A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field a

Question

A tightly wound 100 turns coil of radius 10 cm carries a current of 7 A. The magnitude of the magnetic field at the centre of the coil is: (Take permeability of free space as $4\pi 	imes 10^{-7}$ SI units)

Accepted Answer

1.  **Identify Formula:** The magnitude of the magnetic field ($B$) at the center of a circular coil carrying current is given by the formula:
    $$B = \frac{\mu_0 N I}{2R}$$
    where:
    *   $\mu_0$ is the permeability of free space ($4\pi 	imes 10^{-7} 	ext{ T m/A}$)
    *   $N$ is the number of turns
    *   $I$ is the current
    *   $R$ is the radius
2.  **Substitute Values:**
    *   $N = 100$
    *   $I = 7 	ext{ A}$
    *   $R = 10 	ext{ cm} = 0.1 	ext{ m}$
    $$B = \frac{(4\pi 	imes 10^{-7}) 	imes 100 	imes 7}{2 	imes 0.1}$$
3.  **Calculation:**
    Use $\pi \approx \frac{22}{7}$ for simpler calculation:
    $$B = \frac{4 	imes \frac{22}{7} 	imes 10^{-7} 	imes 100 	imes 7}{0.2}$$
    $$B = \frac{4 	imes 22 	imes 10^{-5}}{0.2}$$
    $$B = \frac{88 	imes 10^{-5}}{0.2}$$
    $$B = 440 	imes 10^{-5} 	ext{ T}$$
    $$B = 4.4 	imes 10^{-3} 	ext{ T}$$
4.  **Convert Units:** Since $10^{-3} 	ext{ T} = 1 	ext{ mT}$:
    $$B = 4.4 	ext{ mT}$$

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