Question : A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?
Doubt by Deepanshi U.
Solution :
We know,
Magnetic Field due to a straight current carrying conductor is given by
B = μI₀/2πr
Here
I = 90 A [Flowing from East to West]
r = 1.5 m [Below]
B = μI₀/2πr
B = 2μI₀/4πr
B = (2 × 90 × 10-7)/1.5
B = 120 × 10-7
B = 1.2 × 10-5 T
By Applying the Fleming's Left Hand Rule, we can easily find that the direction of magnetic field will be in the south direction.
