Two long, straight wires cross each other perpendicularly as...

Two long, straight wires cross each other perpendicularly as shown in Figure P30.61. The wires do not touch. Find (a) the magnitude and (b) the direction of the magnetic field at point $P,$ which is in the same plane as the two wires. (c) Find the magnetic field at a point 30.0 $\mathrm{cm}$ above the point of intersection of the wires along the $z$ axis; that is, 30.0 $\mathrm{cm}$ out of the page, toward you.

Key Concepts

Magnetic Field of a Long Straight Wire

The magnetic field produced by a long, straight, current-carrying wire is given by the formula B = (??I) / (2?r), where ?? is the permeability of free space, I is the current, and r is the perpendicular distance from the wire. This concept originates from Ampère’s law and is fundamental when analyzing the magnetic effect of linear current distributions.

Right-Hand Rule

The right-hand rule is a mnemonic used to determine the direction of the magnetic field around a current-carrying wire. By pointing the thumb of the right hand in the direction of the current, the way the fingers curl gives the direction of the magnetic field lines. This rule is crucial for correctly assigning the direction to the magnetic field vectors.

Superposition Principle

The superposition principle states that when multiple sources produce magnetic fields, the net magnetic field at any point is the vector sum of the individual fields. This principle allows one to calculate complex magnetic field configurations by separately evaluating the contribution from each source.

Vector Addition of Magnetic Fields

Because magnetic fields are vector quantities with both magnitude and direction, combining them requires vector addition. This involves adding the components of each individual field to arrive at the overall magnetic field at a given point, taking into account directional angles provided by rules such as the right-hand rule.

Transcript

00:04 This problem we have to determine the magnetic field at distance 30 cm.

00:10 So b1 at point p will be equals to mu not i divided by 2 pi r that is for the from the first wire and r.

00:23 Okay so now substituting the values mu not is equals to 4 pi multiplied by 10 to the power minus 7 and i1 is equals to 3 ampere and 2 pi and r1 is 30 centimeter okay 30 centimeter that is 30 multiplied 10 to the 4 minus 2 meters so we get b1 p equals to 2 multiplier by 10 to the power minus 6 tesla okay so similarly magnitude of magnetic field due to wire 2 b2 p will be equals to mu not i 2 divide by 2 pi r 2 okay so substituting values mu not is equal to 4 pi multiplied by 10 to the power minus 7 i 2 is 5 amp and 2 pi multiplied by r2 is 50 centimeters so 50 multiplied by 10 to the power minus 2 so from here b2 p comes out to be 2 .5 multiplied by 10 to the 4 minus 6 tesla now if we talk about the direction the b1p will be in the minus k cap direction from the right -hand rule and this will be in the k direction so net magnetic field at point bp will be equal to b1p vector plus b2p vector okay so we get 2 multiplied by 10 to the power minus 6 minus k cap plus 2 .5 multiplied by 10 to the power minus 6 k cap okay so from here we get bp equals to 0 .5 micro tesla okay this is the magnitude of magnetic field for the part a.

02:06 Now in the part b we have to determine the direction.

02:09 So for the part b, the direction will be in the k cap, okay? so the direction for the bp vector will be equals to we can write it as 0 .5 micro tesla k cap.

02:22 So this will be the answer for the part b.

02:25 Or in other words, it can be written as the direction of magnetic field is out of the the plane of paper out of the paper.

02:38 This can also be given as our answer.

02:42 So now moving to the part c in which we have to determine the magnitude of magnetic field at 30 centimeter towards us.

02:51 So magnetic field due to i1 we get b1 equals to mu not i1 divided by 2 pi r...

Please give Ace some feedback