A closely packed coil having 1000 turns has an average radius of 62.8 cm. If the current carried by the wire o

Question

A closely packed coil having 1000 turns has an average radius of 62.8 cm. If the current carried by the wire of the coil is 1 A, the value of the magnetic field produced at the centre of the coil will be nearly: (permeability of free space = $4\pi 	imes 10^{-7}$ H/m)

Accepted Answer

1.  **Identify Formula:** The magnitude of the magnetic field ($B$) at the center of a circular coil with $N$ turns, radius $R$, and carrying current $I$ is given by the formula:
    $$B = \frac{\mu_0 N I}{2R}$$
2.  **Identify Parameters:**
    *   Number of turns ($N$): $1000 = 10^3$
    *   Current ($I$): $1 	ext{ A}$
    *   Radius ($R$): $62.8 	ext{ cm} = 62.8 	imes 10^{-2} 	ext{ m}$
    *   Permeability ($\mu_0$): $4\pi 	imes 10^{-7} 	ext{ T m/A}$
3.  **Calculation:**
    Substitute the values into the formula:
    $$B = \frac{(4\pi 	imes 10^{-7}) 	imes 10^3 	imes 1}{2 	imes (62.8 	imes 10^{-2})}$$
    Recognizing that $2\pi \approx 6.28$, we can write $62.8 = 10 	imes (2\pi)$.
    $$B = \frac{4\pi 	imes 10^{-4}}{2 	imes 10 	imes 2\pi 	imes 10^{-2}}$$
    $$B = \frac{4\pi 	imes 10^{-4}}{40\pi 	imes 10^{-2}}$$
    $$B = \frac{1}{10} 	imes 10^{-2} = 10^{-1} 	imes 10^{-2} = 10^{-3} 	ext{ T}$$
4.  **Conclusion:** The magnetic field is $10^{-3} 	ext{ T}$.

Step-by-Step Solution

Practice All PHYSICS Questions