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What is the magnitude of the equatorial and axial fields due to a bar magnet of length $$5\;{cm}$$ at a distance of $$50\;{cm}$$ from its mid-point? The magnetic moment of the bar magnet is $$0.40\; A m^2$$.
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## QuestionPhysicsClass 12

What is the magnitude of the equatorial and axial fields due to a bar magnet of length $$5\;{cm}$$ at a distance of $$50\;{cm}$$ from its mid-point? The magnetic moment of the bar magnet is $$0.40\; A m^2$$.

$$6.4\times 10^{-7}T$$
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## Solution

Given: Magnetic moment $$\left(M\right) \ = \ 0.40\;Am^2$$
Distance from the mid-point of the bar-magnet $$\left(r\right)=0.5\;m$$
The magnetic field at the equatorial position:
Therefore, $$B \ = \ \dfrac{\mu _0M}{4\pi r^3}$$
Where, $$\mu_0$$permeability of free space  $$= \ 4 \pi \ \times 10^{-7} \ N/A^2$$
$$B \ = \ \dfrac{10^{-7}\times 0.40\;Am^2}{(0.5\;m)^3}$$
$$B \ = \ 3.2\times 10^{-7}T$$
Similarly, the magnetic field at the axial position:
Therefore, $$B' \ = \ \dfrac{\mu _0(2M)}{4\pi r^3}$$
$$B' \ = \ 2\times 3.2\times 10^{-7}T$$
$$B' \ = \ 6.4\times 10^{-7}T$$
Hence, the answer is $$6.4\times 10^{-7}T$$