A compass needle whose magnetic moment is $60\ amp imes m^{2}$ 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 $1.2 imes 10^{- 3}N imes m$. What is the declination at this place

Question

A compass needle whose magnetic moment is $60\ amp 	imes m^{2}$ 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 $1.2 	imes 10^{- 3}N 	imes m$. What is the declination at this place

Dr. Abubacker · Accepted Answer

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

Q. A compass needle whose magnetic moment is $60\ amp \times m^{2}$ 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

Explanation

Reviewed by Dr. Abubacker

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