Question

The radius of a circular wire is 0.25 m and the current in wire is 5 A. What is the magnitude of
magnetic field at the centre of circular wire?​

Answer 1

Concept: Magnetic field is defined as the area around a magnet or a current carrying object under which magnetic force can be felt.

When a circular wire is carrying current then it applies a magnetic field at the centre with a magnitude equal to u₀ I / 2R

Given:

Radius of circular wire= 0.25m

Current flowing in wire = 5 A

To find:

Magnetic field at the centre of the circular wire

Solution:

Using Biot-savart's law, the magnetic field at the centre of a circular wire B = u₀ I/ 2 R, where u₀ = magnetic permeability

So,

B = 4π × 10^-⁷ × 5/ 2×0.25

B = 20 × 3.14 × 10^-⁷ / 0.5

B = 62.8 × 10^-⁷ / 0.5

B = 125.6 × 10^-⁷

B = 1.256 × 10^-⁵ T

Hence, the magnetic field at the centre of a circular wire is 1.256 × 10^-⁵ T

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Answer 2

Explanation:

The radius of circular wire is 0.25m and the current is 5A , find the magnitude of magnetic field at the centre of the circular wire .

Answer :

The magnitude of magnetic field at the centre of the circular wire is

$12.57$ × $10^{-6}T$

Given That :

Radius of the circular wire = 0.25m

Current wire = 5 A

To find :

Magnitude of magnetic field at the centre of the circular wire

Solution :

We know that ,

Magnetic field at the centre of a circular wire

By using Biot-savart's law,

The magnetic field at the centre of a circular wire B = µo I / 2 R, where µo is the magnetic permeability = 4π × $10^{-7}$  T m/A.

Hence ,

B = (4π × $10^{-7}$ × 5 ) / (2 × 0.25)

B = (4π × $10^{-7}$ × 5 ) / 0.5

B = (4π ×$10^{-7}$ × 5 × 10) / 5

B = (4π × $10^{-7}$ × 10)

B = 4π ×$10^{-6}$

B = $12.57$ × $10^{-6}$

Hence, the magnitude of magnetic field at the centre of the circular wire is $12.57$ × $10^{-6}T$

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