Advertisements
Advertisements
प्रश्न
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.
उत्तर
`ointvecB*vecdl = mu_0I_(enclosed)`
`(I_(enclosed))/(pia^2) = I/(pir^2)`
`I_(enclosed) = I r_2/a^2`
`vecB*vecdl = Bdl (because costheta = 1)`
`therefore ointBdl = mu_0Ir_2/a^2`
`Bointdl = mu_0Ir^2/a^2`
`B(2pir) = mu_0 Ir^2/a^2`
`B = (mu_0)/(2pi) I/a^2 r `
(ii) For r > a
From Ampere’s circuital law,
`ointvecB*vecdl = mu_0 l_(enclosed)`
`vecB*vecdl =Bdt cos theta`
`theta = 0`
`I_(enclosed) = I`
`oint Bdl = mu_0I`
`Bointdl = mu_0I`
`B(2pir) = mu_0I`
`B= (mu_0)/(2pi) I/r`
APPEARS IN
संबंधित प्रश्न
Obtain an expression for magnetic induction along the axis of the toroid.
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, 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 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 ______.
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.
When current flowing through a solenoid decreases from 5A to 0 in 20 milliseconds, an emf of 500V is induced in it.
- What is this phenomenon called?
- Calculate coefficient of self-inductance of the solenoid.
