(C) Given: Magnetic moment $M = 16 \ Am^2$,distance $r = 20 \ cm = 0.2 \ m$. The constant $\frac{\mu_0}{4\pi} = 10^{-7} \ T \cdot m/A$. For an axial p

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$4 \times 10^{-4} \ T$ and $2 \times 10^{-4} \ T$

(C) Given: Magnetic moment $M = 16 \ Am^2$,distance $r = 20 \ cm = 0.2 \ m$. The constant $\frac{\mu_0}{4\pi} = 10^{-7} \ T \cdot m/A$. For an axial point: $B_{\text{axial}} = \frac{\mu_0}{4\pi} \frac{2M}{r^3} = 10^{-7} \times \frac{2 \times 16}{(0.2)^3} = 10^{-7} \times \frac{32}{0.008} = 10^{-7} \times 4000 = 4 \times 10^{-4} \ T$. For an equatorial point: $B_{\text{equatorial}} = \frac{\mu_0}{4\pi} \frac{M}{r^3} = 10^{-7} \times \frac{16}{(0.2)^3} = 10^{-7} \times \frac{16}{0.008} = 10^{-7} \times 2000 = 2 \times 10^{-4} \ T$. Thus,the values are $4 \times 10^{-4} \ T$ and $2 \times 10^{-4} \ T$.

Class 12 Physics Magnetism and Matter Magnetic field due to magnetic dipole and Dipole in Magnetic Field and Poential Energy and Work Done

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The magnetic moment of a short magnet is $16 \ Am^2$. The magnetic induction at a point $20 \ cm$ away from its midpoint on $(i)$ axial point and $(ii)$ equatorial point respectively,will be:

A
$2 \times 10^{-4} \ T$ and $10^{-4} \ T$
B
$3 \times 10^{-4} \ T$ and $2 \times 10^{-4} \ T$
C
$4 \times 10^{-4} \ T$ and $2 \times 10^{-4} \ T$
D
none of these

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Magnetic field due to magnetic dipole and Dipole in Magnetic Field and Poential Energy and Work Done

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The magnetic moment of a short magnet is $16 \ Am^2$. The magnetic induction at a point $20 \ cm$ away from its midpoint on $(i)$ axial point and $(ii)$ equatorial point respectively,will be:

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