Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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:

3.14 × 10⁻⁷ T

6.28 × 10⁻⁷ T

3.14 × 10⁻⁴ T

6.28 × 10⁻⁴ T

Step-by-Step Solution

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').
Identify Parameters: Radius ( $R$ ): $100 \text{ cm} = 1 \text{ m}$ Distance ( $x$ ): $1 \text{ m}$ Current ( $I$ ): $\sqrt{2} \text{ A}$ Permeability ( $\mu_0$ ): $4\pi \times 10^{-7} \text{ T m/A}$
Calculation: Substitute the values into the formula: $B = \frac{(4\pi \times 10^{-7}) \times (\sqrt{2}) \times (1)^2}{2(1^2 + 1^2)^{3/2}}$ The denominator term $(R^2 + x^2)^{3/2} = (1 + 1)^{3/2} = (2)^{3/2} = (2^1 \times 2^{1/2}) = 2\sqrt{2}$ . $B = \frac{4\pi \times 10^{-7} \times \sqrt{2}}{2 \times (2\sqrt{2})}$ $B = \frac{4\pi \times 10^{-7} \times \sqrt{2}}{4\sqrt{2}}$ Canceling common terms ( $4\sqrt{2}$ ): $B = \pi \times 10^{-7} \text{ T}$ Since $\pi \approx 3.14$ : $B = 3.14 \times 10^{-7} \text{ T}$

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