Two wires are wrapped over a wooden cylinder to form two co-axial loops carrying currents i1 and i2 . If i2 = 8i1 , the value of x for B = 0 at the origin O is

B1=B2 ⇒ μ04π 2πi1R2 R2+R23/2 =μ04π 2i2πR2 x2+R23/2Given i2 = 8i1Solve to get x= 7 R

Two wires are wrapped over a wooden cylinder to form two co-axial loops carrying currents i1 and i2 . If i2 = 8i1 , the value of x for B = 0 at the origin O is

B1=B2 ⇒ μ04π 2πi1R2 R2+R23/2 =μ04π 2i2πR2 x2+R23/2Given i2 = 8i1Solve to get x= 7 R

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Two wires are wrapped over a wooden cylinder to form two co-axial loops carrying currents i₁ and i₂ . If i₂ = 8i₁ , the value of x for B = 0 at the origin O is

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$(\sqrt{7} - 1) R$

$\sqrt{5} R$

$\sqrt{3} R$

$\sqrt{7} R$

answer is D.

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Detailed Solution

$\begin{array}{l} B_{1} = B_{2} \Rightarrow \frac{μ_{0}}{4 π} \frac{2 {πi}_{1} R^{2}}{{(R^{2} + R^{2})}^{3 / 2}} = \frac{μ_{0}}{4 π} \frac{2 i_{2} {πR}^{2}}{{(x^{2} + R^{2})}^{3 / 2}} \\ Given i_{2} = 8 i_{1} \\ Solve to get x = \sqrt{7} R \end{array}$