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Chapter 4: 27E. (page 983)

Are Motional emfs a Practical Source of Electricity? How fast (in \({\bf{m}}/{\bf{s}}\) and \({\bf{mph}}\)) would a \({\bf{5}}.{\bf{00}} - {\bf{cm}}\) copper bar have to move at right angles to a \({\bf{0}}.{\bf{650}} - {\bf{T}}\) magnetic field to generate \({\bf{1}}.{\bf{50}}\,{\bf{V}}\) (the same as a AA battery) across its ends? Does this seem like a practical way to generate electricity?

Short Answer

Expert verified

\(46.153\,{\rm{m/s}}\,\,{\rm{or}}\,\,{\rm{103}}{\rm{.24}}\,{\rm{mph}}\)

Step by step solution

01

Identification of given data

Length of copper wire \(L = 5.00\,{\rm{cm}}\,\,{\rm{or}}\,\,0.05\,{\rm{m}}\)

Magnetic field \(B = 0.650\,{\rm{T}}\)

Induced motional emf \(\varepsilon = 1.50\,{\rm{V}}\)

02

Significance of motional emf

When a conductor moves through a magnetic field, it creates a motional emf. It is expressed as,

\(\varepsilon = vBL\) 鈥�(i)

Where, \(v\) is the velocity, \(B\) is the magnetic field and \(L\) is the length of the conductor

03

 Determining the velocity of copper bar

Rearranging the equation (i)

\(v = \frac{\varepsilon }{{BL}}\)

Substitute all the values in above equation

\(\begin{aligned}{}v = \frac{{1.50\,{\rm{V}}}}{{\left( {0.650\,{\rm{T}}} \right) \times \left( {0.05\,{\rm{m}}} \right)}}\\ = 46.153\,{\rm{m/s}}\,\,{\rm{or}}\,\,{\rm{103}}{\rm{.24}}\,{\rm{mph}}\end{aligned}\)

Hence, the velocity of copper bar is \(46.153\,{\rm{m/s}}\,\,{\rm{or}}\,\,{\rm{103}}{\rm{.24}}\,{\rm{mph}}\)

For generating electricity by using motional emf, there should be large magnetic field requirement of magnetic field in the region. So it is not possible to generate electricity by using the concept of motional emf

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Most popular questions from this chapter


An electron at pointAinFig. E27.15has a speedv0of1.41脳106m/sFind (a) the magnitude and direction of the magnetic field that will cause the electron to follow the semicircular path fromAtoB, and (b) the time required for the electron to move fromAtoB.

In the circuit shown in Fig. E26.20, the rate at which R1 is dissipating electrical energy is 15.0 W. (a) Find R1 and R2. (b) What is the emf of the battery? (c) Find the current through both R2 and the 10.0 惟 resistor. (d) Calculate the total electrical power consumption in all the resistors and the electrical power delivered by the battery. Show that your results are consistent with conservation of energy.

You connect a battery, resistor, and capacitor as in Fig. 26.20a, where R = 12.0 鈩� and C = 5.00 x 10-6 F. The switch S is closed at t = 0. When the current in the circuit has a magnitude of 3.00 A, the charge on the capacitor is 40.0 x 10-6 C. (a) What is the emf of the battery? (b) At what time t after the switch is closed is the charge on the capacitor equal to 40.0 x 10-6 C? (c) When the current has magnitude 3.00 A, at what rate is energy being (i) stored in the capacitor, (ii) supplied by the battery

The heating element of an electric dryer is rated at 4.1 kW when connected to a 240-V line. (a) What is the current in the heating element? Is 12-gauge wire large enough to supply this current? (b) What is the resistance of the dryer鈥檚 heating element at its operating temperature? (c) At 11 cents per kWh, how much does it cost per hour to operate the dryer?

An idealized ammeter is connected to a battery as shown in Fig.

E25.28. Find (a) the reading of the ammeter, (b) the current through the4.00惟

resistor, (c) the terminal voltage of the battery.

Fig. E25.28.
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