The magnitude of the equatorial magnetic field due to a bar magnet of length $2$ cm at a distance of $1$ m from its mid-point is (magnetic moment of the bar magnet is $0.60 A m^{2}$ )

5.0 \times 10^{- 5}

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6.0 \times 10^{- 8}

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7.0 \times 10^{- 7}

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8.0 \times 10^{- 8}

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Solution

The correct option is A $6.0 \times 10^{- 8}$ T
The magnetic field of the equatorial point is

$B_{E} = \frac{μ_{0} m}{4 π r^{3}}$

Here, $\frac{μ_{0}}{4 π} = 10^{- 7}, r = 1 m$ and $m = 0.60 A m^{2}$

$∴= \frac{10^{- 7} \times 0.60}{(1)^{3}} = 6.0 \times 10^{- 8} T$

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The magnitude of the equatorial magnetic field due to a bar magnet of length 2cm at a distance of 1m from its mid-point is (magnetic moment of the bar magnet is 0.60Am2)

The correct option is A 6.0×10−8TThe magnetic field of the equatorial point isBE=μ0m4πr3Here, μ04π=10−7,r=1m and m=0.60Am2∴=10−7×0.60(1)3=6.0×10−8T

The magnitude of the equatorial magnetic field due to a bar magnet of length $2$ cm at a distance of $1$ m from its mid-point is (magnetic moment of the bar magnet is $0.60 A m^{2}$ )

The correct option is A $6.0 \times 10^{- 8}$ T
The magnetic field of the equatorial point is

$B_{E} = \frac{μ_{0} m}{4 π r^{3}}$

Here, $\frac{μ_{0}}{4 π} = 10^{- 7}, r = 1 m$ and $m = 0.60 A m^{2}$

$∴= \frac{10^{- 7} \times 0.60}{(1)^{3}} = 6.0 \times 10^{- 8} T$