The magnetic induction at a distance of $20 c m$ from centre of a short magnetic dipole having magnetic moment $0.5 A m^{2}$ on its axis is:

1.57 \times 10^{- 5} T

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1.25 \times 10^{- 5} T

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1.57 \times 10^{- 4} T

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12.5 \times 10^{- 5} T

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Solution

The correct option is A $1.57 \times 10^{- 5} T$
Formula,

$B = \frac{μ_{0} (2 M)}{(4 π) r^{3}}$

$= \frac{10^{- 7} (2 \times 0.5)}{(2 \times 10^{- 2})^{3}}$

$= 1.25 \times 10^{- 5} T$

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The magnetic induction at a distance of 20 cm from centre of a short magnetic dipole having magnetic moment 0.5 Am2 on its axis is:

The correct option is A 1.57×10−5 TFormula,B=μ0(2M)(4π)r3=10−7(2×0.5)(2×10−2)3=1.25×10−5T

The magnetic induction at a distance of $20 c m$ from centre of a short magnetic dipole having magnetic moment $0.5 A m^{2}$ on its axis is:

The correct option is A $1.57 \times 10^{- 5} T$
Formula,

$B = \frac{μ_{0} (2 M)}{(4 π) r^{3}}$

$= \frac{10^{- 7} (2 \times 0.5)}{(2 \times 10^{- 2})^{3}}$

$= 1.25 \times 10^{- 5} T$