First slide

Magnetic elements of Earth

Question

A uniform magnetic needle is suspended from its center by a thread. Its upper end is now loaded with a mass of 50 mg when the needle becomes horizontal. If the strength of each pole is 98.1 ab-amp.cm & g is ${981 cm/s}^{2}$ , then the vertical component of the earth’s magnetic field is

Moderate

A

0.50 gauss
B

0.25 gauss
C

0.05 gauss
D

0.005 gauss

Solution

Taking moments about ‘O’
$M g l = m B_{v} l + m B_{v} l$
$\Rightarrow M g = 2 m B_{v} \to (1)$
Given, $M = 50 \times 10^{- 3} g$
$g = 981 c m / s^{2}$
$m = 98.1 a b - a m p . c m$
Then from equation (1)
$50 \times 10^{- 3} \times 981 = 2 \times 98.1 \times B_{v}$
$B_{v} = 0.25 gauss$ .

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