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Question: A compass needle whose magnetic moment is \(60{\text{A}}{{\text{m}}^{\text{2}}}\) pointing geographi...

A compass needle whose magnetic moment is 60Am260{\text{A}}{{\text{m}}^{\text{2}}} pointing geographical north at a certain place where the horizontal component of earth’s magnetic field is 40×10−6Wbm - 240 \times {10^{ - 6}}{\text{Wb}}{{\text{m}}^{{\text{ - 2}}}} experiences a torque of 1.2×10−3Nm1.2 \times {10^{ - 3}}{\text{Nm}} . the declination of the place is?

A. 200{20^0}
B. 450{45^0}
C. 600{60^0}
D. 300{30^0}

Explanation

Solution

Hint: The measure of force that can cause an object to rotate about an axis is known as Torque. It can also be defined as the rotational equivalent of linear force and it causes objects to accelerate in linear kinematics.

Formula used:
Torqueτ=mBsin⁡δ\tau = mB\sin \delta
Where, ‘m’ is magnetic moment, B is the magnetic field, δ\delta is the declination.

Complete step by step answer:
We know that when a magnetic needle (magnetic dipole) is placed in a uniform magnetic field it experiences a torque. If the needle is free to oscillate then it oscillates simple harmonically. Consider a small compass needle of magnetic moment ‘m’ and moment of inertia ‘I’ which is allowed to oscillate in the magnetic field ‘B’ then,
Let us consider the torque on the needle that is given by,τ=m→×B→\tau = \overrightarrow m \times \overrightarrow B
τ=mBsin⁡δ\tau = mB\sin \delta
A freely suspended magnetic needle will lay in the magnetic meridian and the north pole of the needle will point towards the magnetic north pole (MN) and vice versa.

Given, in stable equilibrium condition, a compass needle points along the magnetic north and experiences no torque.
When it turns through declination δ\delta , it experiences a torque. That is given by,
τ=mBsin⁡δ\tau = mB\sin \delta
Here δ\delta is declination, meaning the angle between the magnetic meridian and geographic meridian at a place.
∴sin⁡δ=τmB\therefore \sin \delta = \dfrac{\tau }{{mB}}
sin⁡δ=1.2×10−360×40×10−6=0.5\sin \delta = \dfrac{{1.2 \times {{10}^{ - 3}}}}{{60 \times 40 \times {{10}^{ - 6}}}} = 0.5
δ=sin⁡−1(0.5)=300\delta = {\sin ^{ - 1}}\left( {0.5} \right) = {30^0}
Hence, the correct option is (D).

Additional information:
The pole of the earth’s magnet near the geographical north pole of the earth is called the magnetic North pole.
The pole of earth’s magnet near the geographical south pole is called Magnetic south pole.
The straight line passing through the geographical North and south poles of the earth are called its geographical axis.
The straight line passing through the magnetic North and South pole of the earth is called its Magnetic axis.
The vertical plane passing through the magnetic axis is called the magnetic meridian.
The vertical plane passing through the geographical axis is called a geographical meridian.
A great circle on earth’s surface perpendicular to the magnetic axis is called the magnetic equator.

Note: Three elements are responsible for the direction and magnitude of the earth’s magnetic field. They are magnetic declination, angle of dip or magnetic inclination and the horizontal component of the earth’s magnetic field.