A long, straight, current-carrying wire runs across the floor in a lab. At a distance of 0.250.25 meters from the wire, the magnetic field due to the wire is found to be 2.75~ \mu \text{T}2.75 μT and directed into the floor. What is the magnitude of the magnetic field at a distance of 0.750.75 meters from that same wire?

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hafsaabdulhai hafsaabdulhai · Answer 1 · 2020-03-23T03:25:46+01:00

Answer:

B= 916.7ηT or 916.7 × 10^-9 or 0.0000009167

Explanation:

Magnetic field due to a current carrying conductor is given by

B= μI/(2πr)

B: magnetic field

μ: permeability of free space

r: perpendicular distance from the wire

For r= 0.25, B = 0.00000275

0.00000275= μI/(2π×0.25)

μI/(2π) = 0.0000006875

For r= 0.75

B= μI/(2πr)

B= 0.0000006875/0.75

B= 916.7ηT or 916.7 × 10^-9 or 0.0000009167

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