A compass needle whose magnetic moment is 60 ,amp × m^2 pointing … - Vidyadip

MCQ

A compass needle whose magnetic moment is $60 \,amp × m^2$ pointing geographical north at a certain place, where the horizontal component of earth's magnetic field is $40\, \mu Wb/m^2$, experiences a torque $1.2 \times {10^{ - 3}}\,N \times m.$ What is the declination at this place......$^o$

✓
$30$
B
$45$
C
$60$
D
$25$

✓

Answer

Correct option: A.

$30$

a
(a) As the compass needle is free to rotate in a horizontal plane and points along the magnetic meridian,
so when it is pointing along the geographic meridian, it will experience a torque due to the horizontal component of earth's magnetic field i.e. $\tau = M{B_H}\sin \theta $
where $\theta$ $=$ angle between geographical
and magnetic meridians called angle of declination
So, $\sin \theta = \frac{{1.2 \times {{10}^{ - 3}}}}{{60 \times 40 \times {{10}^{ - 6}}}} = \frac{1}{2}$ $==>$ $\theta = {30^o}$

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