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CBSE 12th Standard Physics Subject Moving Charges And Magnetism Ncert Exemplar 1 Mark Questions With Solution 2021

Published on: 23/05/2021

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Std: 12th Standard CBSE - Physics

chapter: - NCERT Exemplers

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Questions:

Generic 2

5 x 2 = 10

The horizontal component of the earth's magnetic field at a place is B and angle of dip is \(60^{ 0 }\) . What is the value of vertical component of earth's magnetic field at equator?

A short magnet oscillates with a time period 0.1 s at a place, where horizontal magnetic field is \(24 \ \mu \)T.
A downward current of 18 A is established in a vertical wire 20cm East of the magnet. What will be the new time period of the oscillator?

Three identical bar magnets are rivetted together at centre in the same plane as shown in the figure. This system is placed at rest in a system is placed at rest in a slowly varying magnetic field. It is found that the system of magnets does not show any motion. The North-South poles of one magnet is shown in the figure. Determine the poles of the remaining two.

A bar magnet is cut into two equal parts as shown in the Fig. (a) One part is now kept over the other such that, the P₂is above P₁ as shown in the Fig.(b).
If M is the magnetic moment of the original magnet, what would be the magnetic moment of new combination of magnets so formed?

Among Biot-Savart's law and Coulomb's law which one is angle dependent?

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Answers Key:

Generic 2

5 x 2 = 10

Here, there is no significance of the given value of B and angle of dip as the vertical component of magnetic field at equator is zero.

Initially,
\(T=2\pi \sqrt { \frac { I }{ 3{ B }^{ ' } } } and \ finally, \ { T }^{ ' }=2\pi \sqrt { \frac { I }{ m(B+{ B }^{ ' } } } \)
where, B^' = horizontal magnetic field = \(24\mu \)T
and B = magnetic field due to downward conductor
\(=\frac { { \mu }_{ 0 } }{ 4\pi } .\frac { 2i }{ a } \quad =18\mu \)T
\(\therefore \frac { { T }^{ ' } }{ T } =\sqrt { \frac { { B }^{ ' } }{ B+{ B }^{ ' } } } \Rightarrow \frac { { T }^{ ' } }{ 0.1 } =\sqrt { \frac { 24 }{ 18+24 } } \Rightarrow { T }^{ ' }=0.076\)s

The system of magnets will be in stable equilibrium, if the net force on the system is zero and net torque on it is also zero. It will be possible, if the poles of the remaining two magnets is as shown in the figure.

When the bar magnet is cut into two equal parts, as shown in the Fig. (a), P₁ behaves as N and P₂ behaves as S and magnetic moment of each part of magnet becomes \(\frac { M }{ 2 } \) . When pole P₂ is placed over pole P₁ as shown in Fig. (b), the Magnetic moment of the combination is zero i.e. \(\frac { M }{ 2 } -\frac { M }{ 2 } =0.\)

Biot-Savart's law is angle dependent law.