## Saturday, July 2, 2011

### Answers/ Solution to Part 23 (Electromagnetic Induction)

1.a)A bar magnet is inserted into a coil of wire (see Fig. 1) and a galvanometer needle deflects showing an electric current was produced in the wire. The current is electrical energy. What is the source of this energy?

b) A bar magnet which has a field of 1,000 mx is used for the above exp. If it is inserted into a coil of 30 turns in a hundredth of a second, what is the emf produced? Would it be detectable on a galvanometer that reads from - 5 to 5 mA(milli-amps)? Explain.

c) Using the sketch from the diagram, indicate where magnetic N and S poles would appear in the coil.. Explain the reason for this.

Solution:

1(a). The source of the energy is the mechanical (kinetic) energy associated with the motion of the bar magnet through the coil. This is then transferred (with the aid of the magnet’s B-field) to the electrical energy in the galvanometer which causes the deflection of its needle.

(b) We have for the induced emf: E= {Nφ/ t} x (10^-8)

Where φ = 1,000 mx and N = 30, t = 0.01 s

E = [(30) (1,000 mx)/ 0.01s] (10^-8) = 0.03V

Thus, the deflection for a milli-ammeter ought to be enough provided the resistance of the conducting wires isn’t overly large. By Ohm's Law: V = RI

Assming R = 10 ohms, then: I = V/R = 0.03V/ 10 ohms = 0.003 A = 3 mA (or maximum deflections from -3mA to +3mA)

c) By Lenz’s law a N-pole must form on the end of the coil facing the magnet so that the motion of the magnet is thereby opposed. (Otherwise a much larger ‘single’ magnet would form which increases the motion, and violates the conservation of energy).

2) Explain, on the basis of the experimental results, how it is that an electric generator may be rotated easily on an open circuit but becomes extremely hard to turn when it is connected to a load.

Ans. On an open circuit there is no opposition created because there is no current to oppose it by forming a B-field to oppose the (rotating) motion. Connected to a load, Lenz’s law applies and thus conservation of energy means a B-field is created that opposes the rotation. Another way to say this, is that once a load is connected electrons are forced to flow in the opposite direction and so a “back emf” is induced that opposes the motion of the generator in accord with Faraday’s Law: “The EMF generated is proportional to the rate of change of the magnetic flux”.
In an open circuit no change of flux exists. When the circuit is closed (analogous to closing the open key k in the experiment) a flux emerges and a back emf.

3) An alternative way to assess the magnitude of induced emf is by way of the formal equation: E = B Lv, where B is the magnetic flux density, L is the length of the displaced object and v the relative motion (velocity of moving agent).

Use the above to estimate the induced emf created when a copper bar 30 cm in length is initially perpendicular to a field B of flux density 0.8 weber/m^2 and moves at right angles to the field at a speed of 50 cm/sec.

Soln.

E = B Lv = (0.8 wb/m^2)(0.5m/s) (0.3m) = 0.12V

4) A copper disc of 10 cm radius is allowed to rotate at 20 rotations per second about its axis and with its plane perpendicular to a uniform B-field of flux density 0.6 weber/m^2. Find the potential difference between the disc's circumference and its center. (Hint: The magnetic flux threading a uniform disc is φ = BA where A is the area of the disc.)

Soln.:

t = 0.05s (for one axial rev., e.g. 1/20 s)

B = 0.6 wb/m^2

φ = BA = (0.6 wb/m^2)( π(0.1m)^2) = 0.018 mx

Then: p.d. = φ/ t (rate of change of flux) = 0.018 mx/ 0.05s = 0.38V