Question

Question: Figure (a) shows a wire that forms a rectangle ( $\mathit{W}\mathbf{=}\mathbf{20}\mathit{c}\mathit{m}\mathbf{,}\mathit{H}\mathbf{=}\mathbf{30}\mathit{c}\mathit{m}$) and has a resistance of $\mathbf{5}\mathbf{.}\mathbf{0}\mathit{m}\mathit{\Omega }$. Its interior is split into three equal areas, with magnetic fields $\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{B}}_{\mathbf{1}}}\mathbf{,}\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{B}}_{\mathbf{2}}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\stackrel{\mathbf{\rightharpoonup }}{{\mathbf{B}}_{\mathbf{3}}}$ . The fields are uniform within each region and directly out of or into the page as indicated. Figure (b) gives the change in the z components localid="1661850270268" ${\mathit{B}}_{\mathbf{z}}$ of the three fields with time t; the vertical axis scale is set by localid="1661850263101" ${\mathit{B}}_{\mathbf{s}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\mu T}\mathbf{}\mathbf{and}\mathbf{}{\mathit{B}}_{\mathbf{b}}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{5}{\mathit{B}}_{\mathbf{s}}$, and the horizontal axis scale is set by localid="1661850259349" ${\mathit{t}}_{\mathbf{s}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathit{s}$.

(a) What is magnitude of the current induced in the wire?(b) What is the direction of the current induced in the wire?

Short answer

a) Magnitude of the current induced in the wire is, i = 8.0$\mathrm{\mu A}$

b) Direction of the current induced in the wire is counterclockwise.

Step-by-step solution

Step 1: Given

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, $R=5.0m\Omega =5\times {10}^{-3}\Omega$
5. Vertical scale, $B_s=4.0\mathrm{\mu T}$
6. Vertical scale, $B_b=-2.5B_s=-2.5\times 4.0\mathrm{\mu T}=-10\mathrm{\mu T}$
7. Horizontal scale,$t_s=2.0s$

Determining the concept

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

$\epsilon =-\underset{}{\sum \frac{d{\phi }_B}{dt}}=A\left(\frac{d{B}_1}{dt}+\frac{d{B}_2}{dt}+\frac{d{B}_3}{dt}\right)$

1. Ohm’s law

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

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

(a) Determining the magnitude of the current induced in the wire

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’s law,

$$\begin{gathered} \epsilon =-\underset{}{\sum \frac{d{\phi }_B}{dt}} \\ \epsilon =A\left(\frac{d{B}_1}{dt}+\frac{d{B}_2}{dt}+\frac{d{B}_3}{dt}\right) \end{gathered}$$

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

A = (0.10)(0.20)

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

Therefore,

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

By substituting the values,

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

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

Find the magnitude of the current by using ohm’s law,

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

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

(b) Determining the direction of the current induced in the wire

Consideration of Lenz’s 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’s law and Ohm’s law.