Advertisements
Advertisements
प्रश्न
Using Ampere’s circuital law, obtain the expression for the magnetic field due to a long solenoid at a point inside the solenoid on its axis ?
Advertisements
उत्तर
Solenoid
• It consists of an insulating long wire closely wound in the form of helix.
• Its length is large as compared to its diameter.
• Magnetic field due to RQ and SP path is zero because they are perpendicular to the axis of solenoid. Since SR is outside the solenoid, the magnetic field is zero.
• The line integral of magnetic field induction `vecB` over the closed path PQRS is
`oint_(PQRS) vecB*vecdl =oint_(PQ)vecB*vecdl =BL`
From Ampere’s circuital law,
`oint_(PQRS)vecB*vecdl = mu_0 `× Total current through rectangle PQRS
BL = μ0 × Number of turns in rectangle × Current
BL = μ0nLI
∴ B= μ0nI
APPEARS IN
संबंधित प्रश्न
A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?
Explain Ampere’s circuital law.
A long, cylindrical tube of inner and outer radii a and b carries a current i distributed uniformly over its cross section. Find the magnitude of the magnitude filed at a point (a) just inside the tube (b) just outside the tube.
A solid wire of radius 10 cm carries a current of 5.0 A distributed uniformly over its cross section. Find the magnetic field B at a point at a distance (a) 2 cm (b) 10 cm and (c) 20 cm away from the axis. Sketch a graph B versus x for 0 < x < 20 cm.
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 long straight wire of radius 'a' carries a steady current 'I'. The current is uniformly distributed across its area of cross-section. The ratio of the magnitude of magnetic field `vecB_1` at `a/2` and `vecB_2` at distance 2a is ______.
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.
The given figure shows a long straight wire of a circular cross-section (radius a) carrying steady current I. The current I is uniformly distributed across this cross-section. Calculate the magnetic field in the region r < a and r > a.
