A small bar magnet has a magnetic moment 2.5 A-m2. The magnetic field at a distance 0.1m on its equatorial line will be : (μ0=4π×10−7T−m/A)

B=μ04πMd3=10−7×2.510−3=2.5×10−4T

A small bar magnet has a magnetic moment 2.5 A-m2. The magnetic field at a distance 0.1m on its equatorial line will be : (μ0=4π×10−7T−m/A)

B=μ04πMd3=10−7×2.510−3=2.5×10−4T

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A small bar magnet has a magnetic moment 2.5 A-m². The magnetic field at a distance 0.1m on its equatorial line will be : $(μ_{0} = 4 π \times 10^{- 7} T - m / A)$

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$5 \times 10^{- 4} T$

$2.5 \times 10^{- 4} T$

$2.5 \times 10^{4} T$

$5 \times 10^{4} T$

answer is B.

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$B = \frac{μ_{0}}{4 π} \frac{M}{d^{3}} = 10^{- 7} \times \frac{2.5}{10^{- 3}} = 2.5 \times 10^{- 4} T$

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A small bar magnet has a magnetic moment 2.5 A-m2. The magnetic field at a distance 0.1m on its equatorial line will be : (μ0=4π×10−7T−m/A)

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A small bar magnet has a magnetic moment 2.5 A-m². The magnetic field at a distance 0.1m on its equatorial line will be : $(μ_{0} = 4 π \times 10^{- 7} T - m / A)$