A circular coil of radius 5 × 10-2 m and with 40 turns is carrying a current of 0.25 A. Determine the magnetic field of the circular coil at the center.
Question
A circular coil of radius 5 × 10-2 m and with 40 turns is carrying a current of 0.25 A. Determine the magnetic field of the circular coil at the center.
Solution
The magnetic field at the center of a circular coil can be calculated using Ampere's Circuital Law. The formula is given by:
B = μ0 * I * N / (2 * r)
where:
- B is the magnetic field
- μ0 is the permeability of free space, which is a constant equal to 4π × 10^-7 T m/A
- I is the current flowing through the coil
- N is the number of turns in the coil
- r is the radius of the coil
Substituting the given values into the formula:
B = 4π × 10^-7 T m/A * 0.25 A * 40 / (2 * 5 × 10^-2 m)
Solving the above expression will give the magnetic field at the center of the coil.
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