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2 years ago

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If the magnitude of the magnetic field is 2.50 mT at a distance of 12.6 cm from a long straight current carrying wire, what is t

he magnitude of the magnetic field at a distance of 19.8 cm from the wire

1 answer:

Aleksandr-060686 [28]2 years ago

7 0

Answer:

The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

Explanation:

Given;

first magnetic field at first distance, B₁ = 2.50 mT

first distance, r₁ = 12.6 cm = 0.126 m

Second magnetic field at Second distance, B₂ = ?

Second distance, r₂ = ?

Magnetic field for a straight wire is given as;

$B = \frac{\mu I}{2 \pi r}$

Where:

μ is permeability

B is magnetic field

I is current flowing in the wire

r distance to the wire

$Let \ \frac{\mu I}{2\pi} \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT$

Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

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