Case Study Chapter 5 : Magnetism and Matter
Read the following paragraph and answer the questions :
Consider the experimental set up shown in the figure . This jumping ring expriment is an outstanding demostration of some simple laws of physics. A conducting non-magnetic ring is placed over the verticle core of a solenoid. When current is passed through the solenoid. The ring is thrown off. [CBSE 2023]
Answer the following question.
(i) Explain the reason of jumping of the ring when the switch is closed in the circuit. (1)
(ii) What will happen if the terminals of the battery are reversed and the switch is closed ? Explain. (1)
(iii) Explain the two laws that help us understand this phenomenon. (2)
Solution :
(i) Magnetic flux changes in the coil due to which induced current in copper ring opposes the change in magnetic flux in the coil which causes jumping of the ring when the swich is closed.
(ii) Polarity of current change and direction of induced emf also change.
(iii) There are two laws which play important role
(a) Faraday’s laws
First law : Whenever the amount of flux linked with a circuit changes, an emf is induced in the circuit. This induced emf persists as long as the change in magnetic flux continues.
Second law : Magnitude of the induced emf is equal to the rate of change of magnetic flux.
induced emf, e = – dΦ/dt
(b) Lenz’s law :
This law give us the direction of induced emf. According to this law, the direction of induced emf in a circuit is such that it opposes the change in magnetic flux.
Case Study: 2
Read the following paragraph and answer any four question.
Faraday’s laws
According the Faraday’s first law, whenever the amount of magnetic flux linked with a circuit changes, an emf is induced in it. Induced current is determined by the rate at which the magnetic flux changes.
Mathematically, the magnitude of the induced emf in a circuit is equal to the rate of change of magnetic flux through the circuit.
Induced emf ∝ Rate of change of magnetic flux
(i) On the basis of Faraday’s law, current in the coil is smaller. (1)
(a) When the magnet is pushed towards the coil faster
(b) When the magnet is pulled away the coil faster
(c) Both (a) and (b)
(d) Neither (a) nor (b)
(ii) The flux linked with a circuit is given by Φ = t³ + 3t – 5. The graph between time (X-axis) and induced emf(Y-axis) will be a (1)
(a) Straight line through the origin
(b) Straight line with positive intercept
(c) Straight line with negative intercept
(d) Parabola not through the origin
(iii) Wire loop is rotated in a magnetic field. The frequency of change in the direction of the induced emf is (1)
(a) Once per half revolution
(b) Twice per revolution
(c) Four times per revolution
(d) Both (a) and (b)
(iv) The instantaneous magnetic flux linked with a coil is given by Φ = (5 t³ – 100 t + 20) Wb. The emf induced in the coil at time t = 2 s is (1)
(a) -40 V (b) 40 V
(c) 140 V (d) 300 V
OR
(v) A copper disc of radius 0.1 m is rotated about its centre with 20 rev/s in a uniform magnetic field of 0.2 T with its plane perpendicular to the field. The emf induced across the radius of the disc is (1)
(a) (π/20) V (b) (π/120) V
(c) 20 π mV (d) 40 π mV
Solution :
(i) Answer (b)
Explanation: Current will be smaller, when the magnetic is pulled away from the magnet because change in magnetic flux associated with coil decreases.
(ii) Answer(d)
Explanation: Φ = t³ + 3t – 5
∴ Induced emf, e = – dΦ/dt = -(3 t² + 3)
= -3 t² – 3
At t =0, e = -3 V
Therefore, shape of graph will be a parabola not through the origin (Since, e ∝ t²).
(iii) Answer (d)
Explanation: If a wire loop is rotated in a magnetic field, the frequency of change in the direction of the induced emf is twice per revolution, i.e. once per half revolution.
(iv) Answer (b)
Explanation: Given, Φ = 5 t³ – 100 t + 20
Emf induced, e = – dΦ/dt = – d/dt(5 t³ – 100 t + 20)
= – 15 t² + 100 = – 15 ×(2)² + 100
= 40 V
OR
(v) Answer (d)
Explanation: From Faraday’s law of electromagnetic
Induction, e = – dΦ/dt =- BAN (Since, dt = 1 s)
Given, B = 0.2 T, N = 20, A = π r² = π(0.1)²
Since, e = 0.2 × 20 × π(0.1)² = 40 π mV
Case Study Chapter 2: Electrostatic Potential and Capacitance :
A capacitor is a system of two conductors separated by an insulator. The two conductors have equal and opposite charges with a potential difference between them. The capacitance of a capacitor depends on the geometrical configuration(shape, size and separation) of the system and also on the nature of the insulator separating the two conductors. They are used to store charges. Like resistors capacitors can be arranged in series or parallel or a combination of both to obtain desired value of capacitance. [CBSE 2023]
Solution :- See full solution
Case Study Physics Chapter 3 – Current Electricity
Read the following paragraph and answer the question.
When Deepak studies the electrical circuits and the current flowing through them, he became curious about the range of the currents we come across in daily life. He collected the data and presented in a tabular form as shown below. He then studied the instruments used to detect and measure current, however could not understand the difference between an ammeter and an ideal ammeter and thus went to this teacher for the explanation.
Solution: See full solution
Chapter 8: Electromagnetic waves
