NCERT Solution Physics Class 12 Chapter 4 Moving Charges And Magnetism

NCERT Solution Physics Class 12 Chapter 4 Moving Charges And Magnetism

NCERT Solution Physics Class 12 Chapter 4 Moving Charges And Magnetism all questions and answers. Physics Class 12 4th Chapter Moving Charges And Magnetism exercise solution and experts answer. As one of online learning platforms, we (netex.) are excited to offer the NCERT Solution Physics Class 12 Chapter 4. This solution is designed to help students who are looking to brush up on their physics concepts on Chapter 4 Moving Charges And Magnetism.

 

4.1) 

4.3)

 

4.5)

 

4.7) 

 

4.9) 

 

4.11) 

 

4.14)

 

4.15)

 

4.17)

 

4.18 Answer the following questions:

(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?

(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?

(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

ANSWER-

(a) A charged particle enters the chamber and travels undeflected along a straight path with constant speed. It shows that the force on this charged particle is zero. we know that force on charged moving  particle  is F =qvBsin (θ) where (θ) is angle between velocity and direction of magnetic field.

F = 0 if sin (θ) =0 which gives (θ) = 0 or 1800 Hence initial velocity is either parallel or antiparallel to the direction of magnetic field.

(b) We know that magnetic force changes direction of the velocity of the charged particle but does not change the magnitude. Hence the initial and final velocity will remain same.

(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Due to this force electron will experience a force opposite to direction electric field i.e. from south to north direction. To prevent this motion of electron force should be applied in north to south direction. According Flemings left hand rule magnetic field should be applied in vertically downward direction.

 

In case you are missed :- NCERT Solution for Current Electricity

 

4.19)

 

4.24) A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?

ANSWER-

Magnetic field strength, B = 3000 G = 3000 × 10-4T

Length of the rectangular loop (L) = 10cm = 0.1 m

Width of the rectangular loop (b) = 5cm = 0.05m

Area of the loop, A = L × b = 0.1 × 0.05 = 5 × 10-3 m2

Current in the loop, I = 12 A

Formula for torque is given by,

(a) From figure (a);

Consider anticlockwise current as positive and clockwise as negative.

Magnetic field is along z axis and wire is in yz plane hence area vector is along x axis. Hence angle between A and B is 900 hence from equation (1) ;

∴ τ = 12 x 5 x 10-3x 3000 × 10-4T x sin (90)

∴ τ = 1.8 x 10-2 N-m

By right hand thumb rule this torque is along negative y axis and force is zero as angle between A and B is 900.

 

(b) From figure (b);

Consider anticlockwise current as positive and clockwise as negative.

Magnetic field is along z axis and wire is in yz plane hence area vector is along x axis. Hence angle between A and B is 900 hence from equation (1);

∴ τ = 12 x 5 x 10-3 x 3000 × 10-4T x sin (90)

∴ τ = 1.8 x 10-2 N-m

By right hand thumb rule this torque is along negative y axis and force is zero as angle between A and B is 900.

 

(c) From figure (c);

Consider anticlockwise current as positive and clockwise as negative.

Magnetic field is along z axis and wire is in xz plane hence area vector is along negative y-axis. Hence angle between A and B is 900 hence from equation (1);

∴ τ = 12 x 5 x 10-3 x 3000 × 10-4T x sin (90)

∴ τ = 1.8 x 10-2 N-m

By right hand thumb rule this torque is along negative x axis and force is zero as angle between A and B is 900.

 

(d) From figure (d);

Magnetic field is along z axis and wire is in xz plane hence from equation (1);

\ = 12´ 5´10-3´3000 × 10-4T ´ sin (90)

\ = 1.8 ´ 10-2 N-m

By right hand thumb rule this torque is along positive x axis and force is zero as angle between A and B is 900.

 

(e) From figure (e);

Magnetic field is along z axis and wire is in xy plane hence area vector is along z-axis. Hence angle between A and B is 00.

∴ τ = 12 x 5 x 10-3 x 3000 × 10-4T x sin (0)

∴ τ = 0

Torque is zero hence the force is also zero.

 

(f) From figure (f);

Magnetic field is along z axis and wire is in xy plane hence area vector is along z-axis. Hence angle between A and B is 00.

∴ τ = 12 x 5 x 10-3 x 3000 × 10-4T x sin (0)

∴ τ = 0

Torque is zero hence the force is also zero.

Case (f) is case of stable equilibrium as direction of I, A and B is same.

 

4.25) 

 

4.26) 

 

4.27) 

In case you are missed :- NCERT Solution for Electrostatic Potential And Capacitance

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Updated: April 15, 2023 — 3:17 pm

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