मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान इयत्ता ११

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 - Physics

Advertisements
Advertisements

प्रश्न

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?  

टीपा लिहा
Advertisements

उत्तर

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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Magnetic Field due to a Current - Short Answers [पृष्ठ २४८]

APPEARS IN

एचसी वर्मा Concepts of Physics Vol. 2 [English] Class 11 and 12
पाठ 13 Magnetic Field due to a Current
Short Answers | Q 4 | पृष्ठ २४८

संबंधित प्रश्‍न

State Ampere’s circuital law.


State 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.


Consider the situation described in the previous problem. Suppose the current i enters the loop at the points A and leaves it at the point B. Find the magnetic field at the centre of the loop. 


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. 


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).


Define ampere.


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


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 ______.


Which of the following is the correct definition of ampere?

Ampere’s circuital law is given by _______.


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:


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 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 ______.


Briefly explain various ways to increase the strength of the magnetic field produced by a given solenoid.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×