The magnetic field a distance 2 cm from a long straight current-carrying wire is 2 × 10–5 t. the current in the wire is:

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Cetacea Cetacea · Answer 1 · 2018-08-01T11:04:39+02:00

Current in the wire = 2 A

Explanation:

the magnetic field is given by

B= \frac{\mu i}{2\pi r}

μo= 4π x 10⁻⁷ Tm/A

i= current

r=0.02 m

B = magnetic field= 2 x 10⁻⁵ T

2 x 10⁻⁵= (4π x 10⁻⁷)(i) / (2π*0.02)

i=2 A

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snehashish65 snehashish65 · Answer 2 · 2022-03-22T06:47:13+01:00

The current-carrying wire itself behaves like a magnet, hence the magnitude of current flowing through the wire is 2 A.

The region around a magnet where the magnetic force of attraction or repulsion is experienced is known as the magnetic field.

Given data-

The distance from long straight wire is, r = 2 cm = 0.02 m.

The magnitude of the magnetic field is, [tex]B = 2 \times 10^{-5} \;\rm T[/tex].

The expression for the magnetic field for a current-carrying wire is,

[tex]B=\dfrac{\mu_{0} \times I}{2 \pi r}[/tex]

Here,

[tex]\mu_{0}[/tex] is the magnetic permeability of free space and I is the magnitude of the current. Solving as,

[tex]2 \times 10^{-5} = \dfrac{4 \pi \times 10^{-7} \times I}{2 \pi \times 0.02}\\\\I = \dfrac{0.02 \times 10^{-5}}{10^{-7}}\\\\I = 2 \;\rm A[/tex]

Thus, we can conclude that the magnitude of current in the wire is 2 A.

Learn more about the magnetic field here:

https://brainly.com/question/7802337