A compass needle whose magnetic moment is $$60$$ A $$m2$$ pointing geographical north at a certain place, where the horizontal component of earth's magnetic field is $$40$$μ $$Wb/m2$$ experiences a torque of $$1.2$$ x $$10-3$$ N m. What is the declination at this place?

Question

A compass needle whose magnetic moment is $$60$$ A $$m2$$ pointing geographical north at a certain place, where the horizontal component of earth's magnetic field is $$40$$μ $$Wb/m2$$ experiences a torque of $$1.2$$ x $$10-3$$ N m. What is the declination at this place?

Infinity Learn · Accepted Answer

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, il will experience a torque due to the horizontal component of earth's magnetic field, i.e., τ$$=MBHsin$$ θ where θ $$=$$ angle between geographical and magnetic meridians called angle of declinationSo, $$sin$$θ$$=1.2$$ x $$10-360$$ x $$40$$ x $$10-6=12$$⇒θ$$=30o$$

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

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A compass needle whose magnetic moment is 60 A m2 pointing geographical north at a certain place, where the horizontal component of earth's magnetic field is 40μ Wb/m2 experiences a torque of 1.2 x 10-3 N m. What is the declination at this place?

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Ready to Test Your Skills?

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