Geometry Chapter 5 Test Form 1a Answers

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• [DOWNLOAD] Geometry Chapter 5 Test Form 1a Answers

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• Chapter 5 Test B Geometry Answers

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• Chapter 5 Quiz 1 Glencoe Geometry Answers

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• Chapter 5 - Test - Page 275: 2b

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  Electric Power metres any two Question A long straight current carrying wire passes normally through the centre of a circular loop. If the current through the wire increases, will there be an induced emf in the. Delhi Answer: No,As the magnetic field due to current carrying wire will be in the plane of the circular loop, so magnetic flux will remain zero. Question Predict the polarity of the capacitor in the situation described in the given diagram. A bar magnet is moved in the direction indicated by the arrow between two coils PQ and CD. Predict the direction of the induced current in each coil All India Answer: Induced current flows from P to Q through ammeter; while it flows from D to C through ammeter.

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  What is the direction of induced currents in metal rings 1 and 2 when current I in the wire is increasing steadily? In the figure given, mark the polarity of plates A and B of a capacitor when the magnets are quickly moved towards the coil. All India Answer: The polarity of plate is positive; 1 while that of plate B is negative. Derive an expression for the self-inductance of a long air-cored solenoid of length l and number of turns N. What can be done to have larger deflection in the galvanometer with the same battery? Delhi Answer: i To have larger deflection in the galvanometer with the same battery, coil P has to be moved faster towards S so that rate of change of magnetic flux is more. The magnitude of induced emf is proportional to the rate of change of magnetic flux. A coil Q is connected to low voltage bulb B and placed near another coil P as shown in the figure. When coil Q is moved towards left, magnetic flux linked with Q decreases and may even reduce to zero at some distance.

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  The emf induced may decrease and the bulb B gets dimmer. Two identical loops, one of copper and the other of aluminium, are rotated with the same angular speed in the same magnetic field. Compare i the induced emf and ii the current produced in the two coils. Justify your answer. Explain in which loop do you expect the induced emf to be constant during the passage out of the field region. The magnetic field is normal to the loops. All India Answer: In case of rectangular loop, induced emf will be constant.

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  It is because rate of change of the area of rectangular loop is uniform whereas that of a circular loop is not constant. A current is induced in coil C1 due to the motion of current carrying coil C2. Answer: a To obtain a large deflection, one or more of the following steps can be taken : Use a rod made of soft iron inside the coil C2. Connect the coil to a powerful battery. Move the arrangement rapidly towards the test coil C1 b Replace the galvanometer by a small torch bulb. A magnet is quickly moved in the direction indicated by an arrow between two coils C1 and C2 as shown in the figure. What will be the direction of induced current in each coil as seen from the magnet? Delhi Answer: When magnet moves in given direction, induced current will be clockwise in both the coils as magnet is going away from C1 and moving towards C2 making C1 as S pole and C2 also as S pole according to Lenz rule.

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  So C1 will try to attract and C2 will try to repel the motion of magnet. What are eddy currents? Write any two applications of eddy currents. All India Answer: a Eddy current : Due to change in magnetic flux, if there is induced current in the volume bulk of the material, it is called as eddy currents. It is a necessary evil in an arrangement as it can be used in applications like electric brakes, induction furnaces and dead-beat galvanometers and brings loss of energy with heat production etc. Define self-inductance of a coil. Delhi Answer: The self-inductance of a coil may be defined as the induced emf set up in the coil due to a unit rate of change of current through it.

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  Let I be the current through the inductor L at any instant t The current rises at the rate dlldt, so the induced Work done against the induced emf in small time dt is Total work done in building up the current from 0 to I in This work done is stored as the magnetic field energy U in the inductor Question Define mutual inductance between two long coaxial solenoids. Find out the expression for the mutual inductance of inner solenoid of length l having the radius r1 and the number of turns n1 per unit length due to the second outer solenoid of same length and n2 number of turns per unit length.

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  Delhi Answer: Mutual induction is the phenomenon of inducing emf in a coil due to the rate of change of current in a nearby coil. When a current I2 is set up through S2 outer solenoid , it in turn sets up a magnetic flux through S1. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring. Derive the expression for the self inductance of a long solenoid of cross sectional area A and. If a current I flows through the coil, then the magnetic field inside the coil is almost constant and is given by Magnetic flux linked with each turn, Question A metallic rod held horizontally along east-west direction, is allowed to fall under gravity. Will there be an emf induced at its ends? Yes there will be an emf induced as the horizontal component of field of earth, velocity of the motion of the rod and the length of the rod are all perpendicular to each other. Delhi Answer: ii Current in the loop, Question A metallic rod of length l is rotated at a constant angular speed to, normal to a uniform magnetic field B.

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• Chapter 5 - Test - Page 275: 10a

  Derive an expression for the current induced in the rod, if the resistance of the rod is R. Calculate the i frequency at which the power factor of the circuit is unity ii current amplitude at this frequency iii Q-factor Delhi i Power factor will be unity at resonance, Question Deduce expressions for : i Maximum emf induced in the coil. Delhi Answer: We know that induced emf Question Write its S. Its S. All India Answer: i First law : Whenever the magnetic flux linked with a closed circuit changes, an emf and hence a current is induced in it which lasts only so long as the change in flux is taking place. This phenomenon is called electromagnetic induction.

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  Second law : The magnitude of the induced emf is equal to the rate of change of magnetic flux linked with the closed circuit. Since v, l and B are to be perpendicular to get induce emf, the vertical component alone will contribute to the induction — ve sign signifies that the emf will oppose the change in magnetic flux causing it. State the law that gives the polarity of the induced emf. The polarity of induced e. A constant uniform magnetic field B parallel to the axis is present every where. Using Lorentz force, explain how emf is induced between the centre and the metallic ring and hence obtain the expression for it. Delhi Answer: From these two equations, we get Question A constant and uniform magnetic field B parallel to the axis is present eveywhere.

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  Deduce the expression for the emf induced in the rod. If r is the resistance of the rod and the metallic ring has negligible resistance, obtain the expression for the power generated. Write its SI unit for self-inductance of a coil. All India Answer: Coefficient of self induction. Consider a coil L as shown in Figure. The SI unit of self inductance is henry H. Expression for self-inductance : Consider a long solenoid of length Z and cross-sectional area A having n turns per unit length.

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  Calculate the emf induced between the axle and the rim of the wheel. How will the value of emf be affected if the number of spokes were increased? Total number of turns in the two solenoids are N1 and N2 respectively. All India Answer: i Mutual inductance. The mutual inductance of two coils is numerically equal to the induced emf produced in one coil, when the rate of change of current is unit in the other coil. This work, in fact, converts itself in the form of induced emf. Hence, the phenomenon of electromagnetic induction is in accordance with the law of conservation of energy. The corresponding flux linkage with solenoid S1 is M12 is called the mutual inductance of solenoid S1 with respect to solenoid S2. The magnetic field due to the current I2 in S2 We now consider the reverse case : A current I1, is passed through the solenoid S1 and the flux linkage with coil S2 is, M21 is called the mutual inductance of the solenoid S2 with respect to solenoid S1 The flux due to the current I1 in S1 can be assumed to be confined solely inside S1 since the solenoid are very long.

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  Thus, flux linkage with solenoid S2 is Question Define the term self-inductance of a solenoid. Obtain the expression for the magnetic energy stored in an inductor of self-inductance L to build up a current I through it. All India Answer: Self-inductance: emf is induced in a single isolated coil due to change of flux through the coil by means of varying the current through the same coil. This phenomenon is called self-induction. In this case, flux linkage through a coil of N turns is proportional to the current through the coil and is expressed as where constant of proportionality L is called self-inductance of the coil. It is also called the coefficient of self-induction of the coil. When the current is varied, the flux linked with the coil also changes and an emf is induced in the coil. Using the above equation the induced emf is given by Thus, the self-induced emf always opposes any change increase or decrease of current in the coil. It is possible to calculate the self-inductance for circuits with simple geometries.

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