Two magnets A and B are identical and are arr | vedclass

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(C) The magnetic field due to magnet $A$ (axial position) at point $P$ is $B_1 = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d_1^3}$.The magnetic field due to

Vedclass · Accepted Answer

$(2\cot 	heta )^{1/3}$

Vedclass · Answer

(C) The magnetic field due to magnet $A$ (axial position) at point $P$ is $B_1 = \frac{\mu_0}{4\pi} \cdot \frac{2M}{d_1^3}$.The magnetic field due to magnet $B$ (equatorial position) at point $P$ is $B_2 = \frac{\mu_0}{4\pi} \cdot \frac{M}{d_2^3}$.When the magnetic needle is in equilibrium at an angle $	heta$ with the direction of $B_1$,the resultant magnetic field makes an angle $	heta$ with $B_1$,such that $	an 	heta = \frac{B_2}{B_1}$.Substituting the values,we get $	an 	heta = \frac{(\mu_0 / 4\pi) \cdot (M / d_2^3)}{(\mu_0 / 4\pi) \cdot (2M / d_1^3)} = \frac{d_1^3}{2d_2^3}$.Therefore,$\frac{d_1^3}{d_2^3} = 2 	an 	heta$,which implies $\frac{d_1}{d_2} = (2 	an 	heta)^{1/3}$.Wait,checking the equilibrium condition: The needle aligns with the resultant field. If $	heta$ is the deflection from the axis of magnet $A$ (field $B_1$),then $	an 	heta = B_2 / B_1$. Thus $\frac{d_1^3}{2d_2^3} = 	an 	heta \Rightarrow \frac{d_1}{d_2} = (2 	an 	heta)^{1/3}$.Re-evaluating the provided options and standard convention: If $	heta$ is the angle with the vertical (field $B_2$),then $	an 	heta = B_1 / B_2 = \frac{2d_2^3}{d_1^3} \Rightarrow \frac{d_1^3}{d_2^3} = 2 \cot 	heta \Rightarrow \frac{d_1}{d_2} = (2 \cot 	heta)^{1/3}$. Based on the figure,$	heta$ is the angle with the vertical,so option $C$ is correct.

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