Two short magnetic dipoles m_{1} and m_{2}, e | vedclass

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(A) The magnetic field $B_{1}$ produced by the dipole $m_{1}$ at point $P$ (which is on the equatorial line of $m_{1}$) is given by:$B_{1} = \frac{\mu

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(A) The magnetic field $B_{1}$ produced by the dipole $m_{1}$ at point $P$ (which is on the equatorial line of $m_{1}$) is given by:$B_{1} = \frac{\mu_{0}}{4\pi} \frac{m_{1}}{r^{3}}$Given $m_{1} = 1\, 	ext{Am}^{2}$, $r = 1\, 	ext{m}$, and $\frac{\mu_{0}}{4\pi} = 10^{-7}\, 	ext{T}\cdot	ext{m/A}$.$B_{1} = 10^{-7} 	imes \frac{1}{(1)^{3}} = 10^{-7}\, 	ext{T}$.The direction of $B_{1}$ is opposite to the direction of $m_{1}$.The torque $	au$ experienced by dipole $m_{2}$ is given by $\vec{	au} = \vec{m}_{2} 	imes \vec{B}_{1}$.Since $m_{2}$ is perpendicular to $B_{1}$, the magnitude of the torque is:$	au = m_{2} B_{1} \sin(90^{\circ}) = 1 	imes 10^{-7} 	imes 1 = 10^{-7}\, 	ext{Nm}$.Thus, the value is $1 	imes 10^{-7}\, 	ext{Nm}$.

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