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प्रश्न
Explain Ampere’s circuital law.
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उत्तर
Ampere’s law is the generalisation of Biot-Savart’s law and is used to determine magnetic field at any point due to a distribution of current. Consider a long straight current carrying conductor XY, placed in the vacuum. A steady current ‘I’ flows through it from the end Y to X as shown in the figure
Imagine a closed curve (amperian loop) around the conductor having radius 'r'. The loop is assumed to be made of a large number of small elements each of length `vec(dl)`. Its direction is along the direction of the traced loop.
Le `vecB`be the strength of magnetic field around the conductor. All the scalar products of ` vecB` and `vec(dl)` given the product of `mu_0` and I. It is given by `ointvecB.vec(dl) = ointBlcostheta` where, theta = angle between `vecB` and `vec(dl)`
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संबंधित प्रश्न
Write Maxwell's generalization of Ampere's circuital law. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is `I=varepsilon_0 (dphi_E)/dt,`where ΦE is the electric flux produced during charging of the capacitor plates.
State Ampere’s circuital law.
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.
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?
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 ?
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?
A hollow tube is carrying an electric current along its length distributed uniformly over its surface. The magnetic field
(a) increases linearly from the axis to the surface
(b) is constant inside the tube
(c) is zero at the axis
(d) is zero just outside the tube.
In a coaxial, straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero
(a) outside the cable
(b) inside the inner conductor
(c) inside the outer conductor
(d) in between the tow conductors.
Consider the situation described in the previous problem. Suppose the current i enters the loop at the points A and leaves it at the point B. Find the magnetic field at the centre of the loop.
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.
Sometimes we show an idealised magnetic field which is uniform in a given region and falls to zero abruptly. One such field is represented in figure. Using Ampere's law over the path PQRS, show that such a field is not possible.
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).
State Ampere’s circuital law.
Find the magnetic field due to a long straight conductor using Ampere’s circuital law.
Calculate the magnetic field inside and outside of the long solenoid using Ampere’s circuital law
Ampere’s circuital law is given by _______.
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:
Ampere's circuital law is used to find out ______
A thick current carrying cable of radius ‘R’ carries current ‘I’ uniformly distributed across its cross-section. The variation of magnetic field B(r) due to the cable with the distance ‘r’ from the axis of the cable is represented by:
A thick current carrying cable of radius ‘R’ carries current ‘I’ uniformly distributed across its cross-section. The variation of magnetic field B(r) due to the cable with the distance ‘r’ from the axis of the cable is represented by ______
Two identical current carrying coaxial loops, carry current I in an opposite sense. A simple amperian loop passes through both of them once. Calling the loop as C ______.
- `oint B.dl = +- 2μ_0I`
- the value of `oint B.dl` is independent of sense of C.
- there may be a point on C where B and dl are perpendicular.
- B vanishes everywhere on C.
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.
