NEET-XII-Physics
36: Permanent Magnets
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- Qstn #23A short magnet oscillates in an oscillation magnetometer with a time period of 0.10 s where the earth’s horizontal magnetic field is 24 μT. A downward current of 18 A is established in a vertical wire placed 20 cm east of the magnet. Find the new time period.Ans : Here,
Horizontal component of Earth's magnetic field, BH = 24 × 10-6 T
Time period of oscillation, T1 = 0.1 s
Downward current in the vertical wire, I = 18 A
Distance of wire from the magnet, r = 20 cm = 0.2 m
In the absence of the wire net magnetic field, BH = 24 × 10-6 T
When a current-carrying wire is placed near the magnet, the effective magnetic field gets changed.
Now the net magnetic field can be obtained by subtracting the magnetic field due to the wire from Earth's magnetic field.
B = BH - Bwire
Thus, the magnetic field due to the current-carrying wire `` \left({B}_{wire}\right)`` is given by
B = `` \frac{{\mu }_{0}I}{2\pi r}``
The net magnetic field `` \left(B\right)`` is given by
`` B=24\times {10}^{-6}-\frac{{\,\mathrm{\,\mu \,}}_{0}I}{2\,\mathrm{\,\pi \,}r}``
`` B=24\times {10}^{-6}-\frac{2\times {10}^{-7}\times 18}{0.2}``
`` B=\left(24-10\right)\times {10}^{-6}``
`` B=14\times {10}^{-6}``
Time period of the coil `` \left(T\right)`` is given by
`` T=2\,\mathrm{\,\pi \,}\sqrt{\frac{I}{M{B}_{\,\mathrm{\,H\,}}}}``
`` ``
Let T1 and T2 be the time periods of the coil in the absence of the wire and in the presence the wire respectively.
As time period is inversely proportional to magnetic field,
`` \frac{{T}_{1}}{{T}_{2}}=\sqrt{\frac{B}{{B}_{\,\mathrm{\,H\,}}}}``
`` \Rightarrow \frac{0.1}{{T}_{2}}=\sqrt{\frac{14\times {10}^{-6}}{24\times {10}^{-6}}}``
`` \Rightarrow {\left(\frac{0.1}{{T}_{2}}\right)}^{2}=\frac{14}{24}``
`` \Rightarrow {T}_{2}^{2}=\frac{0.01\times 14}{24}``
`` \Rightarrow {T}_{2}=0.076\,\mathrm{\,s\,}``
Page No 278:
- Qstn #24A bar magnet makes 40 oscillations per minute in an oscillation magnetometer. An identical magnet is demagnetized completely and is placed over the magnet in the magnetometer. Find the time taken for 40 oscillations by this combination. Neglect any induced magnetism.Ans : Given:
Frequency of oscillations of the bar magnet in the oscillation magnetometer, `` \nu `` = 40 min`` -1``
Time period for which the bar magnet is placed in the oscillation magnetometer, T1 = `` \frac{1}{40}\,\mathrm{\,min\,}``
The time period of oscillations of the bar magnet `` \left(T\right)`` is given by
`` T=2\,\mathrm{\,\pi \,}\sqrt{\frac{I}{M{B}_{H}}}``
`` ``
Here,
I = Moment of inertia
M = Magnetic moment of the bar magnet
BH = Horizontal component of the magnetic field
Now, let T2 be the time period for which the second demagnetised magnet is placed over the magnet.
As the second magnet is demagnetised, the combination will have the same values of M and BH as those for the single magnet. However, variation will be there in the value of I on placing the second demagnetised magnet.
`` \therefore \frac{{T}_{\mathit{1}}}{{T}_{\mathit{2}}}\mathit{=}\frac{{I}_{\mathit{1}}}{{I}_{\mathit{2}}}``
`` \Rightarrow \frac{1}{40{T}_{2}}=\sqrt{\frac{1}{2}}``
`` ``
`` \Rightarrow \frac{1}{1600{T}_{2}^{2}}=\frac{1}{2}``
`` \Rightarrow {T}_{2}^{2}=\frac{1}{800}``
`` \Rightarrow {T}_{2}=0.03536\,\mathrm{\,min\,}``
For 1 oscillation,
Time taken = 0.03536 min
For 40 oscillations,
Time taken = 0.03536 × 40
= 1.414 min`` =\sqrt{2}\,\mathrm{\,min\,}``
Page No 278:
- Qstn #25A short magnet makes 40 oscillations per minute when used in an oscillation magnetometer at a place where the earth’s horizontal magnetic field is 25 μT. Another short magnet of magnetic moment 1.6 A m2 is placed 20 cm east of the oscillating magnet. Find the new frequency of oscillation if the magnet has its north poleAns : Here,
Frequency of oscillations, `` \nu ``1 = 40 oscillations/min
Earth's horizontal magnetic field, BH = 25 μT
Magnetic moment of the second magnet, M = 1.6 A-m2
Distance at which another short magnet is placed, d = 20 cm = 0.2 m
- #25-atowards north andAns : For the north pole of the short magnet facing the north, frequency `` \left({v}_{1}\right)`` is given by
`` {v}_{1}=\frac{1}{2\,\mathrm{\,\pi \,}}\sqrt{\frac{M{B}_{H}}{I}}``
`` ``
`` ``
Here,
M = Magnetic moment of the magnet
I = Moment of inertia
BH = Horizontal component of the magnetic field
Now, let B be the magnetic field due to the short magnet.
When the north pole of the second magnet faces the north pole of the first magnet, the effective magnetic field `` \left({B}_{eff}\right)`` is given by
Beffective = BH `` -`` B
The new frequency of oscillations `` \left({v}_{2}\right)`` on placing the second magnet is given by
`` {v}_{\mathit{2}}\mathit{=}\frac{1}{2\,\mathrm{\,\pi \,}}\sqrt{\frac{M\left({B}_{H}\mathit{-}B\right)}{\mathit{1}}}``
`` ``
The magnetic field produced by the short magnet `` \left(B\right)`` is given by
`` B=\frac{{\,\mathrm{\,\mu \,}}_{0}}{4\,\mathrm{\,\pi \,}}\frac{m}{{d}^{\mathit{3}}}``
`` \Rightarrow B=\frac{{10}^{-7}\times 1.6}{8\times {10}^{-3}}=20\,\mathrm{\,\mu T\,}``
`` ``
Since the frequency is proportional to the magnetic field,
`` \frac{{v}_{\mathit{1}}}{{v}_{\mathit{2}}}=\sqrt{\frac{{B}_{H}}{{B}_{H}\mathit{-}B}}``
`` \Rightarrow \frac{40}{{v}_{2}}=\sqrt{\frac{25}{5}}``
`` \Rightarrow \frac{40}{{v}_{2}}=\sqrt{5}``
`` \Rightarrow {v}_{2}=\frac{40}{\sqrt{5}}=17.88``
`` =18\,\mathrm{\,oscillations\,}/\,\mathrm{\,min\,}``
- #25-btowards south.Ans : For the north pole facing the south,
`` {v}_{1}=\frac{1}{2\,\mathrm{\,\pi \,}}\sqrt{\frac{M{B}_{H}}{I}}``
`` \Rightarrow {v}_{2}=\frac{1}{2\,\mathrm{\,\pi \,}}\sqrt{\frac{M\left(B\mathit{+}{B}_{H}\right)}{\mathit{1}}}``
`` \Rightarrow \frac{{v}_{1}}{{v}_{2}}=\sqrt{\frac{{B}_{H}}{B\mathit{+}{B}_{H}}}``
`` \Rightarrow \frac{40}{{v}_{2}}=\sqrt{\frac{25}{45}}``
`` \Rightarrow {v}_{2}=\frac{40}{\sqrt{25/45}}=54\,\mathrm{\,oscillations\,}/\,\mathrm{\,min\,}``