Advertisements
Advertisements
प्रश्न
In order to have a current in a long wire, it should be connected to a battery or some such device. Can we obtain the magnetic due to a straight, long wire by using Ampere's law without mentioning this other part of the circuit?
Advertisements
उत्तर
We can obtain a magnetic field due to a straight, long wire using Ampere's law by mentioning the current flowing in the wire, without emphasising on the source of the current in the wire. To apply Ampere's circuital law, we need to have a constant current flowing in the wire, irrespective of its source.
APPEARS IN
संबंधित प्रश्न
State Ampere’s circuital law.
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?
Obtain an expression for magnetic induction along the axis of the toroid.
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 ?
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.
A thin but long, hollow, cylindrical tube of radius r carries i along its length. Find the magnitude of the magnetic field at a distance r/2 from the surface (a) inside the tube (b) outside the tube.
A long, cylindrical wire of radius b carries a current i distributed uniformly over its cross section. Find the magnitude of the magnetic field at a point inside the wire at a distance a from the axis.
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.
State Ampere’s circuital law.
Calculate the magnetic field inside and outside of the long solenoid using Ampere’s circuital law
A straight wire of diameter 0.5 mm carrying a current of 1 A is replaced by another wire of 1 mm diameter carrying the same current. The strength of the magnetic field far away is ______.
The force required to double the length of a steel wire of area 1 cm2, if it's Young's modulus Y = `2 xx 10^11/m^2` is:
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 concentric and coplanar circular loops P and Q have their radii in the ratio 2:3. Loop Q carries a current 9 A in the anticlockwise direction. For the magnetic field to be zero at the common centre, loop P must carry ______.
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.
