CBSE 12th Standard Physics Subject Moving Charges And Magnetism HOT Questions 3 Mark Questions 2021
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CBSE 12th Standard Physics Subject Moving Charges And Magnetism HOT Questions 3 Mark Questions 2021
12th Standard CBSE
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Reg.No. :
Physics
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A short bar magnet placed with its axis at \(30°\)to a uniform magnetic field of 0.2T experiences a torque of 0.06Nm.
(i) Calculate magnetic moment of the magnet and
(ii) Find out what orientation of the magnet corresponds to a stable equilibrium in the magnetic field.(a) -
An iron sample having mass 8.4 kg is repeatedly taken over cycles of magnetisation and demagnetization at a frequency of 50Hz.It is found that energy of 3.2 J is dissipated as heat in the sample in 30 minutes. If the density of iron is \(7200kg{ m }^{ -3 }\) calculate the value of energy dissipated per unit volume per cycle in the iron sample.
(a) -
To increase the current sensitivity of a moving coil galvanometer by 50%, its resistance is increased so that the new resistance becomes twice its initial resistance. By what factor does its voltage sensitivity change?
(a) -
A galvanometer having 30 divisions has a current sensitivity of \(20\mu \)A per division. It has a resistance of \(25\Omega \). How will you convert it into an ammeter measuring upto I A? How will you convert in this ammeter into a voltmeter reading upto 1 V?
(a) -
Relative permeability of a material, \(\mu\)r = 0.5. Identify the nature of the magnetic material and write its relation to magnetic susceptibility.
(a)
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CBSE 12th Standard Physics Subject Moving Charges And Magnetism HOT Questions 3 Mark Questions 2021 Answer Keys
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(i) Given
\(d B=0.2T,\theta =30°,\tau =0.06Nm\)
\( \tau =MB\sin { \theta }\)
\( \therefore d M=\frac { \tau }{ B\sin { \theta } } =\frac { 0.06 }{ 0.2\times \sin { 60° } } =0.06{ Am }^{ 2 }\)
(ii) Potential energy of magnetic dipole in a uniform magnetic field B is given by
\(U=-MB\cos { \theta } \)
In stable equilibrium, the potential energy is minimum i.e.\(\cos { \theta } =1 \ or \ \theta =0°\)
So for stable equilibrium, the magnet must be aligned with its magnetic moment parallel to the magnetic field. -
Given \(m=8.4kg,v=50Hz,\rho =7200kg{ m }^{ -3 }\)
\(\therefore volume=\frac { m }{ \rho } =\frac { 8.4 }{ 7200 } =1.167\times { 10 }^{ -3 }{ m }^{ 3 }\)
Energy dissipated in 30 minutes = \(3.2\times { 10 }^{ 4 }J\)
Energy dissipated/sec = \(\frac { 3.2\times { 10 }^{ 4 } }{ 30\times 60 } =17.78{ Js }^{ -1 }\)
So energy dissipated/vol/cycle
\(=\frac { 17.78 }{ 1.167\times { 10 }^{ -3 }\times 50 }\)
\(=304.7{ Jm }^{ -3 }{ cycle }^{ -1 }\) -
Given \({ I }_{ S }^{ ' }={ I }_{ S }+\frac { 50 }{ 100 } { I }_{ S }=1.5{ I }_{ S }\)
And \({ R }^{ ' }=2R\)
Initial voltage sensitivity, \({ V }_{ S }^{ ' }=\frac { { I }_{ S } }{ R } \)
New, voltage sensitivity, \({ V }_{ S }^{ ' }=\frac { { I }_{ S }^{ ' } }{ 2R } \)
\(=1.5\frac { { I }_{ S } }{ 2R } =0.75V_{ S }\)
% decrease in voltage sensitivity
\(=\frac { V_{ S }-{ V }_{ S } }{ V_{ S } } \times 100=\frac { V_{ S }-0.75V_{ S } }{ V_{ S } } \times 100=25%\) -
Given current sensitivity
\({ I }_{ S }=20\mu A{ div }^{ -1 }=20\times { 10 }^{ -6 }A{ div }^{ -1 }\)
The galvanometer ha 30 divisions, so current for full-scale deflection
,\({ I }_{ g }=30\times 20\times { 10 }^{ -6 }A=6\times { 10 }^{ -4 }A\)
When the galvanometer is to be converted into an ammeter, thus the value of shunt required
\(S=\frac { { I }_{ g } }{ I-{ I }_{ g } } G=\frac { 6\times { 10 }^{ -7 } }{ 1-6\times { 10 }^{ -4 } } \times 25\) (in parallel)
\(\\ =0.015\Omega \)
When the galvanometer is to be converted into a voltmeter, then the value of resistance R is given by
\(R=\frac { V }{ I } -{ R }_{ A }\)
\(=\frac { 1 }{ 1 } -\frac { GS }{ G+S } =1-\frac { 25\times 0.015 }{ 25+0.015 } \)
\(=1-0.015=0.985\Omega \) (in series) -
Diamagnetic material
\({ \mu }_{ r }=1+{ x }_{ m }\)
