When a bar magnet of magnetic moment $¯ ¯¯¯¯ ¯ M$ is placed in a magnetic field of induction field strength $¯ ¯¯ ¯ B$ , each pole experiences a force of $¯ ¯¯ ¯ F$ then the distance between the South and North pole of the magnet measured inside it is :

M B F

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\frac{M B}{F}

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\frac{F}{M B}

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\frac{F B}{M}

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Solution

The correct option is A $\frac{M B}{F}$
As in the figure , $T o r q u e = F \times a sin θ$
$T o r q u e = M \times B sin θ$
$a = \frac{M B}{F}$

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