Two short bar magnets, each with a magnetic m | vedclass

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(C) The origin is on the axial line of the first magnet $(M_1)$ and on the equatorial line of the second magnet $(M_2)$.Given: $M = 9 	ext{ Am}^2$, $

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$0.1$

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(C) The origin is on the axial line of the first magnet $(M_1)$ and on the equatorial line of the second magnet $(M_2)$.Given: $M = 9 	ext{ Am}^2$, $r = 3 	ext{ cm} = 3 	imes 10^{-2} 	ext{ m}$.Magnetic field due to $M_1$ (axial point) at the origin:$B_1 = \frac{\mu_0}{4 \pi} 	imes \frac{2M}{r^3} = 10^{-7} 	imes \frac{2 	imes 9}{(3 	imes 10^{-2})^3} = 10^{-7} 	imes \frac{18}{27 	imes 10^{-6}} = \frac{2}{3} 	imes 10^{-1} 	ext{ T}$.This field is directed along the positive $X$-axis.Magnetic field due to $M_2$ (equatorial point) at the origin:$B_2 = \frac{\mu_0}{4 \pi} 	imes \frac{M}{r^3} = 10^{-7} 	imes \frac{9}{(3 	imes 10^{-2})^3} = 10^{-7} 	imes \frac{9}{27 	imes 10^{-6}} = \frac{1}{3} 	imes 10^{-1} 	ext{ T}$.Since the magnetic moment of $M_2$ is in the negative $X$-direction, the equatorial field at the origin points in the positive $X$-direction.As both $B_1$ and $B_2$ point in the same direction, the resultant magnetic field is:$B = B_1 + B_2 = \left(\frac{2}{3} + \frac{1}{3}ight) 	imes 10^{-1} 	ext{ T} = 1 	imes 10^{-1} 	ext{ T} = 0.1 	ext{ T}$.

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