Two insulated wires perpendicular to each other in the same plane carry currents as shown in Fig

step 1 Here we have two perpendicular wires which are in the same plan. Let us consider the vertical wi

Two insulated wires perpendicular to each other in the same...

Key Concepts

Magnetic Field of a Current-Carrying Wire

This concept involves understanding that a current flowing through a long, straight wire produces a magnetic field that circulates around the wire. The magnitude of this field at a distance r from the wire is given by the expression B = (??I)/(2?r), where ?? is the permeability of free space and I is the current. The field lines form concentric circles centered on the wire, and their strength decreases with increasing distance from the wire.

Right-Hand Rule

The right-hand rule is a mnemonic used to determine the direction of the magnetic field around a current-carrying conductor. When you point the thumb of your right hand in the direction of the conventional current, the curl of your fingers indicates the direction of the magnetic field lines. This rule is essential for correctly assigning vector directions to the fields produced by individual currents.

Superposition Principle

The superposition principle states that the net magnetic field at any point due to multiple sources is the vector sum of the magnetic fields produced by each individual source. This means that when calculating the overall magnetic field from two or more current-carrying wires, the contributions from each wire must be summed taking into account both their magnitudes and directions.

Transcript

00:02 Here we have two perpendicular wires which are in the same plan.

00:14 Let us consider the vertical wire as wire 1 and the wire which is horizontal taken as wire as point p and point q are at equal distance from wire 1.

00:46 So we have magnitude of magnetic field at point p and point q due to wire 1 is equal to mu not times current through wire 1 divided by 2 pi into distance of point p and point q from the wire that is r1 now we can put different values here mu not is 4 pi times 10 to the power minus 7 tesla meter per ampere and current through wire 1 is 12 ampere divided by 2 pi distance of point p and point q from wire 1 is 15 centimeter that is 15 times 10 to the power minus 2 meter this gives us the magnitude of magnetic field at point p and point q due to wire 1 equal to 1 .6 times 10 to the power minus 5 tesla similarly we can write the magnitude of magnetic field at point p and point q due to wire 2 as b2 equal to mu not into i divided by 2 pi r2 again putting the values here mu not is 4 pi times 10 to the power minus 7 tesla meter per ampere into current through second wire is 10 amp divided by 2 pi into distance of point p and point q from second wire is ant centimeter that is add times 10 to the power minus 2 meter this gives us b2 equal to 2 .5 times 10 to the power minus 5 tesla in part a of this problem we are considering the direction of current in wire 2 is from left to rise now we want to find the net magnetic field at point p and point q due to these both wires.

04:04 First we will take the point p at point p we can find the direction of magnetic field produced by wire 1 and wire 2 by using right -hand rule.

04:31 So by right -hand it is clear that the direction of magnetic field at point p due to wire 1 that is b1 is into the page also due to second wire magnetic field is also into the page so we will have the net magnetic field at point p equal to b1 plus b2 as b1 is equal to 1 .6 times 10 to the power minus 5 tesla plus b2 is 2 .5 times 10 to the power minus 5 tesla this gives us net magnetic field at point p equal to 4 .1 times 10 to the power minus 5 tesla and this is the magnitude of net magnetic field and here direction of net magnetic field is into the page.

06:06 Similarly we can find the magnetic field at point q.

06:23 The wire 1 produces magnetic field at point q that is out of page also the wire 2 produces magnetic field which is out of patch so we have net magnetic field at point q and that is also out of the page which will be equal to b1 plus b2 as we have b1 equal to 1 .6 times 10 to the power minus 5 tesla and we have b2 equal to 2 .5 times 10 to the power minus 5 tesla this gives us the net magnetic field at point q equal to 4 .1 times 10 to the power minus 5 tesla and here that direction of field is out of the page...

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