Question

If the magnitude of the magnetic field is 2.50 mT at a distance of 12.6 cm from a long straight current carrying wire, what is the magnitude of the magnetic field at a distance of 19.8 cm from the wire

Answer 1

The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

Step-by-step 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;

`B = (\mu I)/(2 \pi r)`

Where:

μ is permeability

B is magnetic field

I is current flowing in the wire

r distance to the wire

`Let \ (\mu I)/(2\pi) \ be \ constant; = K\\\\B = (K)/(r) \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =(B_1r_1)/(r_2) \\\\B_2 = (2.5*10^(-3) *0.126)/(0.198) \\\\B_2 = 1.591 *10^(-3)\ T\\\\B_2 = 1.591 \ mT`

Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT