Two identical small bar magnets, each of dipo | vedclass

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(D) The distance between the centers of the magnets is $d_{total} = 10 	ext{ cm}$. The point $P$ is at the midpoint, so the distance from the center

Vedclass · Accepted Answer

$7.59$

Vedclass · Answer

(D) The distance between the centers of the magnets is $d_{total} = 10 	ext{ cm}$. The point $P$ is at the midpoint, so the distance from the center of each magnet to point $P$ is $d = 5 	ext{ cm} = 0.05 	ext{ m}$.
For a small bar magnet, the magnetic field at an equatorial point is given by $B_{eq} = \frac{\mu_0}{4\pi} \frac{M}{d^3}$.
For the left magnet, point $P$ lies on its equatorial line, so $B_1 = \frac{\mu_0}{4\pi} \frac{M}{d^3}$.
For the right magnet, point $P$ also lies on its equatorial line, so $B_2 = \frac{\mu_0}{4\pi} \frac{M}{d^3}$.
Since the axes are perpendicular, the magnetic field vectors $B_1$ and $B_2$ are perpendicular to each other. The net magnetic field is $B_{net} = \sqrt{B_1^2 + B_2^2} = \sqrt{2} B_{eq} = \sqrt{2} \frac{\mu_0}{4\pi} \frac{M}{d^3}$.
Substituting the values: $M = 3\sqrt{5} 	ext{ J/T}$, $d = 0.05 	ext{ m}$, and $\frac{\mu_0}{4\pi} = 10^{-7} 	ext{ T} \cdot 	ext{m/A}$.
$B_{net} = \sqrt{2} 	imes 10^{-7} 	imes \frac{3\sqrt{5}}{(0.05)^3} = \sqrt{2} 	imes 10^{-7} 	imes \frac{3\sqrt{5}}{125 	imes 10^{-6}} = \frac{3\sqrt{10} 	imes 10^{-1}}{125} = \frac{3 	imes 3.162 	imes 0.1}{125} \approx 0.00759 	ext{ T} = 7.59 	imes 10^{-3} 	ext{ T}$.
Thus, $\alpha \approx 7.59$.

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