Answer: magnitude of the magnetic field at a distance of 19.4 cm from the wire=4.29mT
Explanation:
According to Biot-Savart law, A magnetic field generated by a current carrying wire at a distance is represented as
B=μ₀I/ 2πr
B = magnetic field intensity 1000 mT =1T, 6.50mT = 6.50 X 10^-3T
μ₀ =permeability of free space 4π × 10−7 H/m
I = current intensity
r = radius, 100cm = 1m, 12.8 cm= 12.8 x 10^-2m
6.50 X 10^-3 = μ₀ x I/ 2 π X 12.8 X 10^-2
I =6.50 X 10 ^-3 X 2π X X 12.8 X 10^-2/ 4π × 10−7 H/m
I= 4160 A
when the magnetic field is at 19.4 cm from the wire
B=μ₀I/ 2πr
= 4π × 10−7 H/m x4160/ 2π x 19.4 x 10^-2
=0.004288
= 4.29x 10 ^-3T
= 4.29mT
