Advertisements
Advertisements
प्रश्न
Read the following paragraph and answer the questions.
|
Consider the experimental set-up shown in the figure. This jumping ring experiment is an outstanding demonstration of some simple laws of Physics. A conducting non-magnetic ring is placed over the vertical core of a solenoid. When current is passed through the solenoid, the ring is thrown off. |

- Explain the reason for the jumping of the ring when the switch is closed in the circuit.
- What will happen if the terminals of the battery are reversed and the switch is closed? Explain.
- Explain the two laws that help us understand this phenomenon.
Advertisements
उत्तर
- When a switch in a circuit is closed, current flows, causing emf in the circuit. The magnetic field of the solenoid varies continually due to ac. This causes an eddy current to form in the ring. The magnetic field produced by the induced eddy current in the ring, according to Lenz's law, opposes the changing magnetic field of the solenoid. As a result, the two magnetic fields repel one another, causing the ring to leap.
- When the switch is closed and the battery terminals are reversed, the magnetic field is reversed as well because the direction of the current is likewise reversed.
- The two laws that assist us in understanding this phenomenon are Faraday's laws and Lenz's law of electromagnetic induction.
When the induced current is applied in the same direction as the coil, the motion is countered by Lenz's law. The magnet in the loop exerts force to generate the current. To counteract the change, the magnet's current must impose a force on it.
"Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced," asserts Faraday's first law of electromagnetic induction. Similarly, when the conductor circuit is closed, a current is induced, which is referred to as induced current."
APPEARS IN
संबंधित प्रश्न
Electron drift speed is estimated to be of the order of mm s−1. Yet large current of the order of few amperes can be set up in the wire. Explain briefly.
In Ampere's \[\oint \vec{B} \cdot d \vec{l} = \mu_0 i,\] the current outside the curve is not included on the right hand side. Does it mean that the magnetic field B calculated by using Ampere's law, gives the contribution of only the currents crossing the area bounded by the curve?
Using Ampere's circuital law, obtain an expression for the magnetic flux density 'B' at a point 'X' at a perpendicular distance 'r' from a long current-carrying conductor.
(Statement of the law is not required).
Two identical current carrying coaxial loops, carry current I in opposite sense. A simple amperian loop passes through both of them once. Calling the loop as C, then which statement is correct?
In a capillary tube, the water rises by 1.2 mm. The height of water that will rise in another capillary tube having half the radius of the first is:
A solenoid of length 0.6 m has a radius of 2 cm and is made up of 600 turns If it carries a current of 4 A, then the magnitude of the magnetic field inside the solenoid is:
A long solenoid having 200 turns per cm carries a current of 1.5 amp. At the centre of it is placed a coil of 100 turns of cross-sectional area 3.14 × 10−4 m2 having its axis parallel to the field produced by the solenoid. When the direction of current in the solenoid is reversed within 0.05 sec, the induced e.m.f. in the coil is:
Using Ampere’s circuital law, obtain an expression for magnetic flux density ‘B’ at a point near an infinitely long and straight conductor, carrying a current I.
Magnetic fields are produced by:
According to Ampere's Circuital Law, the line integral \[\oint\vec{B}\cdot d\vec{l}\] taken around any closed loop equals:
