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Magnetic Dipole Moment Test - 1

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Magnetic Dipole Moment Test - 1
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  • Question 1
    1 / -0.25
    The reason for Earth's magnetic field is-
    Solution

    CONCEPT:

    • Earth's magnetism: The cause of the magnetic field at earth’s core is still not clear
      • Initially, we assume, the magnetic field was caused by a giant bar magnet placed approximately along the axis of rotation of the earth and deep in the interior
    • But now we consider the reason for the magnetic field to arise due to electrical currents produced by the convective motion of metallic fluids (consisting mostly of molten iron and nickel) in the outer core of the earth.
    • This is known as the dynamo effect.

    EXPLANATION:

    • The magnetic field of the Earth is created due to the circling currents inside the core. This is also called the geomagnetic field. So option 1 is correct.
  • Question 2
    1 / -0.25
    A current loop is similar to a :
    Solution

    CONCEPT:

    • When a circular loop is associated with the current I, it starts to act as a magnet and its magnetic moment is find as given below.

    • Magnetic moment (μ): The magnetic strength and orientation of a magnet or other object that produces a magnetic field.
      • It is a vector quantity associated with the magnetic properties of electric current loops.
      • It is equal to the amount of current flowing through the loop multiplied by the area encompassed by the loop.

    μ = N i A 

    where μ is the magnetic moment, A is the area of the coil, N is no. of turns and I is current in the coil.

    • Its direction is established by the right-hand rule for rotations.

    EXPLANATION:

    • From the above, it is clear that a circular loop is considered as a magnetic dipole, then the dipole moment is the product of current and area.
  • Question 3
    1 / -0.25

    Which of the following statements is correct?

    1. Magnetic line of forces always start from north pole and end at south pole of the magnet.

    2. The magnetic lines of forces never intersect each other

    Choose the correct code:

    Solution

    Explanation:

    Magnetic field and magnetic lines of force: It is the space around a magnetic pole or magnet or current-carrying wire within which its magnetic effect can be experienced is defined as a magnetic field.

    • The magnetic field can be represented with the help of a set of lines or curves called magnetic lines of force or magnetic field lines.

    Properties of magnetic field line:

    1. The magnetic field line is directed from the north pole to the south pole outside and south to the north inside the magnet.
    2. Magnetic field lines are closed and continuous.
    3. Magnetic field lines are more crowded near poles which shows that the strength of the magnetic field is maximum at its poles.
    4. Magnetic field lines never intersect with each other.

    ​The reason being is that if two field lines intersect each other, then there will be two directions of the magnetic field, which is not possible.

  • Question 4
    1 / -0.25
    When a bar magnet of magnetic moment \(\vec{M}\) is placed in a magnetic field \(\vec{B}\), then the torque exerted on the magnet will be:
    Solution

    CONCEPT:

    Bar magnet:

    • A bar magnet consists of two equal and opposite magnetic poles separated by a small distance.

    A magnetic moment or magnetic dipole moment:

    • It represents the strength of the magnet.
    • Mathematically it is defined as the product of the strength of either pole (m) and effective length (2l).

    \(\Rightarrow \vec{M}=m(2\vec{l})\)

    EXPLANATION:

    • We know that when a bar magnet is placed in a magnetic field, a torque acts on it.
    • The torque on the bar magnet is given as,

    \(\Rightarrow \vec{\tau}=\vec{M}\times\vec{B}\)

    • Hence, option 3 is correct.
  • Question 5
    1 / -0.25
    A circular coil of radius 4 cm and of 20 turns carries a current of 3 amperes. It is placed in a magnetic field of intensity of 0.5weber/m2. The magnetic dipole moment of the coil is
    Solution

    Concept:

    If the circular loop is considered a magnetic dipole, then the dipole moment is the product of current and area. 

    Therefore, the magnitude of dipole moment = area × current × number of turns.

    M = NIA

    Where M = Magnetic moment of copper coil, N = No of turns, I = Current flowing through a loop and A = Area of coil

    Calculation:

    Given - Radius of the coil (r) = 4 cm = 4 × 10-2 m, N = 20, I = 3 A and B = 0.5weber/m2

    The magnetic dipole moment of circular coil is given by:

    ⇒ M = NIA

    ⇒ M = 20 × 3 × π(4 × 10-2 )2 

    ⇒ M = 3017.14 × 10-4

    ⇒ M = 0.3 Am-2

  • Question 6
    1 / -0.25
    What is the angle of dip at magnetic poles of earth?
    Solution

    CONCEPT:

    The magnetic Declination (θ):

    • It is the angle between geographic and magnetic meridian planes.
    • Declination at a place is expressed at θ° E or θ° W depending upon whether the north pole of the compass needle lies to the east or to the west of the geographical axis.

    The angle of inclination or Dip (Φ):

    • It is the angle between the direction of the intensity of the total magnetic field of earth and a horizontal line in the magnetic meridian

     

    EXPLANATION:

    • From above it is clear that the angle of inclination is the angle by which the total intensity of the earth’s magnetic field dips or comes out of the horizontal plane. It is different at different places and measured using a “Dip Circle”.
    • At the magnetic equator, the dip needle rests horizontally at an angle of zero degrees while, at the magnetic poles, the magnetic needle rests vertically, at an angle of 90 degrees. Therefore option 3 is correct.
    • At all the other places, the angle of dip is between 0° and 90°.
  • Question 7
    1 / -0.25
    A closely wound solenoid of 2000 turns and the area of cross-section 1.5 × 10–4 m2 carries a current of 2.0 A. It is suspended through its center and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5 × 10–2 tesla making an angle of 30º with the axis of the solenoid. The torque on the solenoid will be
    Solution

    CONCEPT:

    • Solenoid: It is a long coil of wire wrapped in many turns.
      • When a current passes through it, it creates a nearly uniform magnetic field inside. 
      • Solenoids can convert electric current to mechanical action, and so are very commonly used as switches.
    • Magnetic dipole moment: It is a quantity that represents the magnetic strength and orientation of a magnet or any other object that produces a magnetic field.
      • More precisely, a magnetic moment refers to a magnetic dipole moment, the component of the magnetic moment that can be represented by a magnetic dipole.
      • A magnetic dipole is a magnetic north pole and a magnetic south pole separated by a small distance.
    • Torque: It is the measure of the force that can cause an object to rotate about an axis.

    Formula:

    M = NIA

    where, N = number of turns, I = current, A = area of crossection, M = Dipole moment

    \(τ = M × \overrightarrow B sin\theta\)

    where, \(\overrightarrow B \) = magnetic field of flux density.

    CALCULATION:

    Given that, n = 2000 turns, A = 1.5 × 10-4 m2, I = 2 A,  \(\overrightarrow B \) = 5 × 10-2 tesla, θ = 30° 

    As per the above discussion, it's clear that,

    M = N × I × A = 2000 × 2 × 1.5 × 10-4 = 0.6  J/T

    \(\tau = M × \overrightarrow B \sin {30^o} = 0.6 × 5 × {10^{ - 2}} = 0.03 × \sin {30^o} = 0.015Nm\) 

     \(\tau = 1.5 × {10^{ - 2}}Nm\)

    The correct option is 3.

  • Question 8
    1 / -0.25
    Magnetism at the centre of a bar magnet is ______?
    Solution

    CONCEPT:

    • Magnetism: It is a class of physical phenomena that are mediated by magnetic fields.
      • Electric currents and the magnetic moments of elementary particles give to a rise a magnetic field, which acts on other currents and magnetic moments.
    • Bar-Magnet: A bar magnet is a rectangular piece of an object, made up of iron, steel or any other ferromagnetic substance or ferromagnetic composite, that shows permanent magnetic properties.
      • It has two poles, a north and a south pole such that when suspended freely, the magnet aligns itself so that the northern pole points towards the magnetic north pole of the earth.
    • Uses of Bar Magnet. Bar magnets are used as stirrers in the laboratory for magnetic experiments.
      • Electronic devices such as telephones, radios, and television sets use magnets.

    CALCULATION:

    • If we consider a bar magnet of length l, then the magnetic moment is M = 2m × I

    Where m is the magnitude of poles.

    • Considering the distance r from the center of the bar magnet, we can write magnetic field intensity as a product of magnetic field strength and positional feature.

    Therefore, the field intensity at a distance r is given by (r- l2)2.

    • The magnetic field of a bar magnet at an axial point is given by

    \(\Rightarrow B = \frac{{{\mu _o}}}{{4\pi }}\frac{{2Mr}}{{{{\left( {{r^2} - {l^2}} \right)}^2}}}\)

    At the center of the magnet, r = 0, so 

    \(\Rightarrow B = \frac{{{\mu _o}}}{{4\pi }}\frac{{2Mr}}{{{{\left( {{r^2} - {l^2}} \right)}^2}}} = \frac{{{\mu _o}}}{{4\pi }}\frac{{2M \times 0}}{{{{\left( {{r^2} - {l^2}} \right)}^2}}} = 0 \)

    • Hence option 4 is correct.
  • Question 9
    1 / -0.25
    An electron moving around the central nucleus has a magnetic moment equal to ___________, where 'l' is the magnitude of the angular momentum of the circulating electron about the central nucleus. ('e' is charge and 'm' is mass of an electron)
    Solution

    CONCEPT: 

    • If the circular loop is considered as a magnetic dipole, then the dipole moment is the product of current and area. 
    • Therefore, the magnitude of magnetic moment = area × current 

    M = IA

    Where M = Magnetic moment of copper coil, I = Current flowing through a loop and A = Area of coil

    EXPLANATION:

    If e is the charge on an electron revolving in an orbit of radius r with uniform angular velocity ω, then equivalent current is

    \(\Rightarrow I = \frac{charge (e)}{time (T)}\)

    Where T = the period of revolution of electron

    \(\Rightarrow T=\frac{2\pi r}{v}\)

    ∴ The equivalent  current will be 

    \(\Rightarrow I = \frac{ev}{2\pi r}\)

    • Area of the orbit is

    \(\Rightarrow A= \pi r^2\)

    The magnitude of the magnetic moment

    \(\Rightarrow M = \frac{ev}{2\pi r}\times\pi r^2=\frac{evr}{2}\)

    Multiply and divide the above equation by m, we get

    \(\Rightarrow M=\frac{emvr}{2m} = \frac{el}{2m}\)            [∵ l = mvr]

  • Question 10
    1 / -0.25
    The magnetic field strength due to a short bar magnet at a distance r from the mid point of the magnet is B. Find the magnetic field strength at a distance 2r from the mid point of the magnet.
    Solution

    Concept:

    Magnetic dipole:

    A Magnetic Dipole consists of two unlike poles of equivalent strength and is separated by a small distance.

    Magnetic moment:

    It is defined as the product of pole strength and the distance between the poles of a magnet.

    ​⇒ M = m × 2l    

    where M = magnetic moment, m = pole strength and 2l = distance between the poles

    Magnetic field due to a bar magnet on the axial point:

    • Magnetic field due to a bar magnet on the axial point at a distance r from the centre of magnet is given as,

    \(\Rightarrow B=\frac{\mu_o}{4\pi}\frac{2Mr}{(r^2-l^2)^2}\)

    If r >> l

    \(\Rightarrow B=\frac{\mu_o}{4\pi}\frac{2M}{r^3}\)

    Calculation:

    Given r1 = r, B1 = B and r2 = 2r

    We know that magnetic field due to a bar magnet on the axial point at a distance r from the centre of magnet is given as,

    \(\Rightarrow B_1=\frac{\mu_o}{4\pi}\frac{2M}{r_1^3}\)

    \(\Rightarrow B_1=B=\frac{\mu_o}{4\pi}\frac{2M}{r^3}\)     -----(1)

    Similarly, magnetic field due to a bar magnet on the axial point at a distance 2r from the centre of the magnet is given as,

    \(\Rightarrow B_2=\frac{\mu_o}{4\pi}\frac{2M}{r_2^3}\)

    \(\Rightarrow B_2=\frac{\mu_o}{4\pi}\frac{2M}{(2r)^3}\)

    \(\Rightarrow B_2=\frac{\mu_o}{4\pi}\frac{2M}{8r^3}\)

    \(\Rightarrow B_2=\frac{B}{8}\)

    Hence, option 3 is correct.

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