered}$$.

Hence, the emf in the square loop at t=3 s is $5.98脳{10}^{-7}V$.

Direction of induced current in loop :

The net magnetic field due to the current in the wire is into the page. To oppose this field the induced magnetic field must be out of the page. Thus by Lenz鈥檚 law, the direction of the current is counter clockwise.

Hence, direction of induced current in the loop is counter-clockwise.

Therefore, using magnetic flux formula, the net flux through the loop can be found, and by substituting it in the Faraday鈥檚 law, the emf induced in loop can be found. Find the direction of induced current by using Lenz鈥檚 law.