Two long wires lie in an x y plane, and each carries a current in the positive direction of the

step 1 In the given problem, first of all, there are x -axis and y -axis. Then there are two parallel c

Two long wires lie in an x y plane, and each carries a...

Key Concepts

Vector Representation and Unit Vectors

Utilizing vector notation and unit vectors to express magnetic fields provides a clear and precise description of both the magnitude and the direction of the field. This method is essential for combining fields in multiple dimensions and for establishing conditions, such as where the net magnetic field may cancel out.

Magnetic Field from a Long Straight Wire

This concept involves determining the magnetic field generated by a long, current-carrying conductor. The magnetic field at a distance r from the wire is given by B = (??I)/(2?r), where ?? is the permeability of free space and I is the current. This expression is derived using Ampère's Law and is fundamental for analyzing magnetic field distributions in systems involving current-carrying wires.

Right-Hand Rule

The right-hand rule is a method used to determine the direction of the magnetic field produced by a current. By pointing the thumb in the direction of the current, the curled fingers indicate the direction of the magnetic field loops. This rule is essential for correctly assigning the direction to the magnetic field vectors in problems that involve current flow.

Superposition Principle

The superposition principle states that the net magnetic field from multiple sources is the vector sum of the individual magnetic fields produced by each source. This principle is crucial when analyzing systems with more than one current-carrying wire, allowing for the calculation of the overall magnetic field by summing the contributions from each wire.

Magnetic Field Cancellation

Magnetic field cancellation occurs when the magnetic fields from different sources counteract each other, resulting in a net field of zero. By equating the vector sum of the fields to zero, one can solve for the locations or conditions under which the individual contributions cancel, a concept that is particularly useful in the analysis of magnetic field configurations created by multiple currents.

Transcript

00:01 In the given problem, first of all, there are x -axis and y -axis.

00:16 Then there are two parallel current carrying conductors which are lying in x -y plane and parallel to the x -axis.

00:29 The first one is at a distance of 10 cm.

00:38 The origin this is wire number one carrying a current i1 is equal to 6 .00 ampere then another one that is also parallel to y x xx sorry and this is named as two it is carrying a current i2 is equal to 10 .0.

01:09 Ampere first of all the gap of the second conductor from x -axis this is 5 .00 centimeter and that of the first conductor this is given as 10 .0 centimeter.

01:30 Now in the first part of the problem if we have to find net magnetic field at the origin so for origin the distance of origin o from the first conductor that will be 10 .0 centimeter and that of the second conductor will be 5 .00 centimeter and as the current are towards right for both the conductors hence using right right hand thumb rule the directions of magnetic fields at the origin due to both the wires will be same and it will be into the plane of paper.

03:03 Means it will be along negative x x x so the total magnetic field will be given by b1 plus b2 hence using the expression for the magnetic field this is mu not by 4 pi into 2 i1 by r1 plus mu not by 4 pi into 2 i2 by r2 taking this mu not by 4 pi 5 pi 2 as a common out leaving behind i1 by r1 plus i2 by r2 now plugging in all known values this is 10 dash 1 minus 7 tesla meter per ampere multiplied by 2 and for i1 this is 6 ampier divided by r 1 which is 10 centimeter or 10 into 10 x per minus 2 plus for i 2 this is 10 ampere and for r 2 this is 5 centimeter or 5 into 10 dash 4 minus 2 meter finally here it becomes 2 into 10 dash par minus 5 if we take this 10 dash 4 minus 2 as a common out in denominator then it will become 10 dash far plus 2 in numerator so finally this is 2 into 10 dash the power minus 5 and in the bracket 10 will be the lcm then this is 6 plus 20 tesla or finally we can say this is 26 into 2 means this is 52 into 10 rich per minus 6 tesla or in vector form as it is along negative x x x so this is b bar is equal to minus 5 .2 into 10 dash bar minus 5 k kab which becomes the answer for the first part of this problem in the second part of the problem here we have to find a point at which the magnet field will be 0 and as the current in both of the conductors are in the same direction so magnetic field will be 0 definitely it will be 0 at any point between them so b will be 0 at any point between the wires so suppose it is at a distance of y from the wire 2 as the the net magnetic field is 0 hence the magnitude of the two magnetic fields will be same if this point is p then b at p due to 1 should be equal to b at p due to 2 so for b at p due to 1 the magnitude will be mu not upon 4 pi into 2 i 1 by 10 minus y here this distance in the figure this is is supposed to be y from the origin.

07:05 So it's a distance from the first wire will remain n minus y.

07:12 So here it is 10 minus y.

07:14 And this distance is y is from not from wire 2, this is from origin.

07:21 Now and this equals to b at p due to 2 which is mu not upon 4 pi into 2 i 2 by y minus 5 canceling this canceling 2 and putting the values of i1 and i 2 finally we get this y to be equal to 8 .1 25 centimeter which becomes the answer for the second part of the problem the point the distance of point at which the magnetic field is zero when the current in the both of the wires are in the same direction now on reversing the direction of i1 now magnetic field will be 0 at some point above wire 1...

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