Advertisements
Advertisements
प्रश्न
State Ampere’s circuital law.
Advertisements
उत्तर १
Statement: The line integral of the magnetic field `(vecB)` around any closed path is equal to μ0 times
the total current (I) passing through that closed path.
`:.ointvecB.vec(dl)=mu_0I`
उत्तर २
Statement: The line integral of the magnetic field `(vecB)` around any closed path is equal to μ0 times
the total current (I) passing through that closed path.
`:.ointvecB.vec(dl)=mu_0I`
उत्तर ३
Ampere’s law states that the path integral or line integral over a closed loop of the magnetic field produced by a current distribution is given by `oint vec("B") . vec("dl") = µ_0"l"`
where I refers to the current enclosed by the loop.
Ampere’s law is a useful relation that is analogous to Gauss’s law of electrostatics. It is a relation between the tangential component of magnetic field at points on a closed curve and the net current through the area bounded by the curve.
To evaluate the expression for `oint vec("B") . vec("dl")` let us consider a long, straight conductor carrying a current I, passing through the centre of a circle of radius r in a plane perpendicular to the conductor.
According to Biot-Savart law of magnetism, the field has a magnitude `(µ_0"l")/(2pir)` at every point on the circle, and it is tangent to the circle at each point.
The line integral of `vec("B")` around the circle is
`oint vec("B") . vec("dl") = oint(µ_0"I")/(2pir)"dl" = (µ_0"l")/(2pir) oint"dl"`
Since `ointvec("dI") = 2pir` is the circumference of the circle,
Therefore , `oint vec("B") . vec("dl") = µ_0"l"`
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.
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 long straight wire of a circular cross-section of radius ‘a’ carries a steady current ‘I’. The current is uniformly distributed across the cross-section. Apply Ampere’s circuital law to calculate the magnetic field at a point ‘r’ in the region for (i) r < a and (ii) r > a.
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 long, straight wire carries a current. Is Ampere's law valid for a loop that does not enclose the wire, or that encloses the wire but is not circular?
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?
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 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.
Two large metal sheets carry currents as shown in figure. The current through a strip of width dl is Kdl where K is a constant. Find the magnetic field at the points P, Q and R.
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 ______.
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 ______
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.
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.
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.
