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A wire carries a current of 4.2 A at what distance from the wire does the magnetic field have a magnitude of 1.3×10^ -5 t

Sagot :

Onyebuchinnajiidn

Answer:

the distance is 6.46 cm.

Explanation:

Given

current in the wire, I = 4.2 A

magnitude of the magnetic field, B = 1.3 x 10⁻⁵ T

The distance from the wire is determined by using Biot-Savart Law;

[tex]B = \frac{\mu_o I}{2\pi r} \\\\r = \frac{\mu_o I}{2\pi B}[/tex]

Where;

r is the distance from the wire where the magnetic field is experienced

[tex]r = \frac{\mu_o I}{2\pi B}\\\\r = \frac{4\pi \times 10^{-7} \times 4.2 }{2\pi \times 1.3 \times 10^{-5}}\\\\r = 0.0646 \ m\\\\r = 6.46 \ cm[/tex]

Therefore, the distance is 6.46 cm.

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