36: Permanent Magnets : Neet-xii-physics

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NEET-XII-Physics

36: Permanent Magnets

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Qstn #8
A magnetic dipole of magnetic moment 1.44 A m² is placed horizontally with the north pole pointing towards north. Find the position of the neutral point if the horizontal component of the earth’s magnetic field is 18 μT.
Ans : Given:
Magnetic moment of the magnetic dipole, M = 1.44 A-m²
Horizontal component of Earth's magnetic field, B_H= 18 μT
We know that for a magnetic dipole with pole pointing to the north, the neutral point always lies in the broadside-on position.
Let d be the perpendicular distance of the neutral point from mid point of the magnet.
The magnetic field due to the dipole at the broadside-on position (B) is given by
`` \stackrel{\mathit{\to }}{B}=\frac{{\,\mathrm{\,\mu \,}}_{0}M}{4\,\mathrm{\,\pi \,}{d}^{3}}``
`` ``
`` ``
This magnetic field strength should be equal to the horizontal component of Earth's magnetic field at that point, that is, B_Hdue to Earth.
Thus,
`` \stackrel{\mathit{\to }}{B}=\frac{{\,\mathrm{\,\mu \,}}_{0}M}{4\,\mathrm{\,\pi \,}{d}^{3}}``
`` \Rightarrow B=\frac{{10}^{-7}\times 1.44}{{d}^{3}}``
`` \Rightarrow 18\times {10}^{-6}=\frac{{10}^{-7}\times 1.44}{{d}^{3}}``
`` \Rightarrow {d}^{\mathit{3}}=8\times {10}^{-3}``
`` \Rightarrow d=2\times {10}^{-1}=20\,\mathrm{\,cm\,}``
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Qstn #9
A magnetic dipole of magnetic moment 0.72 A m² is placed horizontally with the north pole pointing towards south. Find the position of the neutral point if the horizontal component of the earth’s magnetic field is 18 μT.
Ans : Given:
Magnetic moment of the magnetic dipole, M = 0.72 Am²
Horizontal component of Earth's magnetic field, B_H = 18 μT
Let d be the distance of the neutral point from the south of the dipole.
When the magnet is such that its north pole faces the geographic south of Earth, the neutral point lies along the axial line of the magnet.
Thus, the magnetic field on the axial point of the dipole (B) is given by
`` B=\frac{{\,\mathrm{\,\mu \,}}_{0}}{4\,\mathrm{\,\pi \,}}\frac{2M}{{d}^{3}}``
`` ``
This magnetic field strength should be equal to the horizontal component of Earth's magnetic field.
Thus,
`` \frac{{10}^{-7}\times 2\times 0.72}{{d}^{3}}=18\times {10}^{-6}``
`` \Rightarrow {d}^{3}=\frac{2\times 0.72\times {10}^{-7}}{18\times {10}^{-6}}``
`` \Rightarrow d={\left(\frac{8\times {10}^{-9}}{{10}^{-6}}\right)}^{1/3}``
`` \Rightarrow d=2\times {10}^{-1}\,\mathrm{\,m\,}=20\,\mathrm{\,cm\,}``
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Qstn #10
A magnetic dipole of magnetic moment
0.722 A m2is placed horizontally with the north pole pointing towards east. Find the position of the neutral point if the horizontal component of the earth’s magnetic field is 18 μT.
Ans : Given:
Magnetic moment of magnetic dipole, M `` =0.72\sqrt{2}\,\mathrm{\,A\,}-{\,\mathrm{\,m\,}}^{2}``
Horizontal component of the earth's magnetic field, B_H = 18 μT
Let d be the distance of neutral point from the dipole.
Magnetic field due to the bar magnet `` \left(B\right)`` on the equatorial line of the dipole is given by,
`` B=\frac{{\mu }_{0}}{4\pi }\frac{M}{{d}^{3}}``
`` \Rightarrow \frac{4\,\mathrm{\,\pi \,}\times {10}^{-7}}{4\,\mathrm{\,\pi \,}}\times \frac{0.72\sqrt{2}}{{d}^{3}}=18\times {10}^{-6}``
`` \Rightarrow {d}^{3}=\frac{0.72\times 1.414\times {10}^{-7}}{18\times {10}^{-6}}``
`` \Rightarrow {d}^{3}=0.005656``
`` \Rightarrow d=\sqrt[3]{0.005656}``
`` \Rightarrow d\approx 0.2\,\mathrm{\,m\,}=20\,\mathrm{\,cm\,}``
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Qstn #11
The magnetic moment of the assumed dipole at the earth’s centre is 8.0 × 10²² A m². Calculate the magnetic field B at the geomagnetic poles of the earth. Radius of the earth is 6400 km.
Ans : Given:
Magnetic moment of dipole at Earth's centre, M = 8.0 × 10²² Am²
Radius of Earth, d = 6400 km
The geomagnetic pole is at the end of the position (axial) of Earth.
Thus, the magnetic field at the axial point of the dipole `` \left(B\right)`` is given by
`` B=\frac{{\,\mathrm{\,\mu \,}}_{0}}{4\,\mathrm{\,\pi \,}}\frac{\mathit{2}M}{{d}^{\mathit{3}}}``
`` \Rightarrow B=\frac{{10}^{-7}\times 2\times 8\times {10}^{22}}{{\left(64\right)}^{3}\times {10}^{15}}``
`` \Rightarrow B=6\times {10}^{-5}\,\mathrm{\,T\,}=60\,\mathrm{\,\mu T\,}``
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Qstn #12
If the earth’s magnetic field has a magnitude 3.4 × 10^-5 T at the magnetic equator of the earth, what would be its value at the earth’s geomagnetic poles?
Ans : Given:
Magnetic field at the magnetic equator, B = 3.4 × 10^-5 T
Let M be the magnetic moment of Earth's magnetic dipole and R be the distance of the observation point from the centre of Earth's magnetic dipole.
As the point on the magnetic equator is on the equatorial position of Earth's magnet, the magnetic field at the equatorial point `` \left(B\right)`` is given by
`` B=\frac{{\mu }_{0}}{4\,\mathrm{\,\pi \,}}\frac{M}{{R}^{3}}``
`` \Rightarrow \frac{{\mu }_{0}}{4\,\mathrm{\,\pi \,}}\times \frac{M}{{R}^{\mathit{3}}}=3.4\times {10}^{-5}``
`` \Rightarrow M=\frac{3.4\times {10}^{-5}\times {\,\mathrm{\,R\,}}^{3}\times 4\,\mathrm{\,\pi \,}}{4\,\mathrm{\,\pi \,}\times {10}^{-7}}``
`` \Rightarrow M=3.4\times {10}^{2}{R}^{3}``
`` ``
As the magnetic field on Earth's geomagnetic poles lies on the axial point of the magnetic dipole, the magnetic field at the axial point `` \left({B}_{1}\right)`` is given by
`` {\stackrel{\mathit{\to }}{B}}_{1}=\frac{{\,\mathrm{\,\mu \,}}_{0}}{4\,\mathrm{\,\pi \,}}\times \frac{2M}{{R}^{3}}``
`` \Rightarrow {B}_{1}=\frac{{10}^{-7}\times 2\times 3.4\times {10}^{2}\times {R}^{3}}{{R}^{3}}``
`` \Rightarrow {B}_{1}=6.8\times {10}^{-5}\,\mathrm{\,T\,}``
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Qstn #13
The magnetic field due to the earth has a horizontal component of 26 μT at a place where the dip is 60°. Find the vertical component and the magnitude of the field.
Ans : Given:
Horizontal component of Earth's magnetic field, B = 26 μT
Angle of dip, δ = 60°
The horizontal component of Earth's magnetic field `` \left({B}_{H}\right)`` is given by
B_H = Bcosδ
Here,
B = Total magnetic field of Earth
On substituting the respective values, we get
`` 26\times {10}^{-6}=B\times \frac{1}{2}``
`` \Rightarrow B=52\times {10}^{-6}=52\,\mathrm{\,\mu T\,}``
`` ``
The vertical component of Earth's magnetic field `` \left({B}_{y}\right)`` is given by
`` {B}_{y}\mathit{=}B\,\mathrm{\,sin\,}\delta ``
`` \Rightarrow {B}_{y}=52\times {10}^{-6}\times \frac{\sqrt{3}}{2}``
`` \Rightarrow {B}_{y}=44.98\,\mathrm{\,\mu T\,}=45\,\mathrm{\,\mu T\,}``
Page No 277:

Qstn #14
A magnetic needle is free to rotate in a vertical plane which makes an angle of 60° with the magnetic meridian. If the needle stays in a direction making an angle of
tan-1 23with the horizontal, what would be the dip at that place?
Ans : Given:
Angle made by the magnetic meridian with the plane of rotation of the needle, `` \theta =60°``
Angle made by the needle with the horizontal, `` {\delta }_{1}={\,\mathrm{\,tan\,}}^{-1}\left(\frac{2}{\sqrt{3}}\right)``
If `` \delta `` is the angle of dip, then
`` \,\mathrm{\,tan\,}{\delta }_{1}=\frac{\,\mathrm{\,tan\,}\delta }{\,\mathrm{\,cos\,}\theta }``
`` \Rightarrow \,\mathrm{\,tan\,}\delta =\,\mathrm{\,tan\,}{\delta }_{1}\,\mathrm{\,cos\,}\theta ``
`` \Rightarrow \,\mathrm{\,tan\,}\delta =\,\mathrm{\,tan\,}\left({\,\mathrm{\,tan\,}}^{-1}\frac{2}{\sqrt{3}}\right)\,\mathrm{\,cos\,}60°``
`` \Rightarrow \,\mathrm{\,tan\,}\delta =\frac{2}{\sqrt{3}}\times \frac{1}{2}=\frac{1}{\sqrt{3}}``
`` \Rightarrow \delta =30°``
Page No 278:

Qstn #15
The needle of a dip circle shows an apparent dip of 45° in a particular position and 53° when the circle is rotated through 90°. Find the true dip.
Ans : Given:
Apparent dip shown by the needle of the dip circle, δ₁ = 45°
Dip shown by the needle of the dip circle on rotating the circle, δ₂= 53°
True dip `` \left(\delta \right)`` is given by
Cot² δ = Cot² δ₁ + Cot² δ₂
`` \Rightarrow ``Cot²δ = Cot² 45° + Cot² 53°
`` \Rightarrow ``Cot²δ = 1.56
`` \Rightarrow ``Cot²δ = 1.56
`` \Rightarrow ``δ = 38.6° ≈ 39°
Page No 278:

Qstn #16
A tangent galvanometer shows a deflection of 45° when 10 mA of current is passed through it. If the horizontal component of the earth’s magnetic field is B_H = 3.6 × 10^-5 T and radius of the coil is 10 cm, find the number of turns in the coil.
Ans : Given:
Horizontal component of Earth's magnetic field, B_H = 3.6 × 10^-5 T
Deflection shown by the tangent galvanometer, θ = 45°
Current through the galvanometer, I = 10 mA = 10^-2 A
Radius of the coil, r = 10 cm = 0.1 m
Number of turns in the coil, n = ?
We know,
`` {B}_{\,\mathrm{\,H\,}}\,\mathrm{\,tan\,}\theta =\frac{{\,\mathrm{\,\mu \,}}_{0}In}{\mathit{2}r}``
`` \Rightarrow n=\frac{{B}_{\,\mathrm{\,H\,}}\,\mathrm{\,tan\,}\theta \times 2r}{{\,\mathrm{\,\mu \,}}_{0}\,\mathrm{\,I\,}}``
`` \Rightarrow n=\frac{3.6\times {10}^{-5}\times 2\times 1\times {10}^{-1}}{4\,\mathrm{\,\pi \,}\times {10}^{-7}\times {10}^{-2}}``
`` \Rightarrow n=0.57332\times {10}^{3}=573``
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Qstn #17
A moving-coil galvanometer has a 50-turn coil of size 2 cm × 2 cm. It is suspended between the magnetic poles producing a magnetic field of 0.5 T. Find the torque on the coil due to the magnetic field when a current of 20 mA passes through it.
Ans : Given:
Number of turns in the coil, n = 50
Area of the cross section of the coil, A = 2 cm × 2 cm = 2 × 2 × 10^-4 m²
Current flowing through the coil, I = 20 × 10^-3 A
Magnetic field strength due to the presence of the poles, B = 0.5 T
The torque experienced by the coil placed in an external magnetic field `` \left(\tau \right)`` is given by
`` \tau \mathit{=}nI\left(\stackrel{\mathit{\to }}{A}\mathit{\times }\stackrel{\mathit{\to }}{B}\right)``
`` \Rightarrow \tau \mathit{=}nIAB\,\mathrm{\,sin\,}90°``
`` \Rightarrow \tau =50\times 20\times {10}^{-3}\times 4\times {10}^{-4}\times 0.5``
`` \Rightarrow \tau =2\times {10}^{-4}\,\mathrm{\,N\,}-\,\mathrm{\,m\,}``
Page No 278:

Qstn #18
A short magnet produces a deflection of 37° in a deflection magnetometer in Tan-A position when placed at a separation of 10 cm from the needle. Find the ratio of the magnetic moment of the magnet to the earth’s horizontal magnetic field.
Ans : Given:
Deflection in the magnetometer in the given position when placed in the magnetic field of a short magnet, θ = 37°
Separation between the magnet and the needle, d = 10 cm = 0.1 m
Let M be the magnetic moment of the magnet and B_H be Earth's horizontal magnetic field.
According to the magnetometer theory,
`` \frac{M}{{B}_{H}}=\frac{4\,\mathrm{\,\pi \,}}{{\,\mathrm{\,\mu \,}}_{0}}\frac{{\left({d}^{2}-{l}^{2}\right)}^{2}}{2d}\,\mathrm{\,tan\,}\theta ``
`` \,\mathrm{\,For\,}\,\mathrm{\,the\,}\,\mathrm{\,short\,}\,\mathrm{\,magnet\,},``
`` \frac{M}{{B}_{H}}=\frac{4\,\mathrm{\,\pi \,}}{{\,\mathrm{\,\mu \,}}_{0}}\times \frac{{d}^{4}}{2d}\,\mathrm{\,tan\,}\theta ``
`` \Rightarrow \frac{M}{{B}_{H}}=\frac{4\,\mathrm{\,\pi \,}}{4\,\mathrm{\,\pi \,}\times {10}^{-7}}\times \frac{{\left(0.1\right)}^{3}}{2}\times \,\mathrm{\,tan\,}37°``
`` \Rightarrow \frac{M}{{B}_{H}}=0.5\times 0.75\times 1\times {10}^{4}``
`` \Rightarrow \frac{M}{{B}_{H}}=3.75\times {10}^{3}\,\mathrm{\,A\,}-{\,\mathrm{\,m\,}}^{2}/\,\mathrm{\,T\,}``
Page No 278:

Qstn #19
The magnetometer of the previous problem is used with the same magnet in Tan-B position. Where should the magnet be placed to produce a 37° deflection of the needle?
Ans : Given:
Deflection in the magnetometer in the given position when placed in the magnetic field of a short magnet, θ = 37°
Separation between the magnet and the needle, d = 10 cm = 0.1 m
Let M be the magnetic moment of the magnet and B_H be Earth's horizontal magnetic field.
According to the magnetometer theory,
`` \frac{M}{{B}_{H}}=\frac{4\,\mathrm{\,\pi \,}}{{\,\mathrm{\,\mu \,}}_{0}}\frac{{\left({d}^{2}-{l}^{2}\right)}^{2}}{2d}\,\mathrm{\,tan\,}\theta ``
`` \,\mathrm{\,For\,}\,\mathrm{\,the\,}\,\mathrm{\,short\,}\,\mathrm{\,magnet\,},``
`` \frac{M}{{B}_{H}}=\frac{4\,\mathrm{\,\pi \,}}{{\,\mathrm{\,\mu \,}}_{0}}\times \frac{{d}^{4}}{2d}\,\mathrm{\,tan\,}\theta ``
`` \Rightarrow \frac{M}{{B}_{H}}=\frac{4\,\mathrm{\,\pi \,}}{4\,\mathrm{\,\pi \,}\times {10}^{-7}}\times \frac{{\left(0.1\right)}^{3}}{2}\times \,\mathrm{\,tan\,}37°``
`` \Rightarrow \frac{M}{{B}_{H}}=0.5\times 0.75\times 1\times {10}^{4}``
`` \Rightarrow \frac{M}{{B}_{H}}=3.75\times {10}^{3}\,\mathrm{\,A\,}-{\,\mathrm{\,m\,}}^{2}/\,\mathrm{\,T\,}``
= 3.75 × 10³ A-m²/T
Deflection in the magnetometer θ = 37°
From the magnetometer theory in Tan-B position, we have
`` \frac{M}{{B}_{H}}=\frac{4\,\mathrm{\,\pi \,}}{{\,\mathrm{\,\mu \,}}_{0}}{\left({d}^{2}+{l}^{2}\right)}^{3/2}\,\mathrm{\,tan\,}\theta ``
Since for the short magnet l << d, we can neglect l w.r.t. d.
Now,
`` \frac{M}{{B}_{H}}=\frac{4\,\mathrm{\,\pi \,}}{{\,\mathrm{\,\mu \,}}_{0}}{d}^{3}\,\mathrm{\,tan\,}\theta ``
`` \Rightarrow 3.75\times {10}^{3}=\frac{1}{{10}^{-7}}\times {d}^{\mathit{3}}\times 0.75``
`` \Rightarrow {d}^{3}=\frac{3.75\times {10}^{3}\times {10}^{-7}}{0.75}=5\times {10}^{-4}``
`` \Rightarrow d=\sqrt[3]{5\times {10}^{-4}}``
`` \Rightarrow d=0.079\,\mathrm{\,m\,}=7.9\,\mathrm{\,cm\,}``
Magnet will be at 7.9 cm from the centre.
Page No 278:

Qstn #20
A deflection magnetometer is placed with its arms in north-south direction. How and where should a short magnet having M/B_H = 40 A m² T^-1 be placed so that the needle can stay in any position?
Ans : Given:
Ratio, `` \frac{M}{{B}_{H}}=40{\,\mathrm{\,Am\,}}^{2}/\,\mathrm{\,T\,}``
Since the magnet is short, l can be neglected.
So, using the formula for `` \frac{\mathit{M}}{{\mathit{B}}_{\mathit{H}}}`` from the magnetometer theory and substituting all values, we get
`` \frac{M}{{B}_{H}}=\frac{4\,\mathrm{\,\pi \,}}{{\,\mathrm{\,\mu \,}}_{0}}\frac{{d}^{3}}{2}=40``
`` \Rightarrow {d}^{3}=40\times {10}^{-7}\times 2``
`` \Rightarrow {d}^{3}=8\times {10}^{-6}``
`` \Rightarrow d=2\times {10}^{-2}\,\mathrm{\,m\,}=2\,\mathrm{\,cm\,}``
Thus, the magnet should be placed in such a way that its north pole points towards the south and it is 2 cm away from the needle.
Page No 278:

Qstn #21
A bar magnet takes π/10 second the complete one oscillation in an oscillation magnetometer. The moment of inertia of the magnet about the axis of rotation is 1.2 × 10^-4 kg m² and the earth’s horizontal magnetic field is 30 μT. Find the magnetic moment of the magnet.
Ans : Given:
Time taken by the bar magnet to complete one oscillation, T = `` \frac{\,\mathrm{\,\pi \,}}{10}\,\mathrm{\,s\,}``
Moment of inertia of the magnet about the axis of rotation, I = 1.2 × 10^-4 kgm²
Horizontal component of Earth's magnetic field, B_H = 30 μT
Time period of oscillating magnetometer `` \left(T\right)`` is given by
`` T=2\,\mathrm{\,\pi \,}\sqrt{\frac{I}{M{B}_{H}}}``
`` ``
`` ``
Here,
M = Magnetic moment of the magnet
On substituting the respective values, we get
`` \frac{\,\mathrm{\,\pi \,}}{10}=2\,\mathrm{\,\pi \,}\sqrt{\frac{1.2\times {10}^{-4}}{\,\mathrm{\,M\,}\times 30\times {10}^{-6}}}``
`` \Rightarrow {\left(\frac{1}{20}\right)}^{2}=\frac{1.2\times {10}^{-4}}{M\times 30\times {10}^{-6}}``
`` \Rightarrow M=\frac{1.2\times {10}^{-4}\times 400}{30\times {10}^{-6}}``
`` \Rightarrow M=16\times {10}^{2}\,\mathrm{\,A\,}-{\,\mathrm{\,m\,}}^{2}``
`` \Rightarrow M=1600\,\mathrm{\,A\,}-{\,\mathrm{\,m\,}}^{2}``
Page No 278:

Qstn #22
The combination of two bar magnets makes 10 oscillations per second in an oscillation magnetometer when like poles are tied together and 2 oscillations per second when unlike poles are tied together. Find the ratio of the magnetic moments of the magnets. Neglect any induced magnetism.
Ans : Given:
Number of oscillations per second made by the combination of bar magnets with like poles, `` {\nu }_{1}``= 10 s^{`` -1``}
Number of oscillations per second made by the combination of bar magnets with unlike poles, `` {\nu }_{2}`` = 2 s^{`` -1``}
The frequency of oscillations in the magnetometer `` \left(\nu \right)`` is given by
`` \upsilon =\frac{1}{2\,\mathrm{\,\pi \,}}\sqrt{\frac{M{B}_{\,\mathrm{\,H\,}}}{I}}``
When like poles are tied together, the effective magnetic moment is
M = M₁ - M₂
When unlike poles are tied together, the effective magnetic moment is
M = M₁ + `` {M}_{2}``
As the frequency of oscillations is directly proportional to the magnetic moment,
`` \frac{{\upsilon }_{1}}{{\upsilon }_{2}}=\sqrt{\frac{{M}_{1}-M{\hspace{0.17em}}_{2}}{{M}_{1}+{M}_{2}}}``
`` \Rightarrow {\left(\frac{10}{2}\right)}^{2}=\frac{{M}_{1}-{M}_{2}}{{M}_{1}+{M}_{2}}``
`` \Rightarrow \frac{25}{1}=\frac{{M}_{1}-{M}_{2}}{{M}_{1}+{M}_{2}}``
`` \Rightarrow \frac{25+1}{25-1}=\frac{{M}_{1}-{M}_{2}+{M}_{1}+{M}_{2}}{{M}_{1}-{M}_{2}-{M}_{1}-{M}_{2}}``
`` \Rightarrow \frac{26}{24}=\frac{2{M}_{1}}{-2{M}_{2}}``
`` \Rightarrow \frac{{M}_{1}}{{M}_{2}}=-\frac{26}{24}=-\frac{13}{12}``
Hence, the ratio of the effective magnetic moment is `` -\frac{13}{12}``.
Page No 278: