A magnetic needle of pole strength $20 \times 10^{- 4}$ Am is pivoted at it's centre. Its $N - p o l e$ is pulled eastward by a string. The horizontal force required to produce a deflection of $30^{o}$ from the magnetic meridian is (take $B_{H} = 10^{- 4} T)$ is :

Q. The pole strength of

12

cm long bar magnet is

20

A m. The magnetic induction at a point

10

cm away from the centre of the magnet on its axial line is

[\frac{μ_{o}}{4 π} = 10^{- 7} H m^{- 1}]

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The magnetic induction due to a magnetic pole of pole strength 1500Am at a distance of 50cm is

The correct option is D 6×10−4TGiven m=1500Am r=50cm=0.5m B=?B=μ04π×mr2B=10−7×15000.52=10−7×15000.25=10−7×6000=6×10−4Tesla

The magnetic induction due to a magnetic pole of pole strength $1500 A m$ at a distance of $50 c m$ is

The correct option is D $6 \times 10^{- 4} T$
Given
$m = 1500 A m$
$r = 50 c m = 0.5 m$
$B = ?$
$B = \frac{μ_{0}}{4 π} \times \frac{m}{r^{2}}$
$B = 10^{- 7} \times \frac{1500}{{0.5}^{2}} = 10^{- 7} \times \frac{1500}{0.25} = 10^{- 7} \times 6000 = 6 \times 10^{- 4} T e s l a$