Question

A short current element d l ⃗ = (0.500 mm ) j ^ carries a current of 4.70 A in the same direction as d l ⃗ . Point P is located at r ⃗ = ( -0.730 m ) i ^ + (0.390 m ) k ^ . Find the magnetic field at P produced by this current element in T.

Answer 1

Magnetic field at a point P is given by:

We know that

= 4*pi*10^-7

I = 4.70 Amp

dl = current element = (0.500 mm) j

r = (-0.730 m) i + (0.390 m) k

Which gives us

|r| = sqrt ((-0.730)^2 + (0.390)^2)

|r| = 0.828 m

dlxr = (0.500 mm) j x ((-0.730 m) i + (0.390 m) k)

dlxr = -0.500*10^-3*0.730 jxi + 0.500*10^-3*0.390 jxk

jxi = -k & jxk = i

dlxr = -3.65*10^-4 (-k) + 1.95*10^-4 (i)

dlxr = 1.95*10^-4 i + 3.65*10^-4 k

Using above values:

B = 4*pi*10^-7*4.70*(1.95*10^-4 i + 3.65*10^-4 k)/(4*pi*0.828^3)

B = 4*pi*10^-7*4.70*1.95*10^-4/(4*pi*0.828^3) i + 4*pi*10^-7*4.70*3.65*10^-4/(4*pi*0.828^3) k

B = 1.614*10^-10 i + 3.022*10^-10 k

Which gives us

dBx, dBy, dBz = (1.614*10^-10, 0, 3.022*10^-10) T

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