Two short bar magnets of magnetic moments m e | vedclass

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(A) Let the square be $CDEF$. Magnets are placed at corners $D$ and $F$. We need to find the magnetic induction at corner $E$.$1$. The magnet at $F$ i

Vedclass · Accepted Answer

$\frac{\mu_0}{4\pi} \frac{m}{d^3}$

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(A) Let the square be $CDEF$. Magnets are placed at corners $D$ and $F$. We need to find the magnetic induction at corner $E$.$1$. The magnet at $F$ is at a distance $d$ from $E$ along its axis. Thus,$E$ is an axial point for the magnet at $F$. The magnetic induction $B_1$ at $E$ due to the magnet at $F$ is given by:$B_1 = \frac{\mu_0}{4\pi} \frac{2m}{d^3}$This field acts along the direction $EF$ (away from $F$ if $N$ is towards $E$).$2$. The magnet at $D$ is at a distance $d$ from $E$ along its equatorial line. Thus,$E$ is an equatorial point for the magnet at $D$. The magnetic induction $B_2$ at $E$ due to the magnet at $D$ is given by:$B_2 = \frac{\mu_0}{4\pi} \frac{m}{d^3}$This field acts in the direction opposite to the magnetic moment of the magnet at $D$,which is along $FE$.$3$. Since $B_1$ and $B_2$ are in opposite directions,the resultant magnetic induction $B$ at $E$ is:$B = B_1 - B_2 = \frac{\mu_0}{4\pi} \frac{2m}{d^3} - \frac{\mu_0}{4\pi} \frac{m}{d^3} = \frac{\mu_0}{4\pi} \frac{m}{d^3}$

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