The magnetic field on the axis of a circular loop of radius 100 cm carrying current I = √2 A, at a point 1 m a

Question

The magnetic field on the axis of a circular loop of radius 100 cm carrying current I = √2 A, at a point 1 m away from the centre of the loop is given by:

Accepted Answer

1.  **Identify Formula:** The magnitude of the magnetic field ($B$) on the axis of a circular current loop of radius $R$ at a distance $x$ from the center is given by:
    $$B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$$
    (Assuming number of turns $N=1$ as it is described as 'a circular loop').
2.  **Identify Parameters:**
    *   Radius ($R$): $100 	ext{ cm} = 1 	ext{ m}$
    *   Distance ($x$): $1 	ext{ m}$
    *   Current ($I$): $\sqrt{2} 	ext{ A}$
    *   Permeability ($\mu_0$): $4\pi 	imes 10^{-7} 	ext{ T m/A}$
3.  **Calculation:**
    Substitute the values into the formula:
    $$B = \frac{(4\pi 	imes 10^{-7}) 	imes (\sqrt{2}) 	imes (1)^2}{2(1^2 + 1^2)^{3/2}}$$
    The denominator term $(R^2 + x^2)^{3/2} = (1 + 1)^{3/2} = (2)^{3/2} = (2^1 	imes 2^{1/2}) = 2\sqrt{2}$.
    $$B = \frac{4\pi 	imes 10^{-7} 	imes \sqrt{2}}{2 	imes (2\sqrt{2})}$$
    $$B = \frac{4\pi 	imes 10^{-7} 	imes \sqrt{2}}{4\sqrt{2}}$$
    Canceling common terms ($4\sqrt{2}$):
    $$B = \pi 	imes 10^{-7} 	ext{ T}$$
    Since $\pi \approx 3.14$:
    $$B = 3.14 	imes 10^{-7} 	ext{ T}$$

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