91影视

1. Width of the rectangle, W = 20cm = 0.20m
2. Height of the rectangle, H = 30 cm = 0.30M
3. Height of each rectangle, H =$\frac{0.30}{3}\mathrm{m}=0.10\mathrm{m}$
4. Resistance, $R=5.0m惟=5脳{10}^{-3}惟$
5. Vertical scale, $B_s=4.0\mathrm{渭罢}$
6. Vertical scale, $B_b=-2.5B_s=-2.5脳4.0\mathrm{渭罢}=-10\mathrm{渭罢}$
7. Horizontal scale,$t_s=2.0s$

A wire forms a rectangle with resistance 5 ohm and is divided into three parts with magnetic fields$\stackrel{鈬赌}{{\mathrm{B}}_1},\stackrel{鈬赌}{{\mathrm{B}}_2}and\stackrel{鈬赌}{{\mathrm{B}}_3}$ . Calculate the induced emf by using the slope of the lines from Fig 30-44in Faraday鈥檚 law. Find the magnitude and direction of the current in the loop by using Ohm鈥檚 law and Lenz鈥檚 law.

Faraday's law of electromagnetic induction states, 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.

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.

Formulae are as follows:

1. Area of the triangle, A = height$脳$widht
2. Faraday鈥檚 law,

$蔚=-\underset{}{鈭�\frac{d{蠁}_B}{dt}}=A\left(\frac{d{B}_1}{dt}+\frac{d{B}_2}{dt}+\frac{d{B}_3}{dt}\right)$

$\mathrm{i}=\frac{\left|\mathrm{蔚}\right|}{\mathrm{R}}$

Where,${\mathbf{蠒}}_{\mathrm{B}}$is magnetic flux, B is magnetic field, A is area, i is current, R is resistance,饾渶 is emf.

Since, the magnetic flux arises from a dot product of vectors, consider negative sign for the flux $B_{1}and{B}_2$and positive sign for the flux from$B_3$

Find the emf by using Faraday鈥檚 law,

$$\begin{gathered} 蔚=-\underset{}{鈭�\frac{d{蠁}_B}{dt}} \\ 蔚=A\left(\frac{d{B}_1}{dt}+\frac{d{B}_2}{dt}+\frac{d{B}_3}{dt}\right) \end{gathered}$$

Since, the area of triangle is given by, A = height$脳$width

A = (0.10)(0.20)

The slope of the lines from the Fig30-44 gives the corresponding$\frac{dB}{dt}$

Therefore,

$$\begin{gathered} \frac{d{B}_1}{dt}=\frac{4脳{10}^{-6}}{2.0}=2.0脳{10}^{-6} \\ \frac{d{B}_2}{dt}=\frac{2脳{10}^{-6}}{2.0}=1脳{10}^{-6} \\ \frac{d{B}_3}{dt}=\frac{10脳{10}^{-6}}{2.0}=5脳{10}^{-6} \end{gathered}$$

By substituting the values,

$$\begin{gathered} 蔚=\left(0.10\right)\left(0.20\right)\left[\left(2.0脳{10}^{-6}\right)+\left(1脳{10}^{-6}\right)-\left(5脳{10}^{-6}\right)\right] \\ 蔚=-4脳{10}^{-8}V \end{gathered}$$

The minus sign indicates that the effect is dominated by the changes in $B_3$.

Find the magnitude of the current by using ohm鈥檚 law,

$$\begin{gathered} \mathrm{i}=\frac{\left|\mathrm{蔚}\right|}{\mathrm{R}} \\ \mathrm{i}=\frac{\left|-4.0脳{10}^{-8}\right|}{5脳{10}^{-3}} \\ \mathrm{i}=0.8脳{10}^{-5}\mathrm{A} \\ \mathrm{i}=8.0\mathrm{渭础} \end{gathered}$$

Hence, magnitude of the current induced in the wire is,$\mathrm{i}=8.0\mathrm{渭础}$

Consideration of Lenz鈥檚 law leads to the conclusion that the induced current is counterclockwise.

Hence, direction of the current induced in the wire is counterclockwise.

Therefore, the magnitude and direction of the current induced in the wire can be found using Faraday鈥檚 law and Ohm鈥檚 law.