हिंदी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

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 ______. - Physics

Advertisements
Advertisements

प्रश्न

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 ______.

  1. `oint B.dl = +- 2μ_0I`
  2. the value of `oint B.dl` is independent of sense of C.
  3. there may be a point on C where B and dl are perpendicular.
  4. B vanishes everywhere on C.

विकल्प

  • a and b

  • a and c

  • b and c

  • c and d

MCQ
रिक्त स्थान भरें
Advertisements

उत्तर

b and c

Explanation:

Ampere’s law gives another method to calculate the magnetic field due to a given current distribution.


Line integral of the magnetic field `vecB` around any closed curve is equal to μ0 times the net current i threading through the area enclosed by the curve, i.e.

`oint vecB * vec(dl) = mu_0 sumi = mu_0 (i_1 + i_3 - i_2)`

Total current crossing the above area is `(i_1 + i_3 - i_2)`. Any current outside the area is not included in net current. (Outward ⊙ → + ve, Inward ⊗ → – ve)

Applying the Ampere's circuital law, we have

`ointB* dl = i_0 (I - I) = 0` (because current is in opposite sense)

Also, there may be a point on C where B and dl are perpendicular and hence, `oint_c B*dl = 0`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Moving Charges And Magnetism - MCQ I [पृष्ठ २४]

APPEARS IN

एनसीईआरटी एक्झांप्लर Physics [English] Class 12
अध्याय 4 Moving Charges And Magnetism
MCQ I | Q 4.09 | पृष्ठ २४

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

Write Maxwell's generalization of Ampere's circuital law. Show that in the process of charging a capacitor, the current produced within the plates of the capacitor is `I=varepsilon_0 (dphi_E)/dt,`where ΦE is the electric flux produced during charging of the capacitor plates.


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.


Obtain an expression for magnetic induction along the axis of the toroid.


Explain Ampere’s circuital law.


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.


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 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. 


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. 


What is magnetic permeability?


State Ampere’s circuital law.


Find the magnetic field due to a long straight conductor using Ampere’s circuital law.


Calculate the magnetic field inside and outside of the long solenoid using Ampere’s circuital law


Ampere’s circuital law states that ______.

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:


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:


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.

  1. Explain the reason for the jumping of the ring when the switch is closed in the circuit.
  2. What will happen if the terminals of the battery are reversed and the switch is closed? Explain.
  3. Explain the two laws that help us understand this phenomenon.

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.

 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×