$A$ 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?
(A) Current in the power line,$I = 90 \;A$.
Point is located below the power line at a distance,$r = 1.5 \;m$.
The magnetic field $B$ due to a long straight current-carrying wire is given by the formula:
$B = \frac{\mu_0 I}{2 \pi r}$
Substituting the values:
$B = \frac{4 \pi \times 10^{-7} \times 90}{2 \pi \times 1.5}$
$B = \frac{2 \times 10^{-7} \times 90}{1.5} = 1.2 \times 10^{-5} \;T$.
The current is flowing from East to West. According to the Right-Hand Thumb Rule,if you point your thumb in the direction of the current (West),your fingers curl around the wire. Below the wire,the fingers point towards the South. Thus,the direction of the magnetic field is towards the South.
Point is located below the power line at a distance,$r = 1.5 \;m$.
The magnetic field $B$ due to a long straight current-carrying wire is given by the formula:
$B = \frac{\mu_0 I}{2 \pi r}$
Substituting the values:
$B = \frac{4 \pi \times 10^{-7} \times 90}{2 \pi \times 1.5}$
$B = \frac{2 \times 10^{-7} \times 90}{1.5} = 1.2 \times 10^{-5} \;T$.
The current is flowing from East to West. According to the Right-Hand Thumb Rule,if you point your thumb in the direction of the current (West),your fingers curl around the wire. Below the wire,the fingers point towards the South. Thus,the direction of the magnetic field is towards the South.
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