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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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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 \(0.40\; A m^2\).

Answer

\(6.4\times 10^{-7}T\)
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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 \(6.4\times 10^{-7}T\)
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