Advertisements
Advertisements
प्रश्न
Obtain an expression for magnetic induction along the axis of the toroid.
Advertisements
उत्तर
Magnetic induction along the axis of toroid:
The toroid is a solenoid bent into a shape of the hollow doughnut.
Consider a toroidal solenoid of average radius ‘r’ having center carrying the current I. In order to find magnetic field produced at the center along the axis of toroid due to the current flowing through the coil, imagine an Amperial loop of radius ‘r’ and traverse it in the clockwise direction.
According to Ampere’s circuital law,
`ointvecB.vec(dL)=mu_0I`
Here current I flow through the ring as many times time as there are the number of turns. Thus the total current flowing through toroid is N I , where N is the total number of turns.
`:.ointvecB.vec(dL)=mu_0NI" ---------(1)"`
Now, and are in same direction `:.ointvecB.vec(dL)=BointdL`
`:.ointvecB.vec(dL)=B(2pir)" ------(2)"`
Comparing equation (1) and equation (2 )
μ0NI=B(2πr) `:.B=(mu_0NI)/(2pir)` .......(3)
If ‘n’ is the number of turns per unit length of toroid then `n=N/(2pir)`
Substituting this value in equation No (3) we get B = μ0 n I
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?
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.
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).
What is magnetic permeability?
Define ampere.
Calculate the magnetic field inside and outside of the long solenoid using Ampere’s circuital law
Ampere’s circuital law is given by _______.
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:
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 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:
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 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.
Briefly explain various ways to increase the strength of the magnetic field produced by a given solenoid.
