CBSE 12th Standard Physics Subject Moving Charges And Magnetism Ncert Exemplar 3 Mark Questions 2021
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CBSE 12th Standard Physics Subject Moving Charges And Magnetism Ncert Exemplar 3 Mark Questions 2021
12th Standard CBSE
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Reg.No. :
Physics
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A current 10 A is flowing East to West in a long wire kept horizontally in the East-West direction. Find the magnitude and direction of magnetic field in a horizontal plane at a distance of 10 cm North.
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A cyclotron oscillator frequency is 10 MHz. What should be the operating magnetic field for accelerating \(\alpha \) -particle? If the radius of the dees is 50 cm, what is the kinetic energy in MeV of the \(\alpha \)-particle beam produced by the accelerator?
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Two identical magnets with a length 100cm are arranged freely with their like like poles facing in a vertical glass tube.The upper magnet hangs in air above the lower one so that the distance between the nearest poles of the magnet is 3mm. If the pole strength of the pole of these magnets is 6.64 A-m, then determine the force between the two magnets.
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Determine the magnitude of the equatorial fields due to a bar magnet of length 6 cm at a distance of 60 cm from its mid-point. The magnetic moment of the bar magnet is 0.60A-m2
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A Straight solenoid of length 50 cm has 1000 turns per metre and a mean cross-sectional area of 2 x 10-4 m2. It is placed with its axis at 300, with a uniform magnetic field of 0.32 T. Find the torque acting on the solenoid when a current of 2 A is passed through it.
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CBSE 12th Standard Physics Subject Moving Charges And Magnetism Ncert Exemplar 3 Mark Questions 2021 Answer Keys
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Given, current, (I)= 10 A (East to West)
Distance, (r) =10 cm = 10 x 10-2 m
Magnetic field, \(|B|=?\)
The magnitude of magnetic field \(|B|\) for infinite length of
\(wire=\frac { { \mu }_{ 0 }I }{ 2\pi r } \)
\(\Rightarrow \ |B|=\cfrac { 4\pi \times{ 10 }^{ -7 } \times 10 }{ 2\times \pi \times10\times { 10 }^{ -2 } } =2\times{ 10 }^{ -5 }T\)
The direction of magnetic field is given by right hand thumb rule or Maxwell's cork screw rule. So, the direction of magnetic field at point 10 cm North due to flowing current is perpendicularly inwards to the plane of paper. -
Given, V = 10 MHz = 107 Hz
r0 = 50 cm = 0.50 m
m\(\alpha\) = 4.0028 x 1.66 x 10-27kg
= 6.645 x 10-27 kg
q = 2e.= 2 x 1.6 x10 -19 = 3.2 x 10-19C
\(\because \ v=\frac{B q}{2 \pi m_{\alpha}}\)
\(\therefore \ B=\frac{2 \pi m_{\alpha} v}{q}=\frac{2 \times 22}{7} \times \frac{6.645 \times 10^{-27} \times 10^{7}}{3.2 \times 10^{-19}}\)
= 1.305 T
Maximum kinetic energy is given by
\(E_{\max }=\frac{B^{2} q^{2} r^{2}}{2 m_{\alpha}}=\frac{(1.305)^{2} \times(3.2)^{2} \times 10^{-38} \times 0.25}{2 \times 6.645 \times 10^{-27} \times 1.6 \times 10^{-13}} \mathrm{MeV}\)
= 20.5 MeV -
Given, pole strength, m = 6.64 A-m
r = 3 mm = 3 x 10-3 m [ m1 = m2 = m3]
Since, force, F = \(F=\frac { { \mu }_{ 0 } }{ 4\pi } \times \frac { { m }_{ 1 }{ m }_{ 2 } }{ { r }^{ 2 } } \)
\(F=\frac { { \mu }_{ 0 } }{ 4\pi } \times \frac { _{ { m }^{ 2 } } }{ { r }^{ 2 } } \)
\(=\frac { { \mu }_{ 0 } }{ 4\pi } \times \frac { { \left( 6.64 \right) }^{ 2 } }{ { \left( { 3\times }{ 10 }^{ -3 } \right) }^{ 2 } } \)
\(=\frac { { 4\pi \times }{ 10 }^{ -7 } }{ 4\pi } \times \frac { 44.0896 }{ 9\times { 10 }^{ -6 } } ={ 10 }^{ -1 }\times 4.8988\)
F = 0.49N -
Given, magnetic length of bar magnet, 2l = 6 cm
l = 3 cm = 3 x 10-2 m
Distance moment, M = 0.60 A-m2
\(=\frac { { \mu }_{ 0 }M }{ 4\pi { d }^{ 3 } } \)
\(=\frac { 4\pi \times { 10 }^{ -7 }\times 0.60 }{ 4\pi { \times \left( 0.6 \right) }^{ 3 } }\)
B2 = 2.7 x 10-7T -
Given, l = 50cm
Number of turns per metre = 1000
Total number of turns (N) = 1000 x 1/2 = 500
Area, A = 2 x 10-4 m2
Currecnt, I = 2 A
Magnetic field, B = 0.32 T
Torque, = \(MB\sin { \theta } \quad \)
\(= (NIA)B\sin { \theta } \quad \)
= 500 x 2 x (2 x 10-4) x 0.32
\( \tau \) = 0.032 N - m
