6b-b-fields

# 6b-b-fields - Magnetism Problem 1(medium Two infinite wires...

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Unformatted text preview: Magnetism Problem 1 (medium): Two infinite wires are both parallel to the y-axis, and both lie in the xy-plane. Wire 1 carries a current I 1 > 0 in the positive y-direction, and crosses the x-axis at x =- a . Wire 2 carries a current I 2 > 0 in the negative y-direction, and crosses the x-axis at x = a . 1: Find the magnetic field at a point ( x,y ) where (i) x <- a , (ii)- a < x < a , and (iii) x > a . 2: Find the force per unit length exerted on wire 2 by wire 1. 3: A point charge q moves along the x-axis, in the positive direction, at a speed v . At the instant this charge crosses the origin, (0 , 0), what magnetic force does it experience? & ¡ & ¡ & ¢ & ¡ & ¡ & ¢ ¡ £ ¡ ¢ £ Solution: Because both wires lie in the xy-plane, we know that at any point in the xy- plane the magnetic field must point in the z-direction, i.e., either + ˆ k or- ˆ k . In wire 1, located at x =- a , the current I 1 flows in the positive y-direction. Therefore, from the right-hand rule, the magnetic field due to wire 1 at a point ( x,y ) is in the + ˆ k direction if ( x,y ) is to the left of the wire ( x <- a ), and is in the- ˆ k direction if ( x,y ) is to the right of the wire ( x >- a ). The perpendicular distance to wire 1 from ( x,y ) is | x + a | . The magnetic field at ( x,y ) contributed by wire 1 is therefore 1 B 1 ( x <- a,y ) = μ I 1 2 π | x + a | ˆ k B 1 ( x >- a,y ) =- μ I 1 2 π | x + a | ˆ k In wire 2, located at x = a , the current I 2 flows in the negative y-direction. Therefore, from the right-hand rule, the magnetic field due to wire 2 at a point ( x,y ) is in the- ˆ k direction if ( x,y ) is to the left of the wire ( x < a ), and is in the + ˆ k direction if ( x,y ) is to the right of the wire ( x > a ). The perpendicular distance to wire 2 from ( x,y ) is | x- a | . The magnetic field at ( x,y ) contributed by wire 2 is therefore B 2 ( x < a,y ) =- μ I 2 2 π | x- a | ˆ k B 2 ( x > a,y ) = μ I 2 2 π | x- a | ˆ k The total magnetic field is the sum of the contributions from each wire. For x <- a , the total magnetic field at ( x,y ) is B ( x <- a,y ) = B 1 ( x <- a,y ) + B 2 ( x < a,y ) = μ I 1 2 π | x + a | ˆ k- μ I 2 2 π | x- a | ˆ k = μ 2 π I 1 | x + a |- I 2 | x- a | ¶ ˆ k For- a < x < a , the total magnetic field at ( x,y ) is B (- a < x < a,y ) = B 1 ( x >- a,y ) + B 2 ( x < a,y ) =- μ I 1 2 π | x + a | ˆ k- μ I 2 2 π | x- a | ˆ k =- μ 2 π I 1 | x + a | + I 2 | x- a | ¶ ˆ k For x > a , the total magnetic field at ( x,y ) is B ( x > a,y ) = B 1 ( x >- a,y ) + B 2 ( x > a,y ) =- μ I 1 2 π | x + a | ˆ k + μ I 2 2 π | x- a | ˆ k = μ 2 π- I 1 | x + a | + I 2 | x- a | ¶ ˆ k Notice that the magnetic field has no dependence on y , only on x ....
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6b-b-fields - Magnetism Problem 1(medium Two infinite wires...

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