This jumping ring expriment is an outstanding demostration

Case Study Chapter 5 : Magnetism and Matter

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]

(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

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

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 :

Explanation: Current will be smaller, when the magnetic is pulled away from the magnet because change in magnetic flux associated with coil decreases.

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).

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.

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

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