The magnetic field at the centre of a current carrying circular loop of radius R is $B_{1}$ . The magnetic field at a point on its axis at a distance R from the center of the loop is $B_{2}$ . Then the ratio $(B_{1} / B_{2})$ is

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Solution

Magnetic field at centre, $B_{1} = \frac{μ_{0} i}{2 R}$

Magnetic field at axial distance, R is $B_{2} = \frac{μ_{0} I R^{2}}{2 (R^{2} + R^{2})^{\frac{3}{2}}}$

$= \frac{μ_{0} I R^{2}}{2 \times 2 \sqrt{2} R^{3}}$

$∴ \frac{B_{1}}{B_{2}} = 2 \sqrt{2}$

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The magnetic field at the centre of a current carrying circular loop of radius R isB1. The magnetic field at a point on its axis at a distance R from the center of the loop is B2. Then the ratio (B1/B2) is

Magnetic field at centre, B1=μ0i2R Magnetic field at axial distance, R is B2=μ0IR22(R2+R2)32 =μ0IR22×2√2R3 ∴ B1B2=2√2

The magnetic field at the centre of a current carrying circular loop of radius R is $B_{1}$ . The magnetic field at a point on its axis at a distance R from the center of the loop is $B_{2}$ . Then the ratio $(B_{1} / B_{2})$ is

Magnetic field at centre, $B_{1} = \frac{μ_{0} i}{2 R}$

Magnetic field at axial distance, R is $B_{2} = \frac{μ_{0} I R^{2}}{2 (R^{2} + R^{2})^{\frac{3}{2}}}$

$= \frac{μ_{0} I R^{2}}{2 \times 2 \sqrt{2} R^{3}}$

$∴ \frac{B_{1}}{B_{2}} = 2 \sqrt{2}$