A bar magnet of length 3 , cm has points A an | vedclass

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(A) The magnetic field $B$ on the axial line of a short bar magnet at a distance $d$ from its center is given by $B = \frac{\mu_0}{4\pi} \frac{2Md}{(d

Vedclass · Accepted Answer

$8$

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(A) The magnetic field $B$ on the axial line of a short bar magnet at a distance $d$ from its center is given by $B = \frac{\mu_0}{4\pi} \frac{2Md}{(d^2 - l^2)^2}$,where $2l$ is the length of the magnet. Since the distances $d_A = 24 \, cm$ and $d_B = 48 \, cm$ are much larger than the length of the magnet $(2l = 3 \, cm)$,we can use the approximation for a short dipole: $B \propto \frac{1}{d^3}$. Therefore,the ratio of the magnetic fields at points $A$ and $B$ is: $\frac{B_A}{B_B} = \left( \frac{d_B}{d_A} \right)^3 = \left( \frac{48}{24} \right)^3 = (2)^3 = 8$. Thus,the ratio is $8:1$.

$A$ bar magnet of length $3 \, cm$ has points $A$ and $B$ along its axis at distances of $24 \, cm$ and $48 \, cm$ from the center on opposite sides. The ratio of the magnetic fields at these points is:

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