Question

A small current element at the origin has a length of 3.8 mm and carries a current of 1.8 A in the +z direction. Find the magnitude and direction of the magnetic field due to this current element at the point (0, 3.8 m, 4.6 m). B = * ptî + 0 pt î + 0 рт k

Answer 1

Given

l = 1.8 amp

L = 3.8 mm = 3.8 x 10^-3​​​​​​ m

r = (0, 3.8, 4.6 m)

We can use Biot - Savart law

B = μ/4π x (lLsin(θ)/r²)

where sin(θ ) is angle between position vector and current element

For r = 0 i
o re 01 for of [: kxi=j) = 4 x I Lsing = Bay 1.810 x(3.8x 103 m xoi)= | By = oj pt 1 4h oi 1 for r= 318mi [i fxj=-] B = M . Il sind ZL sind - lox 1.8 *(3.8x7ox 318}) 3.8 vöxlexióx14.4 i = .91 -107 = 25.92 X10 xido; =-0.259x103 T Box = -0.2597pT for r= 4.6 m [:: RX&=0] B, - 2 4K Why It sina = 10X1.8X(318x13 kx4.CÊ ) = ORT 4.6 By = GFpt

В = -0, 257 рті + б t орт е
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