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A long, straight wire lies along the z-axis and carries a 4.00-A current in the +z-direction. Find the magnetic field (magnitude and direction) produced at the following points by a 0.500-mm segment of the wire cantered at the origin: (a) x = 2.00 m, y = 0, z = 0; (b) x = 0, y = 2.00 m, z = 0; (c) x = 2.00 m, y = 2.00 m, z = 0; (d) x = 0, y = 0, z = 2.00 m,

Short Answer

Expert verified

(a) the magnetic field (magnitude and direction) produced at the following points by a 0.500 - mm segment of the wire cantered at the origin x=2.00m,y=0,z=0is50.0pTi^

(b) the magnetic field (magnitude and direction) produced at the following points by a 0.500-mm segment of the wire cantered at the origin x=0,y=0,2.00m,z=0is50.0pTi^

(c) the magnetic field (magnitude and direction) produced at the following points by a 0.500 - mm segment of the wire cantered at the origin x=2.00m,y=0,2.00m,z=0is17.7pT-i^+j^

(d) the magnetic field (magnitude and direction) produced at the following points by a segment of the wire cantered at the originx=0,y=0,z=2.00mis0

Step by step solution

01

Direction of at point in the - x direction

Given the field point having co-ordinates (2.00,0,0)msor^=2.00mi^ and

B^=μ04πldl^×r^r2=μ04πldlr2j^=10-7T.m/A4.00A0.500×10-3m2.00m)2j^=50.0pTj^

The direction of at the point in the direction is

02

For the points (0,2.00,0)m

For the field point has the coordinates 0,2.00,0msor^=2.00mj^

Now, B^=μ04πldl^r2=-μ04πlDl^r2l^

=-10-7T.m/A4.00A0.500×10-3m2.002i^=-50.0pTi^

The direction of B^at this point is in the - x direction

03

For xy-planes

For the field point the radial position

r^=2mi^+jandr2=8.00m2B^=μ04πldl^×r^r2=10-7T.m/A4.00A0.500×10-3m8.00m2-i^+j=17.7pT-i^+j^)

The vectorB^ lies in the xy planes and makes an angle of45.0° with the -x direction.

The field point has the coordinates (0,02.00) m , hence the cross products

dl^×r^=dlk^×k^=0andthusB^=0

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