EXERCISE SHORT QUESTIONS
CHAPTER # 15: ELECTROMAGNETIC INDUCTION
Q # 1. Does the inducted emf in circuit depend on the resistance of the circuit? Does the induced current depend on the resistance of the circuit?
Ans. The expression for induced emf is given by
\[\mathbf{\varepsilon = - N}\frac{\mathbf{\mathrm{\Delta}\phi}}{\mathbf{\mathrm{\Delta}t}}\]
The relation shows that the induced emf in a coil only depend upon the rate of change of magnetic flux and number of turns but does not depend upon the resistance of the coil.
As the induced current flowing through a coil is given by:
\[I = \frac{\varepsilon}{R}\]
this expression shows that the value of current depends upon the resistance of the coil. The smaller the value of the resistance of the coil, greater will he the value of current.
Q # 2. A square loop of wire is moving through a uniform magnetic field. The normal to the loop is oriented parallel to the magnetic field. Is emf induced in the loop? Give a reason for your answer.
Ans. The induce emf in a wire is given by:
\[\varepsilon = vBL\sin\theta\]
Where \(\theta\) the angle between ``\(v\)'' and ``\(B\)''.
When normal to the loop is parallel to the field, the velocity vector ``\(v\)'' of side of loop is also parallel to field ``\(B\)'', so \(\theta = 0\). Therefore,
\[\varepsilon = vBL\sin 0\]
\[\Longrightarrow \varepsilon = vBL(0)\]
\[\Longrightarrow \varepsilon = 0\]
Thus, emf induced in the loop is zero.
Q # 3. A light metallic ring is released from above into a vertical bar magnet as shown in the figure. Viewed from above, does the current flow clockwise or anti-clockwise in the ring?
Ans. According to Lenz's law, the direction of the induced current is opposite to the cause which produces it. So, the side of the ring facing north pole of magnet must be north pole of the induced magnetic field.
When viewed from above, the current in the ring is clockwise.
Q # 4. What is the direction of the current through resistor R as shown in the figure? As the switch S is (a) closed (b) open.
Ans. (a) When switch S is closed, then the current in the primary coil increases from zero to maximum. During this time interval, magnetic flux through the secondary coil increases from zero to maximum and induced current produce in it. According to Lenz's law, the current through the secondary should flow in anti-clockwise direction. And current through resistor will be from left to right.
(b) However, if the switch is opened, the induced current through secondary should flow in clockwise direction. So the current through resistor R will flow from right to left.
Q # 5. Does the induced emf always act to decrease the magnetic flux through a circuit?
Ans. The induced emf always opposes the cause that produces it.
- If the magnetic flux through the circuit through the circuit is increasing, then induced emf acts to decrease the magnetic flux.
- If the magnetic flux through the circuit through the circuit is decreasing, then induced emf acts to increase the magnetic flux.
Hence, the induced emf does not always act to decrease the magnetic flux through the circuit.
Q # 6. When the switch in the circuit is closed, a current is established in the coil and the metal ring jumps upward. Why? Describe what would happen to the ring if the battery polarity were reversed?
Ans. When the switch in the circuit is closed, the current is set up in the coil which establish magnetic field in it.
This result in change of magnetic flux through the metallic ring and hence an induced emf is produced in it.
The induced magnetic field in the ring opposes the magnetic field of the coil (according to Lenz's law). Therefore the ring experience a force of repulsion and jumps up.
The same event occurs even if the polarity of the battery is reversed.
Q # 7. Figure shows a coil of wire in the xy-plane with a magnetic field directed along the y-axis. Around which of the three coordinate axes should the coil be rotated in order to generate an emf and a current in the coil?
Ans.
The coil must be rotated along x-axis to get change of magnetic flux and an induced current through it.
If the coil is rotated about y-axis, the flux passing through the coil zero because plane of the coil remains parallel to magnetic field B all the times.
If the coil is rotated about z-axis then no change of magnetic flux takes place through coil.
Hence if the coil is rotated about x-axis, then there is a change of magnetic flux passing through a coil. So only in this case, an emf is induced in the coil.
Q # 8. How would you position a flat loop of wire in a changing magnetic field so that there is no emf induced in the loop?
Ans. If the plane of loop of wire is placed parallel to changing magnetic field i.e., \(\theta = 0\), then no flux through it will change. Hence no emf will be induced through the loop as:
\[\varepsilon = \omega AB\sin\theta\]
\[\varepsilon = \omega AB\sin 0 = \omega AB(0)\]
\[\varepsilon = 0\]
Q # 9. In a certain region, the earth's magnetic field point vertically down. When a plane flies due north, which wing tip is positively charged?
Ans. The magnetic force on electrons in the wing is given by:
\[\mathbf{F} = - e\left( \mathbf{v \times B} \right)\]
When the plane flies due north in the earth magnetic field directed vertically downward, then electrons will experience force in east direction.
Thus west wing-tip of the plane is positively charged.
Q # 10. Show that \(\mathbf{\varepsilon}\) \textbf{and} \(\frac{\mathbf{\mathrm{\Delta}\varphi}}{\mathbf{\mathrm{\Delta}t}}\) have the same units.
Ans. As we know that:
\(\varepsilon = \frac{W}{q}\)
\(\Longrightarrow unit\ of\ \varepsilon = \frac{unit\ of\ Work}{unit\ of\ charge} = \frac{joule}{coulomb} = volt\) -\/-\/-\/-\/-\/-\/-\/-\/-\/- (1)
\(\frac{\mathbf{\mathrm{\Delta}\varphi}}{\mathbf{\mathrm{\Delta}t}}\mathbf{=}\frac{B\mathbf{\mathrm{\Delta}A}}{\mathbf{\mathrm{\Delta}t}}\)
\[\Longrightarrow unit\ of\ \frac{\mathbf{\mathrm{\Delta}\varphi}}{\mathbf{\mathrm{\Delta}t}}\mathbf{=}\frac{(unit\ of\ B)(unit\ of\ \mathrm{\Delta}A)}{unit\ of\ \mathbf{\mathrm{\Delta}t}}\mathbf{=}\frac{\left( NA^{- 1}m^{- 1} \right)\left( m^{2} \right)}{s}\]
\[\Longrightarrow unit\ of\ \frac{\mathbf{\mathrm{\Delta}\varphi}}{\mathbf{\mathrm{\Delta}t}}\mathbf{=}\frac{N\mathbf{\times}m}{A \times s}\]
As \(N\mathbf{\times}m = J\ (joule)\) and \(A \times s = C\ (coulomb)\)
\(\Longrightarrow unit\ of\ \frac{\mathbf{\mathrm{\Delta}\varphi}}{\mathbf{\mathrm{\Delta}t}}\mathbf{=}\frac{joule}{coulomb}\mathbf{=}volt\) -\/-\/-\/-\/-\/-\/-\/-\/-\/- (2)
Hence from (1) and (2), it is proved that both
\(\mathbf{\varepsilon}\)and \(\frac{\mathrm{\Delta}\varphi}{\mathrm{\Delta}t}\)have the same units.
Q # 11. When an electric motor, such as an electric drill, is being used, does it also act as a generator? If so what is the consequences of this?
Ans. When a motor like drill machine is working, its armature (coil) is revolving in a uniform magnetic field by a potential difference V. The revolving armature of the motor experiences change in flux, which produces emf, known as back emf of the motor. Hence, a drill machine (or a motor) is also operating like a generator whose generated emf is known as~\textbf{back emf.
Q # 12. Can a DC motor be turned into a DC generator? What changes are required to be done?
Ans. Yes, a DC motor be turned into a DC generator.
In order to convert DC motor into a DC generator, two changes are to be done:
The magnetic field must be supplied by the permanent magnet and not by electromagnet.
An arrangement to rotate the coil armature should be provided.
Q # 13. Is it possible to change both the area of the loop and the magnetic field passing through the loop and still not have an induced emf in the loop?
Ans.
Q # 14. Can an electric motor be used to drive an electric generator with output from the generator being used to operate the motor?
Ans. No it is not possible. Because if it is possible, it will be a self operating system without getting energy from some external source and this is against the law of conservation of energy.
Q # 15. A suspended magnet is oscillating freely in a horizontal plane. The oscillations are strongly damped when a metal plate is placed under the magnet. Explain why this occurs?
Ans. the oscillating magnet produces change of magnetic flux close to it. The metal plate placed below it experiences the change of magnetic flux. As the result, eddy current are produced inside metal. According to Lenz's law, these eddy current oppose the cause which produce it. So, the oscillations of magnet are strongly damped.
Q # 16. Four unmarked wires emerge from a transformer. What steps would you take to determine the turn ratio?
Ans. By checking continuity of the coils, the coils are separated as primary and secondary coils. An alternating voltage of known value \(V_{P}\) is connected to one coil (primary coil), the output voltage \(V_{S}\) across the ends of the other coil (secondary coil) is measured. The turn ratio of the coil is determined by using relation:
\[\frac{V_{S}}{V_{P}} = \frac{N_{S}}{N_{P}}\]
Q # 17. (a) Can a step-up transformer increase the power level? (b) In a transformer, there is no transfer of charge from the primary to the secondary. How is, then the power transferred?
Ans.
(a). In case of an ideal transformer, the power output is equal to the power input. In actual transformer, due of dissipation of energy in the coil, the output power is always less than input power. Therefore, a step-up transformer can't increase power level.
(b). The two coils of transformer are magnetically linked i.e.,the change of flux through one coil is linked with the other coil.
Q # 18. When the primary of a transformer is connected to AC mains, the current in it (a) Is very small if the secondary circuit is open, but (b) Increases when the secondary circuit is closed. Explain these facts.
Ans.
(a). If the secondary circuit is open, then output power will be zero. Because output power is always slightly smaller than the output power, therefore a very small value of current is being drawn by a primary coil of transformer form AC mains.
(b). When the secondary circuit is closed, the output power increases. To produce this power, transformer will draw large current from the A.C. mains to increase its primary power (V\textsubscript{p}I\textsubscript{p}).
0 comments:
Post a Comment