Question

19. A short magnetic dipole has magnetic moment 0.5 Am-.
Calculate the magnetic induction by it at a distance of
20 cm. from the centre of the magnetic dipole on.
(a) the axis and
(b) the equatorial line of the short magnetic dipole
(Given . = 4 x 10-7 wb/Am)
​

Answer 1

(a) we know, magnetic induction is along the direction of magnetic dipole moment ( the axial line ) is given by,

$\boxed{\bf{B=\frac{\mu_0}{4\pi}\frac{2M}{r^3}}}$

here, M = 0.5 Am² , r = 20cm = 0.2m

and $\mu_0=4\pi\times10^{-7}$

so, $B_{axial}$ = {4π × 10^-7/4π} × {2 × 0.5/(0.2)³}

= 10^-7 × 1/0.008

= 10^-4/8

= 10/8 × 10^-5 wb/m²

= 1.25 × 10^-5 wb/m²

(b) magnetic field is parallel to the magnetic dipole moment and opposite to the direction of dipole moment (on equatorial line) is given by,

$\boxed{\bf{B=\frac{\mu_0}{4\pi}\frac{M}{r^3}}}$

so, $B_{equatorial}$ = (4π × 10^-7/4π) × 0.5/(0.2)³

= 10^-7 × 0.5/0.008

= 0.625 × 10^-5 wb/m²

= 6.25 × 10^-6 wb/m²

Answer 2

Answer:

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