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VashaNatasha [74]
2 years ago
5

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
Physics
1 answer:
Stells [14]2 years ago
7 0

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}.

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<h3>What is magnetic force?</h3>

Magnetic force, attraction or repulsion that arises between electrically charged particles because of their motion.The magnitude of the magnetic force acting on the charge is given by:

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The magnitude of the charge   q=5.7\ \mu C =5.7\times 10^{-6} \ C

The velocity of the charge        v=4.5\times 10^5\ \frac{m}{s}

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