English
Karnataka Board PUCPUC Science Class 11

In Ampere's ∮ → B ⋅ d → l = μ 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

Advertisements
Advertisements

Question

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?  

Short/Brief Note
Advertisements

Solution

In Ampere's law \[\oint \vec{B} . \vec{dl} = \mu_o i\] , i is the total current crossing the area bounded by the closed curve. The magnetic field B on the left-hand side is the resultant field due to all existing currents.   

shaalaa.com
  Is there an error in this question or solution?
Chapter 35: Magnetic Field due to a Current - Short Answers [Page 248]

APPEARS IN

HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 35 Magnetic Field due to a Current
Short Answers | Q 4 | Page 248

RELATED QUESTIONS

State Ampere’s circuital law.


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


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 ?


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.


Consider the situation of the previous problem. A particle having charge q and mass mis projected from the point Q in a direction going into the plane of the diagram. It is found to describe a circle of radius r between the two plates. Find the speed of the charged particle.


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).


State Ampere’s circuital law.


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


Ampere’s circuital law is equivalent to ______.

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 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 ______.

  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.

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.


Which law provides a simpler alternative method for finding the magnetic field \[B\] when the current distribution has symmetry (straight wire, solenoid, toroid)?


Substituting \[B=\frac{\mu_0 I}{2\pi r}\] into \[\oint\vec{B}\cdot d\vec{l}=B(2\pi r)\] for a straight wire yields:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×