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Question: A short bar magnet of magnetic moment \(5.25 \times {10^{ - 2}}J{T^{ - 1}}\) is placed with its axis...

A short bar magnet of magnetic moment 5.25×10−2JT−15.25 \times {10^{ - 2}}J{T^{ - 1}} is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45∘{45^ \circ } with earth’s field on (a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42G0.42G. Ignore the length of the magnet in comparison to the distances involved.

Explanation

Solution

Here first we have to find the distances for the normal bisector and the axis using the equation for the horizontal component of the earth’s magnetism.

Complete step by step answer:
Given,
Resultant magnetic field is inclined at 45∘{45^ \circ }
Magnetic moment, M=5.25×10−2JT−1M = 5.25 \times {10^{ - 2}}J{T^{ - 1}}
Magnitude of the earth’s field, B=0.42G=0.42×10−4TB = 0.42G = 0.42 \times {10^{ - 4}}T
Case (a):
According to question, the resultant field is inclined at 45∘{45^ \circ }.
So,tan⁡θ=BHB2 θ=45∘ tan⁡θ=1=BHB BH=B=μ∘M4πr3 \tan \theta = \dfrac{{{B_H}}} {B}\sqrt 2 \\\ \theta = {45^ \circ } \\\ \tan \theta = 1 = \dfrac{{{B_H}}} {B} \\\ {B_H} = B = \dfrac{{{\mu _ \circ }M}}{{4\pi {r^3}}} \\\
Therefore, 0.42×10−4=10−7×5.25×10−2r3r3=1.25×10−5=125×10−6=˚5×10−2 m=5 cm 0.42 \times {10^{ - 4}} = {10^{ - 7}} \times \dfrac{{5.25 \times {{10}^{ - 2}}}} {{{r^3}}} \\\\{r^3} = 1.25 \times {10^{ - 5}} = 125 \times {10^{ - 6}} \\\r = 5 \times {10^{ - 2}}\,m = 5\,cm \\\
Hence, the distance from the centre of the magnet, the resultant field is inclined at 45∘{45^ \circ } with earth’s field on its normal bisector is 5 cm5\,cm.
For case (b):
At axis of magnet, tan⁡45∘=1=BBH\tan {45^ \circ } = 1 = \dfrac{B}{{{B_H}}}
BH=B=μ∘2M4πr3 ⇒0.42×10−4=10−7×2×5.25×102r3 r3=25×10−5 m=˚6.3 cm {B_H} = B = \dfrac{{{\mu _ \circ }2M}}{{4\pi {r^3}}} \\\ \Rightarrow 0.42 \times {10^{ - 4}} = {10^{ - 7}} \times \dfrac{{2 \times 5.25 \times {{10}^{ 2}}}}{{{r^3}}} \\\ {r^3} = 25 \times {10^{ - 5}}\,m \\\r = 6.3\,cm \\\
Hence, the distance from the centre of the magnet, the resultant field is inclined at 45∘{45^ \circ } with earth’s field on its axis 6.3 cm6.3\,cm.

Additional information:
At the surface of the Earth, the planet’s magnetic field can be applied in multiple ways. There are the elements that are responsible at a given position for the amplitude and orientation of the earth’s magnetic field:
(1) Magnetic declination:
The angle created by the magnetic meridian with the geographic meridian is known as magnetic declination. The geographic meridian here is defined as the plane that passes through the earth’s north and south poles.
(2) The angle of dip or magnetic inclination:
The angle with the horizontal axis produced by the north pole of the needle is known as the angle dip or magnetic inclination.
(3) Horizontal component of the earth’s magnetic field:
The horizontal component is the horizontal orientation component of the magnetic field of the earth in the magnetic meridian.

Note: We have to be careful while calculating the distances for the normal bisector and axis because the formula for horizontal components are different. For the axis the horizontal component is two times the magnetic moment.