Question
Question: A magnet of dipole moment \(2\; Am^2\) is deflected through \(30^{\circ}\) from magnetic meridian. T...
A magnet of dipole moment 2Am2 is deflected through 30∘ from magnetic meridian. The required deflecting torque is (BH=0.4×10−4T):
A.)0.4×10−4 Nm
B.)0.4 Nm
C.)0.2×10−4 Nm
D.)None of these
Solution
Hint: The torque required can be calculated by the cross product of the magnet dipole moment and horizontal component of magnetic meridian as there is no presence of any other magnetic field.
Formula used:
Torque, τ=μ×BH, where μ is the magnets dipole moment and BH is the horizontal component of Earth’s magnetic field.
Complete step by step solution:
We have been given that the dipole moment of the magnet, μ=2Am2 and the horizontal component of the Earth’s magnetic field is BH=0.4×10−4T.
It is also given that the angle between the magnet and the magnetic meridian is 30∘. So, we can illustrate the condition as follow
Now, we can find the torque generated by using the formula τ=μ×BH=μBHsinθ, where μ is the magnets dipole moment, BH is the horizontal component of Earth’s magnetic field and θ is the angle between the magnet and the magnetic meridian.
Therefore, τ=2×0.4×10−4sin30∘=20.8×10−4=0.4×10−4Nm
Hence, option a is the correct answer.
Additional information:
An imaginary line that connects the magnetic north and south poles of the Earth is called a magnetic meridian. A compass needle when suspended freely on the Earth’s surface align itself parallel to the magnetic meridian. although there is some hindrance in their alignment due to the east to west longitude which become complete geodesic and thus magnetic declination comes into picture.
Note:
The orientation of the magnet or say its inclination changes as we move from equator to the poles and is called the angle of dip. This dip is caused by both horizontal and vertical components of the magnetic field. At poles, this dip is 90∘ and at the equator, the dip is 0∘.