Question

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.

Answer 1

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