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NCERT Solutions for CBSE Class 12 Physics ‒ Chapter 5: Magnetism and Matter (Part I)

May 26, 2017 17:30 IST

    Download NCERT Solutions for Class 12 Physics ‒ Chapter 5: Magnetism and Matter

    NCERT Solutions for Class 12th Physics - Chapter 5: Magnetism and Matter are available here. These solutions are also available for download in PDF format. Some important topics of this chapter are: Bar Magnet, Magnetic Field lines, Bar magnet as an equivalent solenoid, Dipole in Uniform Magnetic Field, Magnetism and Gauss’s Law, Earth’s Magnetism, Magnetisation and Magnetic Intensity, Magnetic Properties of Materials, Permanent Magnets, and Electromagnets. Most of the questions given in Chapter 5 of 12th Physics NCERT textbook are based on these topics. This chapter has large number of problems so, we have provided solutions in two parts. This is Part I, here you will find solutions from question number 5.1 to 5.8.

    NCERT Solutions for Class 12 Physics ‒ Chapter 5: Magnetism and Matter question number 5.1 to 5.8. are given below:

    Question5.1: Answer the following questions regarding earth’s magnetism:

    (a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field

    Solution (a):

    The three quantities are: horizontal component of earth’s magnetic field, magnetic declination and angle of dip.

    (b) The angle of dip at a location in southern India is about 18o. Would you expect a greater or smaller dip angle in Britain?

    Solution (b):

    Great, because it increases from equator to poles.

    (c) If you made a map of magnetic field lines at Melbourne in

    Australia, would the lines seem to go into the ground or come out of the ground?

    Solution (c):

    Coming out of the ground.

    (d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole?

    Solution (d):

    Compass can point in any direction at poles.

    (e) The earth’s field, it is claimed, roughly approximates the field due to a dipole of magnetic moment 8 × 1022 J T–1 located at its centre. Check the order of magnitude of this number in some way.

    Solution (e):

    By the formula:

    B = (μo m)/(4 π r3)

    Substituting m = 8 × 1022 J/T

    r = 6.4 × 106 m

    Thus B = 0.3G.

    (f) Geologist’s claim that besides the main magnetic N-S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all?

    Solution (f):

    Because of local magnetic field due to magnetic deposits or heavy solar flares.

    CBSE Class 12th Physics Notes: All Chapters

    Question5.2: Answer the following questions:

    (a) The earth’s magnetic field varies from point to point in space.

    Does it also change with time? If so, on what time scale does it change appreciably?

    Solution (a):

    Yes it changes with time. The time scale is of order of few hundred years.

    (b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why?

    Solution (b):

    As molten iron is not ferromagnetic.

    (c) The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ‘battery’ (i.e., the source of energy) to sustain these currents?

    Solution (c):

    The answer is uncertain. Radioactivity may be one reason.

    (d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such distant past?

    Solution (d):

    Analysis of rock magnetism gives the clue.

    (e) The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion?

    Solution (e):

    The extra activities like solar winds from space affect magnetism of earth.

    (f) Interstellar space has an extremely weak magnetic field of the order of 10–12 T. Can such a weak field be of any significant consequence? Explain.

    Solution (f):

    The interstellar distances are very great, and even the small minute field can affect the passage of charged particles.

    CBSE Class 12 NCERT Solutions

    Question5.2: Answer the following questions:

    (a) The earth’s magnetic field varies from point to point in space.

    Does it also change with time? If so, on what time scale does it change appreciably?

    Solution (a):

    Yes it changes with time. The time scale is of order of few hundred years.

    (b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why?

    Solution (b):

    As molten iron is not ferromagnetic.

    (c) The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ‘battery’ (i.e., the source of energy) to sustain these currents?

    Solution (c):

    The answer is uncertain. Radioactivity may be one reason.

    (d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such distant past?

    Solution (d):

    Analysis of rock magnetism gives the clue.

    (e) The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion?

    Solution (e):

    The extra activities like solar winds from space affect magnetism of earth.

    (f) Interstellar space has an extremely weak magnetic field of the order of 10–12 T. Can such a weak field be of any significant consequence? Explain.

    Solution (f):

    The interstellar distances are very great, and even the small minute field can affect the passage of charged particles.

    Question5.3: A short bar magnet placed with its axis at 30o with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10–2 J. What is the magnitude of magnetic moment of the magnet?

    Solution5.3:

    Given:

    θ = 30o

    B = 0.25 T

    T = 4.5x10‒2 J

    Torque is given by

    T = m B sin θ

    Substitution yields

    4.5 × 10‒2 = m (0.25) (0.5)

    Therefore magnetic moment,

    m = 0.36 J / T.

    Question5.4: A short bar magnet of magnetic moment m = 0.32 JT–1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its

    (a) stable and

    (b) Unstable equilibrium?

    What is the potential energy of the magnet in each case?

    Solution5.4:

    Given:

    m = 0.32 J / T

    B = 0.15 T

    The bar is free to rotate in the plane of the field. It will be in stable position in which the net torque is zero. Since the field is uniform, there will be no net force.

    The angle between the field and the moment should be either zero or 180 degrees. This implies that it can be parallel or antiparallel to field.

    (a) For stable equilibrium, magnetic moment m should be parallel to applied magnetic field.

    (b) For unstable equilibrium, magnetic moment should be antiparallel to applied magnetic field.

    Now, calculating potential energy of the magnet in each case.

    For stable equilibrium

    Potential energy

    U = -mB

    Putting the values

    U = ‒ 0.32 x 0.15

    U = ‒ 0.048 J

    For unstable equilibrium

    U = mB = 0.32 x 0.15 = +0.048 J.

    Question5.5: A closely wound solenoid of 800 turns and area of cross section2.5 × 10–4 m2 carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

    Solution5.5:

    Given:

    N = 800

    A = 2.5x10-4 sq. m

    I = 3 A

    The solenoid acts like a bar magnet along the axis, determined by flow of current.

    The magnetic moment associated

    m = NIA = 0.6 J / T.

    Question5.6: If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

    Solution5.6:

    Here:

    Horizontal magnetic field, B = 0.25 T, θ = 30o

    Torque is experienced by the solenoid given by

    Torque = m B sin θ

    = 0.6 × 0.25 × 0.5

    = 0.075 Nm.

    Question5.7: A bar magnet of magnetic moment 1.5 J T–1 lies aligned with the direction of a uniform magnetic field of 0.22 T.

    (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment:

    (i). normal to the field direction,

    (ii). Opposite to the field direction?

    (b) What is the torque on the magnet in cases (i) and (ii)?

    Solution5.7:

    Given:

    m = 1.5 J / T

    B = 0.22 T

    The magnet is aligned with the direction of uniform magnetic field, and the net force is zero.

    (a)

    (i). θ at normal to the field direction is 90.

    The work required is given by, U = mB

    Substituting the values

    U = 1.5 x 0.22 or U = 0.33 J

    (ii). Opposite to the field direction = 180

    So, the work required will be double as calculated above. Thus, it is 0.66 J.

    (b)The value of torque can be determined by the formula

    T = m. B. sin θ

    Considering the both cases

    (i) θ = 90o

    T = m B sin 90o

    = 1.5 x 0.22 x 1

    = 0.33 Nm

    (ii) θ = 180

    T = m B sin180 = zero N.

    Question5.8: A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10–4 m2, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.

    (a) What is the magnetic moment associated with the solenoid?

    (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10–2 T is set up at an angle of 30o with the axis of the solenoid?

    Solution5.8:

    Given:

    N = 2000

    A = 1.6 x 10‒4 m2

    I = 4 A

    (a) Magnetic moment is given by

    m = NIA

    Solving by putting values, m = 1.28 Am2

    (b) The question mentions that the given magnetic field is uniform in nature. In uniform magnetic field, the net force is always zero, irrespective of the magnitude.

    Now the torque on the solenoid is given by

    T = m B sinϴ

    Given, m = 1.28 Am2 as calculated above, B = 0.075 T, θ = 30o

    Therefore,

    T = 0.048 Nm.

    Download NCERT Solutions for Class 12 Physics ‒ Chapter 5: Magnetism and Matte in PDF format

    DISCLAIMER: JPL and its affiliates shall have no liability for any views, thoughts and comments expressed on this article.

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