The magnetic field at the centre of current carrying circular loop is B1. The magnetic field at a distance of 3 times radius of the given circular loop from the center on its axis is B2. The value of B1/B2 will be

B1=μ0i2R B2=μ0iR22 (R2+ (3R)2)3/2=μ0iR22*8R3=μ0i8*2R=B18 ⇒B1B2=8:1

The magnetic field at the centre of current carrying circular loop is B₁. The magnetic field at a distance of $\sqrt[]{3}$ times radius of the given circular loop from the center on its axis is B₂. The value of B₁/B₂ will be

Option 1 -  9 : 4

Option 2 - 12 : <math><mrow><mroot><mrow><mn>5</mn></mrow><mrow></mrow></mroot></mrow></math>

Option 3 - 8 : 1

Option 4 - 5 : <math><mrow><mroot><mrow><mn>3</mn></mrow><mrow></mrow></mroot></mrow></math>

4 Views|Posted 6 months ago

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Answered by

Vishal Baghel | Contributor-Level 10

6 months ago

Correct Option - 3

Detailed Solution:

$B_{1} = \frac{μ_{0} i}{2 R}$

$B_{2} = \frac{μ_{0} i R^{2}}{2 {(R^{2} + {(\sqrt[]{3} R)}^{2})}^{3 / 2}} = \frac{μ_{0} i R^{2}}{2 * 8 R^{3}} = \frac{μ_{0} i}{8 * 2 R} = \frac{B_{1}}{8}$

$\Rightarrow \frac{B_{1}}{B_{2}} = 8 : 1$

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