Two very long straight conducting wires are perpendicular to...

Two very long straight conducting wires are perpendicular to the x-y plane. At point P, the magnetic field is the sum of B1 and B2, due respectively to the currents in wire 1 and wire 2, with |B1| = 1.442x10^-4 T and |B2| has a magnitude of 8.010x10^-5 T. Both are illustrated by arrows in the figure below. y (cm) 10 9 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 x (cm) wire1 wire2 p Indicate the direction of the currents in the wires, +z (out of the page) or -z (into the page). In wire 2, the current is in the ... direction. In wire 1, the current is in the ... direction. Calculate the magnitude of the current in wire1. Calculate the magnitude of the total B-field at P.

Transcript

00:01 Hi, here in this given problem this is the position of wire 1 and approximately here this is wire 2.

00:14 This is the observation point, common observation point.

00:20 If we join the observation point from the cross -sectional views of the two wires, then here this is one of the magnetic field given so obviously it will be due to wire 1 so let it be b1 and another magnetic field this is the smaller one this is b2 here this distance r1 this distance r2 the magnitude of b1 this is 1 .442 into 10 -dh to the power minus 4 tesla and the magnitude of b2 this is 8 .010 into 10 to the power minus 5 tesla so b2 is less than b1 hence we conclude that this arrow the smaller arrow the shorter arrow this should be for b2 okay now there are two parts in this given problem in the first part of the problem we have to find the directions of current in these two wires which are parallel to each other and parallel perpendicular to xy plane.

01:41 So using right hand thumb rule, direction of current i1 in wire 1, that should be a long, positive z -axis means outward and that of i 2 in wire 2 this is a long negative x axis means inward so these two are the answers for the first part of the problem then in the second part of the problem we have to find the the magnitudes of these currents i1 and i2 so here this is i1 outward and i2 here that is inward along negative x xxis now with the help of graph counting the blocks we conclude r1 is a hypotenuse of a right angle to triangle whose base is equal to five centimeters and perpendicular that is equal to 6 cm so that is the expression for the hypotenuse means this is square root of 61 centimeter or we can say r1 is equal to 7 .8 into 10 dash to the power minus 2 meter similarly r2 is the hypotenuse of a right angle to triangle whose base that is three three square and high perpendicular its perpendicular will be one two three four five six seven that is seven square of seven centimeters so this is nine plus means it comes out to be equal to square root of 58 cm means this is 7 .6 centimeter or 7 .6 into 10 reach to power minus 2 meters.

04:48 So using an expression for this magnetic field due to a straight wire of infinite length that is mu not upon 4x using by 10th and severed sloth that is given as mu not 5 pi into 2 i by r we get an expression for the current and that is i is equal to b into r divided by 2 times of nu not upon 4 pi so i 1 will be given by for b this is 1 .442 into 10 dash to the power minus 4 into 7 .8 into 10 dash to the power minus 2 for the value of r1, then 2 into 10 to the power minus 7...

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