Question

A wire is bent into three circular segments, each of radius r=10 cm, as shown in Figure. Each segment is a quadrant of a circle, ab lying in the xy plane, bc lying in the yz plane, and ca lying in the zx plane. (a) If a uniform magnetic field$\overrightarrow{\mathbf{B}}$points in the positive x direction, what is the magnitude of the emf developed in the wire when B increases at the rate of 3.0 mT/s? (b) What is the direction of the current in segment bc?

Short answer

a) Magnitude of the emf developed in the wire when increases at the rate of 3mT/s is,$2.{410}^{-5}\mathrm{V}$

b) The direction of the current in segment bc is from c tob

Step-by-step solution

Step 1: Given

i) Radius is, r=10 cm

ii) Rate of increase of magnetic field is,$\frac{\mathrm{dB}}{\mathrm{dt}}=3\times {10}^{-3}\mathrm{T}/\mathrm{s}$

Determining the concept

First, find the expression for magnetic flux and then substitute it in the equation for Faraday’s law. By substituting the given values, get the emf developed in the wire. Using Lenz’s law, find the direction of the induced current in the wire segment.

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 which opposes the motion.

Formulae are as follow:

$$\begin{gathered} \varphi =\int \overrightarrow{B}.d\overrightarrow{A} \\ \epsilon =-\frac{d\varphi }{dt} \end{gathered}$$

Where,$\Phi$is magnetic flux, B is a magnetic field, A is an area,𝜀 is emf.

(a) Determining the magnitude of the emf developed in the wire when increases at the rate of 3 mT/s

Themagnetic field is along +x axis, i.e.,

$\overrightarrow{B}=B\stackrel{⏜}{i}$

So the magnetic flux only passes through the area on yz plane.

The area on the yz plane is one-fourth of the area ofthecircle.

The magnetic flux through the loop is given by,

$$\begin{gathered} {\varphi }_B=\int \overrightarrow{B}.d\overrightarrow{A} \\ {\varphi }_B=\int BdA\cos \theta \end{gathered}$$

Where,$\theta$is the angle betweenthemagnetic field vector and the area vector.

Asthemagnetic field and area vector is in the same direction,$\cos \theta =\cos 0=1$

Since the magnetic field is uniform, it is taken out of the integral.

${\varphi }_B=B\int dA$

$\int dA=\frac{1}{4}\pi r^2$

${\varphi }_B=\frac{1}{4}\pi r^{2}B$

The magnitude of emf is given by Faraday’s law.

$\left|\epsilon \right|=\left|\frac{d{\varphi }_B}{dt}\right|$

Thus,

$$\begin{gathered} \left|\mathrm{\epsilon }\right|=\left|\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{1}{4}{\mathrm{\tau \tau r}}^2\mathrm{B}\right)\right|=\frac{{\mathrm{\tau \tau r}}^2}{4}\left|\frac{\mathrm{dB}}{\mathrm{dt}}\right| \\ \left|\mathrm{\epsilon }\right|=\frac{1}{4}\mathrm{\tau \tau }{\left(0.10\mathrm{m}\right)}^2\left(3\times {10}^{-3}\frac{\mathrm{T}}{\mathrm{s}}\right) \\ \left|\mathrm{\epsilon }\right|=2.4\times {10}^5\mathrm{V} \end{gathered}$$

Hence, magnitude of the emf developed in the wire when increases at the rate of 3mT/s is,$2.4\times {10}^{-5}\mathrm{V}$

(b) Determining the direction of the current in segment bc

Since, the magnetic field is along positive x axis, the induced field is in its opposite direction. Thus by Lenz’s law, the current in the coil must flow from point c to point b

Hence, the direction of the current in segmentbcis from c tob

Therefore, find the magnetic flux by using Faraday’s law and the direction of the current by using Lenz’s law.