A horizontal overhead power line carries a current of 90 A in the east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Hint: we are not given with the length of the current-carrying wire. The current direction is mentioned. We need to find the magnitude and direction of the magnetic field due to the current 1.5 m below the line. We can use the formula for the magnetic field due to an infinite current-carrying conductor.

Complete step by step answer:We know the magnetic field due to an infinitely long current-carrying wire at some perpendicular distance from it is given by \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2I}{r}\]
Here r is the perpendicular distance from the wire and I is the current in the wire
I= 90A
r= 1.5m
substituting the values, we get,
\[B=\frac{{{\mu }_{0}}}{4\pi }\frac{2I}{r}={{10}^{-7}}\times \frac{2\times 90}{1.5}\]
B= \[1.2\times {{10}^{-5}}T\]
Now we have to found the magnitude of the field and now we need the direction. Current flows from east to west, the point is below the wire. Using the right-hand thumb rule, the direction of B is southward.

Additional Information: Right-hand rule states that, if the thumb of the right hand is in the direction of the current flow then, the curl fingers show the direction of the magnetic field.

Note:we need to remember that the magnetic field is a vector quantity and we have to use right-hand thumb rule carefully while determining the direction of the field. Had it been the direction of the current opposite, the direction of B would have been changed to north direction.