A bar magnet 20 cm in length is placed with its south pole towards geographic north. The neutral points are situated at a distance of 40 cm from centre of the magnet. If horizontal component of earth's field $3.2 \times 10^{- 5} T$ , then pole strength of magnet is :

5 Am

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10 Am

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45 Am

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20 Am

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Solution

The correct option is C 45 Am
As the neutral points are on the axial line, so using $\frac{μ_{o}}{4 π} \frac{2 M r}{(r^{2} - l^{2})^{2}} = B_{H}$
Given: $2 l = 20 c m = 0.2 m; B_{H} = 3.2 \times 10^{- 5} T; r = 0 .4 m$
$10^{- 7} \times \frac{2 M (0.4)}{({0.4}^{2} - {0.1}^{2})^{2}} = 3.2 \times 10^{- 5} ⟹ M = 9$
Now, $m (2 l) = M$
Thus, $m (0.2) = 9 ⟹ m = 45$
Hence, pole strength of magnet is $45 A m$

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A bar magnet 20 cm in length is placed with its south pole towards geographic north. The neutral points are situated at a distance of 40 cm from centre of the magnet. If horizontal component of earth's field 3.2×10−5T, then pole strength of magnet is :

The correct option is C 45 AmAs the neutral points are on the axial line, so using μo4π2Mr(r2−l2)2=BHGiven: 2l=20cm=0.2m ;BH=3.2×10−5T; r=0.4m10−7×2M(0.4)(0.42−0.12)2=3.2×10−5⟹M=9Now, m(2l)=MThus, m(0.2)=9⟹m=45Hence, pole strength of magnet is 45Am

A bar magnet 20 cm in length is placed with its south pole towards geographic north. The neutral points are situated at a distance of 40 cm from centre of the magnet. If horizontal component of earth's field $3.2 \times 10^{- 5} T$ , then pole strength of magnet is :