MCQs on Magnetism including EAMCET 2009 (Medical) Question

Some multiple choice questions on magnetism have already been posted on this site. You can access them by clicking on the label ‘magnetism’ below this post. Today we will discuss a few more multiple choice questions on magnetism.

(1) The period of oscillation of a magnetic needle in a magnetic field is T. If an identical bar magnetic needle is tied at right angles to it to form a cross (fig), the period of oscillation in the same magnetic field will be

(a) 2^1/4T

(b) 2^1/2T

(c) 2T

(d) T√3

(e) T/2

The period of oscillation (T) of the single magnetic needle is given by

T = 2π√(I/mB) where ‘I’ is the moment of inertia of the magnetic needle about the axis of rotation, ‘m’ is the magnetic dipole moment of the needle and ‘B’ is flux density of the magnetic field.

When two magnetic needles are tied together to form a cross, the moment of inertia becomes 2I and the magnitude of the magnetic dipole moment becomes √(m² + m²) = m√2.

[Note that magnetic dipole moment is a vector quantity. Two identical vectors (each of magnitude m) at right angles will yield a resultant magnitude m√2].

The resultant magnetic moment will be directed along the bisector of the angle between the axes of the individual magnets since the magnets are identical. In the absence of a deflecting torque, the resultant dipole moment vector will align along the applied magnetic field B. On deflecting from this position, the system will oscillate with period T₁ given by

T₁ = 2π√(2I/mB√2) = 2π√(I√2/mB) = 2^1/4T

(2) Three identical magnetic needles each L metre long and of dipole moment m ampere metre are joined as shown without affecting their magnetisation. At points B and C unlike poles are in contact. The dipole moment of this system is

(a) m

(b) 2m

(c) 3m

(d) 3m/2

(e) 5m/2

The distance (AD) between the ends of the compound magnet is 2L. Since the pole strength is m/L, the dipole moment of the compound magnet is (m/L)2L = 2m

(3) A magnet of length L and moment M is cut into two halves (A and B) perpendicular to its axis. One piece A is bent into a semicircle of radiur R and is joined to the other piece at the poles as shown in the figure below:

Assuming that the magnet is in the form of a thin wire initially, the moment of the resulting magnet is given by

(1) M/2π

(2) M/π

(3) M(2 + π)/2π

(4) Mπ/(2 + π)

The above question appeared in EAMCET 2009 (Medicine) question paper.

The distance between the poles of the resulting magnet is (L/2) + 2R

Since the semicircular portion of radius R is made of the magnetised wire of length L/2, we have L/2 = πR so that R = L/2π and 2R = L/π

Therefore, length of the resulting magnet (L/2) + (L/π)

The pole strength (p) of the magnet is given by

p = M/L

Therefore, the dipole moment of the resulting magnet = Pole strength×Length = (M/L)[ (L/2) + (L/π)] = M(2 + π)/2π

Solution to MCQ on Magnetism

In the post dated 2^nd July 2007, two questions on magnetism were given without solution. These questions with solution are given below as promised in the post:

(1) A thin, non-conducting rod of length ‘L’ is uniformly charged to have a linear charge density ‘λ’ from its mid point to one end (The other half of the rod is uncharged). The rod is rotated about an axis perpendicular to its length and passing through the other end with angular velocity ‘ω’. The magnetic dipole moment of the rotating rod is

(a) zero (b) λωL³/3 (c) λωL³/6

(d) 5λωL³/46 (e) 7λωL³/48

The magnetic dipole moment of a current loop is iA where ‘i’ is the current and ‘A’ is the area of the loop. The rotation of the rod OP with linear charge density λ from its middle to the end P is equivalent to many current loops of varying area. Considering the current loop made by the charge λdr on a small length dr of the rod at distance ‘r’ from the end O, the magnetic moment contributed by it is (λdr/T)×πr² where T is the period of rotation of the rod.

Since T = 2π/ω, the above magnetic moment is ½ (λωr²dr).

The magnetic dipole moment (m) due to the rotation of the entire rod is obtained by integrating the above between the limits L/2 and L.

Therefore, m = ∫[½(λωr²dr) = ½(λω)×[L³/3 – (L/2)³/3] = 7λωL³/48.

(2) A magnetic needle oscillates in a horizontal plane with a period ‘T’ at a place where the angle of dip is 30º. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian, its period will be

(a) T (b) T/√3 (c) T×√3 (d) T×3^1/4 (e) T×3^–1/4

When the magnetic needle is made to oscillate in the horizontal plane, the restoring force required for the motion is provided by the horizontal component (B_h) of the earth’s magnetic field B so that we have

T = 2π√(I/mB_h) where ‘I’ is the moment of inertia of the needle (about the axis of angular oscillation, which is through its centre of mass and is perpendicular to its length.) and ‘m’ is its magnetic dipole moment.

When the needle is made to oscillate in the vertical plane, the restoring force required for the motion is prvided by the vertical component (B_v) of the earth’s magnetic field B so that we have the new period

T’ = 2π√(I/mB_v).

Dividing the first equation by the second, T/T’ = √(B_v/B_h) from which

T’ = T/√(B_v/B_h).

But, B_v/B_h = tanθ where ‘θ’ is the angle of dip (which is 30º here) so that

T’ = T/√(tan30º) = T/√(1/√3) = T×3^1/4.

Multiple Choice Questions on Magnetism

I give you two questions (MCQ) from magnetism without solution. See whether you can work out these within five minutes. I’ll be back shortly with the solution.

(1) A thin, non-conducting rod of length ‘L’ is uniformly charged to have a linear charge density ‘λ’ from its mid point to one end (The other half of the rod is uncharged). The rod is rotated about an axis perpendicular to its length and passing through the other end with angular velocity ‘ω’. The magnetic dipole moment of the rotating rod is

(a) zero (b) λωL³/3 (c) λωL³/6

(d) 5λωL³/46 (e) 7λωL³/48

(2) A magnetic needle oscillates in a horizontal plane with a period ‘T’ at a place where the angle of dip is 30º. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian, its period will be

(a) T (b) T/√3 (c) T×√3 (d) T×3^1/4 (e) T×3^–1/4

MCQ on Oscillation of Magnets

The period (T) of angular oscillations of a magnet suspended freely in a magnetic field of flux density ‘B’ is given by
T = 2π √(I/ mB) where ‘I’ is the moment of inertia of the magnet about the axis of the angular motion and ‘m’ is the magnetic dipole moment. You will find questions based on this relation in your entrance examination question papers. Here is a question which is meant also for checking your grasp of the moment of inertia and the vector property of the magnetic dipole moment:
Two identical magnets are placed one above the other and tied together so that their like poles are in contact. The period of oscillation of this combination (on suspending horizontally using torsionless suspension fibre) in a horizontal magnetic field is 2 s. What will be the period of oscillation if the magnets are placed one above the other such that they are mutually perpendicular and bisect each other?
(a) 2 s (b) √2 s (c) 2√2 s (d) 2^¼ s (e) 2×2^¼ s

The period of oscillation is given by T = 2π √(I/ mB) with usual notations.
In the both cases, the moment of inertia is twice that of one magnet. In the first case, the net dipole moment is twice that of one magnet since the like poles are pointing in the same direction. In the second case, the net dipole moment is √2 times the moment of one magnet since the moment vectors at right angles get added.
The equations for the period in the two cases are therefore
T1 = 2π √(2I/ 2mB) = 2π √(I/ mB) = 2 s (as given in the question) and
T2 = 2π √(2I/ √2 mB) = 2π √(√2 I/ mB) = T1 √(√2) = 2×2^¼ s.
The following MCQ appeared in EAMCET (Med) A.P.2003 question paper:
The period of oscillation of a magnet at a place is 4 seconds. When it is remagnetised so that the pole strength becomes 4 times the initial value, the period of oscillation in seconds is
(a) ½ (b) 1 (c) 2 (d) 4

The period of oscillation is given by T = 2π √(I/ mB) with usual notations. The period is therefore inversely proportional to the square root of the dipole molent ‘m’. When the pole strength is made 4 times the initial value, the dipole moment is made 4 times the initial value so that the period becomes half the initial value. The correct option is 2 seconds.

Multiple Choice Questions on Magnetism

The following question which appeared in the Kerala Engineering Entrance test paper of 2006 has been popular among question setters for long:
A magnetized wire of magnetic moment M and length L is bent in the form of a semicircle of radius ‘r’. The new magnetic moment is
(a) M (b) M/2π (c) M/π (d) 2M/π (e) zero
The pole strength of the magnet, p = M/L. The pole strength of the magnet is unchanged, but the moment is changed since the poles come closer on bending the wire, thereby changing the magnetic length of the magnet from L to L'
We have L' = 2r = 2L/π so that the new magnetic moment = pL' = (M/L) ×(2L/π) = 2M/π [Option (d)].
Consider now the following M.C.Q.:
In a hydrogen atom the electron is making 6.6×10¹⁵ revolutions per second around the nucleus in an orbit of radius 0.528 Å. The equivalent magnetic dipole moment is approximately ( in Am²)
(a)10^-10 (b) 10^-15 (c) 10^-23 (d) 10^-25 (e) 10^-17
The orbiting electron is equivalent to a circular current loop, whose magnetic dipole moment is given by M = IA where I is the equivalent current and A is the area of the loop. Therefore, M = (q/T)×(πr²) = qf πr² where q is the electronic charge, T is the orbital period, r is the orbital radius and f is the frequency of revolution of the electron.
Thus M = 1.6×10^-19×6.6×10¹⁵×π×(5.28×10^-10)² . This yields a value nearly 10^-23 [Option (c)].
Let us now consider the following question:
Two short magnets of dipole moments M and 2M are arranged on the table so that the axial line of the weaker magnet and the equatorial line of the stronger magnet are coinciding. If the separation between the magnets is 2d, what is the magnetic flux density midway between these magnets? Ignore the earth’s magnetic field.
(a) μ₀M/4πd³ (b) 3μ₀M/4πd³ (c) (μ₀M/4πd³)√3 (d) (μ₀M/4πd³)√5 (e) (μ₀M/4πd³) √8
At the point midway between the magnets, the axial field (μ₀/4π)(2M/d³) of the magnet of moment M and the equatorial field (μ₀/4π)(2M/d³) of the magnet of moment 2M are acting at right angles so that the net field there is √2×(μ₀/4π)(2M/d³) = μ₀M/4πd³)√8. The correct option therefore is (e).