Consider a circular loop of radius R. the value of BPBO is 1x then x is _________________.

BP=μ0IR22[R2+x2]3/2=μ0IR22[R2+15R2]3/2=μ0IR22.16R2.4R=μ0I128RBO=μ0I2RBPBO=164

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Consider a circular loop of radius R. the value of $\frac{B_{P}}{B_{O}}$ is $\frac{1}{x}$ then x is _________________.

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answer is 64.

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Detailed Solution

$B_{P} = \frac{μ_{0} I R^{2}}{2 {[R^{2} + x^{2}]}^{3 / 2}} = \frac{μ_{0} I R^{2}}{2 {[R^{2} + 15 R^{2}]}^{3 / 2}} = \frac{μ_{0} I R^{2}}{2.16 R^{2} .4 R} = \frac{μ_{0} I}{128 R}$

$B_{O} = \frac{μ_{0} I}{2 R}$

$\frac{B_{P}}{B_{O}} = \frac{1}{64}$

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