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Ronch [10]
3 years ago
13

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
Physics
1 answer:
Aleksandr-060686 [28]3 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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saveliy_v [14]

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6 0
4 years ago
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jonny [76]

Answer:

The electric field value is 240 N/C

Explanation:

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6 0
3 years ago
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kakasveta [241]
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8 0
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4 0
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