Figure E28.38 shows, in cross section, several conductors that carry currents through the plane

Figure E28.38 shows, in cross section, several conductors...

Key Concepts

Ampere's Law

Ampere’s Law is a fundamental relation in magnetostatics stating that the line integral of the magnetic field B around any closed loop is equal to the permeability of free space multiplied by the net current enclosed by the loop. This law provides the basis for computing magnetic fields in systems with high symmetry and for relating current configurations to measurable magnetic fields in closed paths.

Enclosed Current

The concept of enclosed current refers to the net current that passes through the surface spanned by a closed loop. When using Ampere’s Law, one must carefully account for the contributions of each current, including their directions, since currents flowing in opposite directions can partially or entirely cancel each other. Determining the correct enclosed current is essential for accurately applying Ampere’s Law.

Right-Hand Rule and Sign Convention

The right-hand rule is used to determine the direction of the magnetic field generated by a current and to assign the proper sign to currents when calculating the net current enclosed by a loop. In the context of Ampere’s Law, the orientation of the path (e.g., counterclockwise) combined with the right-hand rule helps establish whether particular currents contribute positively or negatively to the line integral of the magnetic field.

Line Integral of the Magnetic Field

The line integral of the magnetic field along a closed path, expressed as ?B · dl, quantitatively measures the total 'circulation' of the magnetic field around that loop. It is a central quantity in Ampere’s Law and, when equated to ?0 times the enclosed current, it reflects how the configuration and geometry of current-carrying conductors influence the magnetic field in their vicinity.

Transcript

00:02 Hi, in the given problem there are three current carrying conductors which have been kept perpendicular to the plane of paper and few of them are carrying current which is coming out of the plane of paper and few others are carrying current which is getting into the plane of paper.

00:20 So first of all here this i1 here the value of this i1 is 4 .0 ampere and this cross represents that the current is getting into the plane of paper.

00:33 Nearby it another current carrying conductor carrying the current i2 and here this dot represents that the current is coming out of the plane of paper and the value of this i2 is 6 .0 amp finally the third conductor carrying the current i3 which is also coming out perpendicularly to the plane of paper and having the value i 3 is equal to 2 .0 ampere.

01:04 Now there are few loops the closed paths enclosing these current carrying conductors so first of all the path a here this is path a which is enclosing none of this current carrying conductor then there is a path b which is enclosing the current carrying conductor carrying current i1 only then there is a path c which is enclosing the two conductors carrying current i1 and i2 and finally there is a path closed path c which is enclosing sorry this is path d which is enclosing all the current carrying conductors in it now we have to find the value of line integral of magnetic field along these paths for which we use amper's circuit a law which says the line integral of magnetic field along a closed path is equal to mu not times the current threading that loo and here the value of this mu not i is positive if the integration is being made counterclockwise and it is very negative if the integration is being made clockwise or we can say if the current using right -hand screw rule if the current is getting into the plane of paper then using right -hand screw rule the integration will be made clockwise integration will be clockwise and this b dot dl will come out to be negative while if the current is coming out then the integration will be made counterclockwise and that integration will be taken as positive so starting for loop a as it is carrying none of the current means the current passing through it is zero hence the line integral of magnetic field b dot dl along a will come out to be mu note times the current which is zero hence it will come out to be 0 then for loop b there is only one current passing through this loop b which is i 1 and which is getting into the plane of paper so it should come out to be negative and hence this b dot d l is equal to mu not times minus you not times the current passing through it means this is i1 and i1 is having a value of 4 .0amper.

04:45 So using the value of mu not which is 4 pi into 10 dash to the power minus 7 tesla meter per ampere multiplied by 4 amp.

04:58 Finally this line integral of magnetic field along the closed loop b gets a value of minus 5 .02 into 10 dash per minus 6 tesla into meter which is the answer the second answer for this given problem then for loop c this time there are two currents and closed in this loop c so this b dot d l along the closed loop c this is mu not times i 3 sorry i 2 which will be positive as the current is coming out and minus mu not i1 as the current is getting in.

05:48 So taking this mu not as a common out and putting its value, which is 4 pi means 4 into 3 .14 into 10 dash per minus 7, tesla meter per ampere multiplied by i 2 minus i 1 means this is 6 ampere minus 4 ampere, canceling this ampier...

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