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

`1.5 * 10^(-5) \mathrm{~T}`.

Step-by-step explanation:

Power carried by the line
`=P=450 \mathrm{MW}=450 * 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=(P)/(\Delta V)`

Magnetic field by the current carrying wire is given as

`B=\left((\mu)/(4 \pi)\right)\left((2 i)/(r)\right)`

Inserting the values

`B=\left(10^(-7)\right)\left((2(1500))/((20))\right) \\ B=1.5 * 10^(-5) \mathrm{~T}`

Conclusion:

Thus, the magnetic field comes out to be
`1.5 * 10^(-5) \mathrm{~T}`.