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Zina [86]
3 years ago
15

A long, straight wire lies along the z-axis and carries a 4.20-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 centered at the origin.
Physics
1 answer:
lana [24]3 years ago
8 0

Complete question:

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 centered 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.

Answer:

a) 5.25*10^-1^1 T;

b) 5.25*10^-^1^1 T;

c) 1.86*10^-^1^1 T;

d) 0

Explanation:

Given

I = 4.20A

P = 0.500mm

a) r= (2.00m) i

∆T*r = (0.500*10^-^3)(2.00)= 1.00*10^-^3m^2;

B = [(1.00*10^-^7 T.m/A)(4.20A)(1*10^-^3m^2)] / (2.00)^3;

(B= 5.25*10^-^1^1T) j

b) r = (2.00m) j

∆T*r = (0.500*10^-^3)(2.00)= 1.00*10^-^3m^2;

B = [(1.00*10^-^7 T.m/A)(4.20A)(1*10^-^3m^2)] / (2.00)^3;

(B= 5.25*10^-^1^1T) i

c) r = (2.00m) (i+j)

(i/j) r = \sqrt{2} (2.00m);

∆T*r = (0.500*10^-^3)(2.00)= 1.00*10^-^3m^2;

B = [(1.00*10^-^7 T.m/A)(4.20A)(1*10^-^3m^2)] / (\sqrt{2})(2.00)^3;

(B= 1.86*10^-^1^1T)(i-j)

d) r = (2.00m) k

∆T*r = (0.500*10^-^3)(2.00) k*k = 0;

B = 0

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