find the magnitude of magnetic field at the centre of one quarter of a circular coil of radius 4 cm if current of 9A flows through it

To find the magnitude of the magnetic field at the center of one quarter of a circular coil, we can use the Biot-Savart Law, which describes the magnetic field generated by a current-carrying conductor. In this case, we have a quarter of a circular coil with a radius of 4 cm and a current of 9 A flowing through it.

Understanding the Magnetic Field from a Circular Coil

The magnetic field at the center of a full circular coil can be calculated using the formula:

• B = (μ₀ * I) / (4 * π * R)

Where:

• B = magnetic field strength
• μ₀ = permeability of free space (approximately 4π × 10^-7 T·m/A)
• I = current in amperes
• R = radius of the coil in meters

Calculating for One Quarter of the Coil

Since we only have a quarter of the circular coil, the magnetic field at the center will be one-fourth of the magnetic field produced by the entire coil. First, we need to convert the radius from centimeters to meters:

• Radius, R = 4 cm = 0.04 m

Now, substituting the values into the formula for the full circular coil:

• B (full coil) = (4π × 10^-7 T·m/A * 9 A) / (4 * π * 0.04 m)

Let's simplify this step by step:

• B (full coil) = (9 × 10^-7 T·m/A) / (0.16 m)
• B (full coil) = 5.625 × 10^-6 T

Now, since we only need the magnetic field for one quarter of the coil:

• B (quarter coil) = (1/4) * B (full coil)
• B (quarter coil) = (1/4) * 5.625 × 10^-6 T
• B (quarter coil) = 1.40625 × 10^-6 T

Final Result

The magnitude of the magnetic field at the center of one quarter of the circular coil is approximately:

• B ≈ 1.41 μT

This value indicates the strength of the magnetic field generated by the current flowing through the quarter coil. Understanding how the magnetic field behaves in different configurations is crucial in fields like electromagnetism and electrical engineering.