Consider a tightly wound 100 turn coil of radius 10 cm carrying a current of 1 A. What is the magnitude of the magnetic field at the center of the coil?

The equation for the magnetic field is given by:

[tex] B = \frac{\mu_0 n i}{2 a} [/tex]

where [tex] \mu_0 [/tex] is the permeability of free space, [tex] n [/tex] is the number of turns, [tex] i [/tex] is the current, and [tex] a [/tex] is the radius of the coil.

The magnitude of the magnetic field at the center of a tightly wound coil with 100 turns and a radius of 10 cm carrying a current of 1 A is approximately 6.28 × 10^{-4} T. This value is calculated using the formula for the magnetic field in a coil. The calculation involves substituting the known values into the provided formula.
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Answered by Anonymous | 2025-07-04

To find the magnitude of the magnetic field at the center of the coil, we need to use the given formula:
B = 2 a μ 0 ​ ni ​
Let's break down the components of the formula:

μ 0 ​ : This is the permeability of free space, which is a constant value: [ \mu_0 = 4\pi \times 10^{-7} \text{ T}

n : The number of turns in the coil. In this case, it is 100 turns.

i : The current flowing through the coil, which is 1 A.

a : The radius of the coil, which is given as 10 cm. We need to convert this into meters for consistency in units: a = 0.10 m


Now, plug these values into the formula:
B = 2 ⋅ 0.10 m 4 π × 1 0 − 7 T ⋅ 100 ⋅ 1 A ​
Let's calculate:
B = 0.20 4 π × 1 0 − 7 × 100 ​ B = 0.20 4 π × 1 0 − 5 ​ B = 2 π × 1 0 − 5 T
Therefore, the magnitude of the magnetic field at the center of the coil is approximately:
B ≈ 6.28 × 1 0 − 5 Tesla

Answered by danjohnbrain | 2025-07-06