Question

Calculate the size of the magnetic field 20 m below a high voltage power line. The line carries 450 MW at a voltage of 300,000 V. Group of answer choices 0.237 T 0.0237 T 0.474 T 2.37 T 0.237 J

Answer 1

Answer:

$1.5 \times 10^{-5} \mathrm{~T}$.

Explanation:

Power carried by the line $=P=450 \mathrm{MW}=450 \times 10^{6} \mathrm{~W}$

Voltage across the line Volts

Current flowing in the line =i

Size of magnetic field =B

Distance from the line

Formula Used:

Current flowing is given as

$i=\frac{P}{\Delta V}$

Magnetic field by the current carrying wire is given as

$B=\left(\frac{\mu}{4 \pi}\right)\left(\frac{2 i}{r}\right)$

Inserting the values

 $B=\left(10^{-7}\right)\left(\frac{2(1500)}{(20)}\right) \\ B=1.5 \times 10^{-5} \mathrm{~T}$

Conclusion:

Thus, the magnetic field comes out to be $1.5 \times 10^{-5} \mathrm{~T}$.