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.
Advertisements
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
State Ampere’s circuital law.
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?
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).
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 ______.
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 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.
In the sign convention for Ampere's law, if the fingers of the right hand curl along the direction of integration (\[d\vec{l}\]), the thumb points:
According to the right-hand rule sign convention for Ampere's law, currents flowing in the direction of the thumb are taken as ______ and currents flowing opposite to it are taken as ______.
