Answer:
The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
Explanation:
Given;
first magnetic field at first distance, B₁ = 2.50 mT
first distance, r₁ = 12.6 cm = 0.126 m
Second magnetic field at Second distance, B₂ = ?
Second distance, r₂ = ?
Magnetic field for a straight wire is given as;
[tex]B = \frac{\mu I}{2 \pi r}[/tex]
Where:
μ is permeability
B is magnetic field
I is current flowing in the wire
r distance to the wire
[tex]Let \ \frac{\mu I}{2\pi} \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT[/tex]
Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT
