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Byju's Answer

Standard XII

Physics

Magnetic Moment

A pole of pol...

Question

A pole of pole strength $80 A m$ is placed at a point at a distance $20 c m$ on the equatorial line from the centre of a short magnet of magnetic moment $20 A m^{2}$ . The force experienced by it is

8 \times 10^{- 2} N

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2 \times 10^{- 2} N

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16 \times 10^{- 2} N

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64 \times 10^{- 2} N

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Solution

The correct option is B $2 \times 10^{- 2} N$
$p = 80 A m$
$d = 20 c m .$
$m = 20 A m^{2}$
$B_{2} = \frac{μ_{0} m}{4 π d^{3}}$
$= \frac{4 π \times 10^{- 7} \times 20}{4 π (0.2)^{3}}$
$B = 2.5 \times 10^{- 4} T$
$F = P B$
$= 80 \times 2.5 \times 10^{- 4}$
$F = 0.02 N$

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A pole of pole strength 80Am is placed at a point at a distance 20cm on the equatorial line from the centre of a short magnet of magnetic moment 20Am2. The force experienced by it is

The correct option is B 2×10−2Np=80Amd=20cm.m=20Am2B2=μ0m4πd3=4π×10−7×204π(0.2)3B=2.5×10−4TF=PB=80×2.5×10−4F=0.02N

A pole of pole strength $80 A m$ is placed at a point at a distance $20 c m$ on the equatorial line from the centre of a short magnet of magnetic moment $20 A m^{2}$ . The force experienced by it is

The correct option is B $2 \times 10^{- 2} N$
$p = 80 A m$
$d = 20 c m .$
$m = 20 A m^{2}$
$B_{2} = \frac{μ_{0} m}{4 π d^{3}}$
$= \frac{4 π \times 10^{- 7} \times 20}{4 π (0.2)^{3}}$
$B = 2.5 \times 10^{- 4} T$
$F = P B$
$= 80 \times 2.5 \times 10^{- 4}$
$F = 0.02 N$