Permanent Magnets
Questions for Short Answer
1. Can we have a single north pole? A single south pole?
ANSWER: No, we can not have a single north pole or south pole. Magnetic poles always occur in pairs. Even if a magnet is broken into pieces, each and every piece will have a north and a south pole.
2. Do two distinct poles actually exist at two nearby points in a magnetic dipole?
ANSWER: No, two distinct poles can not actually exist at two nearby points. In a magnetic dipole, two distinct poles are located near its ends.
3. An iron needle is attracted to the ends of a bar magnet but not to the middle region of the magnet. Is the material making up the ends of a bar magnet different from that of the middle region?
ANSWER: No, the material making up the ends and middle region of a bar magnet is not different. When an iron needle is brought near a magnet, each pole at the ends of the magnet induces an opposite pole in the needle. When the needle is brought near the end of the magnet, the pole at this end induces the opposite pole in the needle. The pole at the far end has less influence. Hence, the needle is attracted towards the nearer end. But when the needle is brought near the middle of the magnet, the poles at each end have equal and opposite influences on the needle. Thus, the forces cancel each other, and the needle is not attracted.
4. Compare the direction of the magnetic field inside a solenoid with that of the field there if the solenoid is replaced by its equivalent combination of a north pole and a south pole.
ANSWER: The end of the solenoid where current is anticlockwise when seen from outside, acts as the magnetic north pole of a magnet, and the other end as the south pole. So, the magnetic field inside the solenoid is from the apparent south pole to the apparent north pole.
When the solenoid is replaced by equivalent north and south poles, the magnetic fields remain the same.
5. Sketch the magnetic field lines for a current-carrying circular loop near its center. Replace the loop with an equivalent magnetic dipole and sketch the magnetic field lines near the center of the dipole. Identify the difference.
ANSWER: In a current-carrying circular loop, the magnetic lines of force are directed perpendicular to the loop and coming out towards the viewer who sees the anticlockwise current. So, if we consider one side of the loop as the north pole and the other side as the south pole, then the direction of the fields near the center is from south to north.  |
| Diagram for Q-5 |
Now replace the loop with a magnetic dipole. As we see in the diagram, the magnetic lines of force emerge from the N-pole, and in a smooth curve, they enter the S-pole. So near the center, the direction of the field is from the north to the south, which is just opposite to the case of the loop.
6. The force on a north pole, F =mB, is parallel to field B. Does it contradict our earlier knowledge that a magnetic field can exert forces only perpendicular to itself?
ANSWER: The earlier knowledge that a magnetic field can exert forces only perpendicular to itself is related to a moving electrical charge. Here, the force on a north pole, F =mB, is directed on a magnetic charge m. It is just like an electric charge in an electric field. So, though our earlier knowledge seems to be contradictory in this case, the entity in the magnetic field is different.
7. Two bar magnets are placed close to each other with their opposite poles facing each other. In the absence of other forces, the magnets are pulled towards each other and their kinetic energy increases. Does it contradict our earlier knowledge that magnetic forces cannot do any work and hence cannot increase the kinetic energy of a system?
ANSWER: The earlier knowledge that magnetic forces cannot do any work was related to the magnetic force on a moving charged particle, where the magnetic force is always perpendicular to the direction of movement. The force here can only change the direction of movement and cannot increase the speed. Thus, the kinetic energy cannot be increased.
The given case is not the case of moving charged particles. Here, the magnetic force is on another magnet. The movement of the magnet is in the direction of the force; hence, the work is done and the kinetic energy is increased.
So, yes, it contradicts, but the condition is different.
8. The magnetic scalar potential is defined as
U(r₂) -U(r₁) =-∫r2r₁(B.dl).
Apply this equation to a closed curve enclosing a long straight wire. The RHS of the above equation is then -µₒi by Ampere's law. We see that U(r₂) ≠U(r₁) even when r₂ =r₁. Can we have a magnetic scalar potential in this case?
ANSWER: We apply Ampere's law for the magnetic field due to current-carrying wires. The magnetic scalar potential is calculated at a point at distance r due to a pole of pole strength m and is equal to µₒm/4πr. So, in this case, we can not have a magnetic scalar potential.
9. Can the Earth's magnetic field be vertical at a place? What will happen to a freely suspended magnet at such a place? What is the value of dip here?
ANSWER: Yes, the Earth's magnetic field is vertical above the magnetic north and south poles of the Earth.
A freely suspended magnet will hang vertically at these places. Above the north magnetic pole of the Earth, the north pole of the magnet will be down towards the north magnetic pole of the Earth. And just reverse at the south magnetic pole of the Earth.
The value of dip above the north pole will be 90°, and above the south pole, it will be -90°.
10. Can the dip at a place be zero? 90°?
ANSWER: Yes, the dip at the magnetic equator is zero. Here, a freely suspended magnet will rest horizontally.
The dip at the north magnetic pole of the Earth is 90°. Here the freely suspended magnet will rest vertically with its north pole pointing towards the magnetic north pole of the Earth.
11. The reduction factor K of a tangent galvanometer is written on the instrument. The manual says that the current is obtained by multiplying this factor by tanθ. The procedure works well at Bhubaneswar. Will the procedure work if the instrument is taken to Nepal? If there is some error, can it be corrected by correcting the manual, or will the instrument have to be taken back to the factory?
ANSWER: The reduction factor K of the tangent galvanometer is fixed only for a place because it is proportional to the horizontal component of the Earth's magnetic field, which varies from place to place. So this procedure that works well at Bhubneswar will not work well in Nepal if the same reduction factor K is used.
The error can be corrected by correcting the manual with the actual reduction factor in Nepal. The actual reaction factor can be obtained by passing a known current through the instrument and measuring the angle.
OBJECTIVE-I
1. A circular loop carrying a current is replaced by an equivalent magnetic dipole. A point on the axis of the loop is in
(a) end-on position
(b) broadside-on position
(c) both
(d) none.
ANSWER: (a).
Explanation: A point on the magnetic axis of a bar magnet is called an "end-on" position.
2. A circular loop carrying a current is replaced by an equivalent magnetic dipole. A point on the loop is in
(a) end-on position
(b) broadside-on position
(c) both
(d) none.
ANSWER: (b).
Explanation: A position on a perpendicular bisector of the bar magnet is called the "broadside-on" position.
3. When a current in a circular loop is equivalently replaced by a magnetic dipole,
(a) The pole strength m of each pole is fixed
(b) The distance d between the poles is fixed
(c) The product 'md' is fixed
(d) None of the above.
ANSWER: (c).
Explanation: When a loop is equivalently replaced by a magnetic dipole, that means the dipole moment of both is the same. So the dipole moment of the loop iA = md is fixed.
4. Let r be the distance of a point on the axis of a bar magnet from its center. The magnetic field at such a point is proportional to
(a) 1/r
(b) 1/r²
(c) 1/r³
(d) None of the above.
ANSWER: (d).
Explanation: The magnetic field at a point at a distance r on end-on position is
B =µₒ*2Mr/{4π(r²-l²)²}
Which is none of the above.
5. Let r be the distance of a point on the axis of a magnetic dipole from its center. The magnetic field at such a point is proportional to
(a) 1/r
(b) 1/r²
(c) 1/r³
(d) None of the above.
ANSWER: (c).
Explanation: For a magnetic dipole, the magnetic length 2l may be considered very small in comparison to r, i.e., r>>l. Neglecting l in the expression for the magnetic field at the end-on position,
B =µₒ*2Mr/4πr⁴
=µₒ*2M/4πr³.
So, B ∝ 1/r³.
6. Two short magnets of equal dipole moments M are fastened perpendicularly at their centers (figure 36-Q1). The magnitude of the magnetic field at a distance 'd' from the center on the bisector of the right angle is
(a) µₒM/4πd³
(b) µₒ√2M/4πd³
(c) µₒ2√2M/4πd³
(d) µₒ2M/4πd³  |
| Figure for Q-6 |
ANSWER: (c).
Explanation: These short magnets can be considered magnetic dipoles with dipole moment M each. For a point P, at distance r from the dipoles (as shown in the diagram below), the radial strength of the field due to each dipole
B' =µₒ*2Mcos45°/4πr³
=µₒ2M/(4√2πr³)
=µₒM/(2√2πr³)  |
| Diagram for Q-6 |
The strength of the component perpendicular to OP is B". B" is equal in magnitude due to each dipole but opposite in direction. So they cancel each other. Thus, only radial components remain in the same direction. Hence, the net field at P
B =B'+B' =2B'
=2*(µₒM/2√2πr³)
=µₒM/(√2πr³)
=µₒ√2M/2πr³
=µₒ2√2M/4πr³
(multiplying numerator and denominator by 2)
7. The magnetic meridian is
(a) a point
(b) a line along north-south
(c) a horizontal plane
(d) a vertical plane.
ANSWER: (d).
Explanation: The Magnetic meridian is a plane passing through a point and the magnetic North and South poles.
8. A compass needle that is allowed to move in a horizontal plane is taken to a geomagnetic pole. It
(a) will stay in the north-south direction only
(b) will stay in the east-west direction only
(c) will become rigid, showing no movement
(d) will stay in any position.
ANSWER: (d).
Explanation: At the magnetic pole, the direction of the magnetic field is vertical. But the needle can move only in the horizontal plane. The horizontal component of the magnetic field here is zero. Hence, there will be no effect of the magnetic field on the compass needle, and it will stay in any position.
9. A dip circle is taken to the geomagnetic equator. The needle is allowed to move in a vertical plane perpendicular to the magnetic meridian. The needle will stay
(a) In a horizontal direction only
(b) In the vertical direction only
(c) in any direction except vertical and horizontal
(d) In any direction, it is released.
ANSWER: (d).
Explanation: At the Geomagnetic equator, the direction of the magnetic field is horizontal in the magnetic meridian. So the component of the magnetic field in a plane perpendicular to the magnetic meridian is zero. So the needle of the dip circle that is free to move in this perpendicular plane will stay in any direction it is released.
10. Which of the following four graphs may best represent the current-deflection relation in a tangent galvanometer?  |
| The figure for Q-10 |
ANSWER: (c).
Explanation: In a tangent galvanometer, the relation between the current and deflection is given as,
i =K*tanθ
Where K is a constant.
For very small θ, tanθ varies nearly linearly. As θ approaches π/2, even a small increase in θ increases tanθ by a great amount. And near π/2, tanθ tends to infinity.
So in the tangent galvanometer, the deflection-current graph will be a straight line initially, and with the increase in deflection, the graph will curve and become asymptotic to the current axis near π/2. Only graph (c) fulfills the condition.
11. A tangent galvanometer is connected directly to an ideal battery. If the number of turns in the coil is doubled, the deflection will
(a) increase
(b) decrease
(c) remain unchanged
(d) Either increase or decrease.
ANSWER: (c).
Explanation: The magnetic field created by the coil at the center is
B =µₒin/2r
When the number of turns is kept double, i.e. =2n, the resistance of the coil is also doubled. It results in reducing the current to half, i.e., i/2. Now, the magnetic field of the coil is
µₒ(i/2)(2n)/2r =µₒin/2r =B.
Thus, the magnetic field remains unchanged, and so will be the deflection.
12. If the current is doubled, the deflection is also doubled in
(a) a tangent galvanometer
(b) A moving coil galvanometer
(c) both
(d) none.
ANSWER: (b).
Explanation: In a tangent galvanometer, the current is proportional to the tangent of the deflection. So, doubling the current will not double the deflection.
In a moving coil galvanometer, the current is directly proportional to the deflection. Hence, when we double the current, the deflection will also double. Option (b) is correct.
13. A very long bar magnet is placed with its north pole coinciding with the center of a circular loop carrying an electric current i. The magnetic field due to the magnet at a point on the periphery of the wire is B. The radius of the loop is a. The force on the wire is
(a) very nearly 2πaiB perpendicular to the plane of the wire
(b) 2πaiB in the plane of the wire
(c) πaiB along the magnet
(d) zero.
ANSWER: (a).
Explanation: Since the bar magnet is long, the direction of the magnetic field on the periphery due to the north pole placed at the center of the loop will be very nearly radially outward. So the angle between the length element dl and B will be 90°. The force on the length element dl is the cross product,
dF =i dlxB
Clearly, the direction of the force is perpendicular to the plane of the loop. Integrating over the whole loop, the magnitude of the force is,
F =i(2πaB)*sin90° =2πaiB.
OBJECTIVE-II
1. Pick the correct options.
(a) A Magnetic field is produced by electric charges only.
(b) Magnetic poles are only mathematical assumptions having no real existence.
(c) A north pole is equivalent to a clockwise current, and a south pole is equivalent to an anticlockwise current.
(d) A bar magnet is equivalent to a long, straight current.
ANSWER: (a), (b).
Explanation: A Magnetic field is produced by moving charges only. Even in permanent magnets, the magnetic properties come from the moving charges at the atomic level. (a) is correct.
The magnetic poles are mathematical assumptions used to study the magnetic effects. And we call a positive magnetic charge the North Pole and a negative magnetic charge the South Pole. Option (b) is also correct.
(c) is not correct because a north pole is equivalent to an anticlockwise current, and a south pole is equivalent to a clockwise current.
Option (d) is also not correct because a bar magnet is not equivalent to a long, straight current because the magnetic fields produced by them are different.
2. A horizontal circular loop carries a current that looks clockwise when viewed from above. It is replaced by an equivalent magnetic dipole consisting of a south pole S and a north pole N.
(a) The line SN should be along the diameter of the loop.
(b) The line SN should be perpendicular to the plane of the loop.
(c) The south pole should be below the loop.
(d) The north pole should be below the loop.
ANSWER: (b), (d).
Explanation: If a circular loop is replaced by an equivalent magnetic dipole, then the SN line should be perpendicular to the plane of the loop. It is because the north pole is towards the side from which the current in the loop is anticlockwise, and the south pole is to that side from which the current in the loop is viewed clockwise. Hence, only the options (b) and (d) are correct.
3. Consider a magnetic dipole kept in the north-south direction. Let P₁, P₂, Q₁, Q₂ be four points at the same distance from the dipole towards the north, south, east, and west of the dipole, respectively. The directions of the magnetic field due to the dipole are the same at
(a) P₁ and P₂
(b) Q₁ and Q₂
(c) P₁ and Q₁
(d) P₂ and Q₂
ANSWER: (a), (b).
Explanation: The points P₁ and P₂ are equidistant from the dipole on both sides of its axis. At P₁, the direction of the field will be from the dipole towards the north pole. Similarly, at P₂, the direction of the field will also be towards the north and the south end of the dipole. Option (a) is correct.
Q₁ and Q₂ are symmetrical points on both sides of the perpendicular bisector of the dipole. Hence, the directions of the magnetic field will also be the same at these points. The direction will be from north to south. Hence the option (b) is also correct.
The directions of the field at points in options (c) and (d) are opposite to each other. Hence, not correct.
4. Consider the situation of the previous problem. The directions of the magnetic field due to the dipole are opposite at
(a) P₁ and P₂
(b) Q₁ and Q₂
(c) P₁ and Q₁
(d) P₂ and Q₂
ANSWER: (c), (d).
Explanation: As we can see in the diagram below,  |
| Diagram for Q-4 |
At points P₁ and Q₁, the directions of the magnetic field are opposite. Also, at points P₂ and Q₂, the directions are opposite. Hence, only options (c) and (d) are correct.
5. To measure the magnetic moment of a bar magnet, one may use
(a) a tangent galvanometer
(b) a deflection galvanometer if the Earth's horizontal field is known
(c) An oscillation magnetometer if the Earth's horizontal field is known
(d) Both deflection and oscillation magnetometers if the Earth's horizontal field is not known.
ANSWER: (b), (c), (d).
Explanation: To measure the magnetic moment of a bar magnet, we can measure M/BH using a deflection magnetometer or MBH using an oscillation magnetometer if we know BH. Hence, options (b), (c), and (d) are correct, not option (a).
EXERCISES
1. A long bar magnet has a pole strength of 10 A-m. Find the magnetic field at a point on the axis of the magnet at a distance of 5 cm from the north pole of the magnet.
ANSWER: Pole strength, m =10 A-m. Distance of the point from the pole, r =5 cm =0.05 m. Hence, the magnetic field,
B =(µₒ/4π)*m/r²
=(10⁻⁷)*10/(0.05)² T
=4x10⁻⁴ T
Outward along the axis from the north pole.
2. Two long bar magnets are placed with their axes coinciding in such a way that the north pole of the first magnet is 2 cm from the south pole of the second. If both the magnets have a pole strength of 10 A-m, find the force exerted by one magnet on the other.
ANSWER: Pole strength m =10 A-m, r =2 cm =0.02 m. The field at 2 cm from a pole,
B = (µₒ/4π)(m/r²)
=(10⁻⁷)*10/0.02² T
=2500x10⁻⁶ T
=2.5x10⁻³ T
Hence, the force exerted on the other magnet's pole,
F =mB
=10*2.5x10⁻³ N
=2.5x10⁻² N.
3. A uniform magnetic field of 0.20x10⁻³ T exists in space. Find the change in the magnetic scalar potential as one moves through 50 cm along the field.
ANSWER: The strength of the uniform magnetic field B =0.20x10⁻³ T. The change in the magnetic scalar potential is given as
V(r') -V(r) =-∫B.dr
=-B∫dr
{Because the direction of Band dr is the same and B is uniform. The limit of integration is from r =0 to r' =50 cm =0.50 m}
=-Br'
=-0.20x10⁻³*0.50 T-m
=-0.1x10⁻³ T-m.
So the magnetic scalar potential decreases by 0.1x10⁻³ T-m.
4. Figure (36-E1) shows some of the equipotential surfaces of the magnetic scalar potential. Find the magnetic field B at a point in the region.
 |
| The figure for Q-4 |
ANSWER: The direction of the magnetic field in the region at any point will be perpendicular to the equipotential surfaces. The strength of the field is,
B =-dV/dl
=-(-0.1x10⁻⁴ T-m)/(10xSin30°/100 m)
=1.0x10⁻⁴/(½) T
=2.0x10⁻⁴ T
5. The magnetic field at a point, 10 cm away from a magnetic dipole, is found to be 2.0x10⁻⁴ T. Find the magnetic moment of the dipole if the point is (a) in an end-on position of the dipole and (b) in a broadside-on position of the dipole.
ANSWER: (a) In an end-on position of the dipole, the magnetic field is given as,
B =(µₒ/4π)(2M/r³), where Mis the magnetic moment of the dipole.
So,
2x10⁻⁴ =(10⁻⁷){2M/(0.10)³}
→M =10⁷*10⁻⁴*10⁻³ A-m² =1.0 A-m²
(b) In the broadside-on position of a dipole, the magnetic field is given as,
B =(µₒ/4π)(M/r³), so
2x10⁻⁴ =10⁻⁷*M/(0.10)³
→M =2x10⁻⁴*10⁷*10⁻³ A-m²
=2.0 A-m².
6. Show that the magnetic field at a point due to a magnetic dipole is perpendicular to the magnetic axis if the line joining the point with the center of the dipole makes an angle of tan⁻¹(√2) with the magnetic axis.
ANSWER: See the figure below. θ is the angle between the line joining the point with the center of the dipole and the magnetic axis, while α is the angle between the magnetic field B and the line joining the point with the center of the dipole.  |
| Figure for Q-6 |
When the magnetic field B is perpendicular to the magnetic axis, the sum,
α+θ =π/2
→α =π/2 -θ
Also in general, tan α =(tanθ)/2
→tan(π/2 -θ) =(tanθ)/2
→Cotθ =(tanθ)/2
→tan²θ =2
→tanθ =√2
→θ =tan⁻¹(√2).
7. A bar magnet has a length of 8 cm. The magnetic field at a point at a distance of 3 cm from the center in the broadside-on position is found to be 4x10⁻⁶ T. Find the pole strength of the magnet.
ANSWER: Length of the bar magnet,
2l =8 cm =0.08 m
→l =0.04 m.
Distance of the point from the center of the magnet, d = 3 cm = 0.03 m.
The magnetic field B =4x10⁻⁶ T, (given).
But B = (µₒ/4π)M/(d²+l²)³'²
→M =B(d²+l²)³'²/(µₒ/4π)
=4x10⁻⁶(0.03²+0.04²)³'²*10⁷ A-m²
=40*0.05³ A-m²
=0.005 A-m²
→m*2l =0.005 A-m²
→m =0.005/0.08 A-m
=0.0625 A-m
=6.25x10⁻² A-m.
8. A magnetic dipole of magnetic moment 1.44 A-m² is placed horizontally with the north pole pointing towards the north. Find the position of the neutral point if the horizontal component of the Earth's magnetic field is 18 µT.
ANSWER: Horizontal component of Earth's magnetic field, BH =18 µT.
The neutral point will be at a point distance d away in the broad-on position where the magnetic field due to the dipole (B') and BH are equal in magnitude but opposite in direction. See the diagram below.
Magnetic field in the broad-on position of a dipole at a distance d from the center of the magnet,
B' =(µₒ/4π)*M/d³
→d³ =(µₒ/4π)*M/B'
=(10⁻⁷)*1.44/(18x10⁻⁶)
=0.008 m³
→d =0.20 m =20 cm
Such points will be all around the dipole, making a circle. So the neutral point is at a distance of 20 cm in the plane bisecting the dipole.
9. A magnetic dipole of magnetic moment 0.72 A-m² is placed horizontally with the north pole pointing towards the south. Find the position of the neutral point if the horizontal component of the Earth's magnetic field is 18 µT.
ANSWER: The direction of horizontal component of the Earth's magnetic field is towards the north. The direction of magnetic field due to the dipole is opposite in direction, i.e., towards the south will be in end-on position, south of the dipole. The neutral point will be at that point south of the dipole where the magnitude is equal to the horizontal component of the Earth's magnetic field but opposite in direction.
B' =(µₒ/4π)*(2M/d³)
→d³ =(µₒ/4π)*(2M/B')
=(10⁻⁷)*2*0.72/(18x10⁻⁶) m³
=0.008 m³
→d =0.20 m =20 cm.
So the neutral point in this case will be 20 cm south of the dipole.
10. A magnetic dipole of magnetic moment 0.72√2 A-m² is placed horizontally with the north pole pointing towards the east. Find the position of the neutral point if the horizontal component of the Earth's magnetic field is 18 µT.
ANSWER: Since the dipole is in east-west position, the magnetic field opposite to the horizontal component of Earth's magnetic field will be at an angle θ from the axis of the dipole. Suppose the magnitude of the magnetic field due to the dipole is equal to BH at the neutral point at a distance d from it. See the diagram below.  |
| Diagram for Q-10 |
In such case α+θ =90°,
→α = 90°-θ. Also,
tanα =(tanθ)/2
→tan(90°-θ) =(tanθ)/2
→cotθ =(tanθ)/2
→1/tanθ =(tanθ)/2
→tan²θ =2
→tanθ =√2.
So, θ =tan⁻¹√2.
The magnitude of the magnetic field,
B' =(µₒ/4π)M√(1+3.cos²θ)/d³ ----(i)
cosθ = 1/secθ =1/√(1+tan²θ)
=1/√(1+2) =1/√3.
And B' =BH =18 µT ==18x10⁻⁶ T.
Now from (i)
d³=(10⁻⁷)*0.72√2*√(1+1)/18x10⁻⁶ m³
=0.008 m³.
→d =0.20 m =20 cm.
So the neutral point is at 20 cm from the center of the dipole, and it is tan⁻¹√2 south of east.
11. The magnetic moment of the assumed dipole at the earth's center is 8.0x10²² A-m². Calculate the magnetic field B at the geomagnetic poles of the earth. The radius of the earth is 6400 km.
ANSWER: The geomagnetic poles are at 6400 km from the center of the earth on the magnetic axis. So we have to find the magnetic field on the end-on position of the dipole at a distance r =6.4x10⁶ m.
Hence the required field,
B =(µₒ/4π)*2M/r³
=2*(10⁻⁷)*8x10²²/(6.4x10⁶)³ T
=6.0x10⁻⁵ T
=60x10⁻⁶ T
=60 µT.
12. If the earth's magnetic field has a magnitude 3.4x10⁻⁵ T at the magnetic equator of the earth, what would be its value at the earth's geomagnetic poles?
ANSWER: The magnetic equator and the geomagnetic poles are at the same distance from the center of the earth that is equal to the distance of the earth's radius R. The magnetic equator is in a broadside-on position and the field is given as B =(µₒ/4π)M/R³.
The poles are at the end-on position and the field is given as B' =(µₒ/4π)*2M/R³.
So the field at the poles is double the field at the equator. Given the magnitude of the field at the magnetic equator B=3.4x10⁻⁵ T. Hence the magnitude of the field at the geomagnetic poles,
B' =2B
=2x3.4x10⁻⁵ T
=6.8x10⁻⁵ T.
13. The magnetic field due to the earth has a horizontal component of 26 µT at a place where the dip is 60°. Find the vertical component and the magnitude of the field.
ANSWER: The horizontal component of the earth's magnetic field,
BH =B.cosδ
Where B is the magnetic field of the earth and δ is the dip at that place. Hence the magnitude of the earth's field
B =BH/cosδ
=26/cos60° µT
=26*2 µT
=52 µT.
Also the vertical component of the magnetic field
BV =B.sinδ
=52*sin60° µT
=52*√3/2 µT
=45 µT.
14. A magnetic needle is free to rotate in a vertical plane which makes an angle of 60° with the magnetic meridian. If the needle stays in a direction making an angle of tan⁻¹(2/√3) with the horizontal, what would be the dip at that place?
ANSWER: Angle between magnetic meridian and the plane,
α =60°.
Apparent dip, δ' =tan⁻¹(2/√3)
→tanδ' =(2/√3)
The true dip at that place δ is given as,
cotδ' =cotδ*cosα
→Cotδ =cotδ'/cosα
=1/(tanδ'*cosα)
=1/(2/√3*cos60°)
=√3
→tanδ =1/√3 = tan30°
→δ =30°.
Otherwise, you can understand it as below:-
Effective horizontal component of the magnetic field
BH' = BHcos60° =BH/2
Apparent dip δ' is given by
tanδ' =Bv/BH' =2Bv/BH =2tanδ
(where δ is the true dip).
→2/√3 =2tanδ
→tanδ =1/√3
→δ =30°.
15. The needle of a dip circle shows an apparent dip of 45° in a particular position and 53° when the circle is rotated through 90°. Find the true dip.
ANSWER: Here, apparent dip δ' =45°. Suppose the plane of the dip circle makes an angle α with the magnetic meridian. Now if the true dip is δ then,
cotδ' =cotδ cosα --------(i)
When the dip circle is rotated through 90°, the angle between the plane of the dip circle and the meridian=90°-α. So the apparent dip now δ" is given as,
cotδ" =cotδ cos(90°-α) =cotδ sinα --(ii)
Squaring and adding (i) and (ii),
cot²δ =cot²δ' +cot²δ"
→cot²δ =cot²45°+cot²53°
=1² +(3/4)²
=5²/4²
→cotδ =5/4
→tanδ =4/5 =tan39°
→δ = 39°.
16. A tangent galvanometer shows a deflection of 45° when 10 mA of current is passed through it. If the horizontal component of the earth's magnetic field is BH = 3.6x10⁻⁵ T and the radius of the coil is 10 cm, find the number of turns in the coil.
ANSWER: Radius of the coil r =0.10 m. Deflection θ =45°. Current i =10 mA =0.01 A. Horizontal component of earth's magnetic field BH =3.6x10⁻⁵ T. If n is the number of turns in the coil, then the current is given as,
i =2rBH*tanθ/(µₒn)
→n =2*0.10*3.6x10⁻⁵/(4πx10⁻⁷*0.01)
=0.573x10³
=573.
17. A moving-coil galvanometer has a 50-turn coil of size 2 cm x 2 cm. It is suspended between the magnetic poles producing a magnetic field of 0.5 T. Find the torque on the coil due to the magnetic field when a current of 20 mA passes through it.
ANSWER: Number of turns, n =50.
Area of the coil A =0.02x0.02 m²
→A =4x10⁻⁴ m²
Magnetic field between the magnetic poles,
B = 0.5 T
Current in the coil, i =20 mA =0.02 A
Hence the torque on the coil,
𝛤 =niAB
=50*0.02*4x10⁻⁴*0.5 N-m
=2x10⁻⁴ N-m.
18. A short magnet produces a deflection of 37° in a deflection magnetometer in Tan-A position when placed at a separation of 10 cm from the needle. Find the ratio of the magnetic moment of the magnet to the earth's horizontal magnetic field.
ANSWER: Deflection θ =37°. Distance of the magnet from the needle d =10 cm =0.10 m
In the Tan-A position, the needle is in the end-on position of the magnet. Hence the magnetic field produced by the magnet at the needle,
B =(µₒ/4π)*2Md/(d²-l²)²
Since the magnet is short, so l<<d, we can neglect l. Thus
B =(µₒ/4π)*2M/d³
In Tan-A position, B and BH are perpendicular to each other and,
B = BH tanθ
→(µₒ/4π)*2M/d³ =BH tanθ
→M/BH =½(4π/µₒ)d³tanθ
=½(10⁷)*(0.10)³*tan37° A-m²/T
=(½*¾)x10⁴ A-m²/T
=3.75x10³ A-m²/T.
19. The magnetometer of the previous problem is used with the same magnet in the Tan-B position. Where should the magnet be placed to produce a 37° deflection of the needle?
ANSWER: In the Tan-B position, the needle is in a broadside-on position. The magnetic field by the magnet at needle is given as
B =(µₒ/4π)M/(d²+l²)³'²
For the short magnet we ignore l, thus
B =(µₒ/4π)M/d³
Since B and BH are perpendicular to each other,
B =BH tanθ
(µₒ/4π)M/d³ =BH tanθ
d³ =(µₒ/4π)(M/BH)/tan37°
=(10⁻⁷)(3.75x10³)/(¾) m³
=5x10⁻⁴ m³
→d =0.079 m =7.9 cm
So the magnet should be placed at 7.9 cm from the center in Tan-B position to produce a deflection of 37° in the needle.
20. A deflection magnetometer is placed with its arm in the north-south direction. How and where should a short magnet having M/BH =40 A-m²/T be placed so that the needle can stay in any position?
ANSWER: The horizontal component of the earth's magnetic field has a direction along the arm of the magnetometer i.e. from south to north. The condition to stay the needle in any position can be achieved if the net field in the horizontal direction is zero at the center and it can be done by putting the magnet in the north-south direction on the arm so that the north pole of the magnet faces the magnetic south direction. The needle is in the end-on position of the short magnet. The distance d is adjusted to have,
B =BH
but opposite in direction.
(µₒ/4π)*2M/d³ =BH
→d³ =2(µₒ/4π)(M/BH)
=2(10⁻⁷)(40) m³
=8x10⁻⁶ m³
d =2x10⁻² m = 0.02 m
→d =2 cm.
So the short magnet should be placed 2.0 cm from the needle, north pole pointing towards the south.
21. A bar magnet takes π/10 seconds to complete one oscillation in an oscillation magnetometer. The moment of inertia of the magnet about the axis of rotation is 1.2x10⁻⁴ kg-m² and the earth's horizontal magnetic field is 30 µT. Find the magnetic moment of the magnet.
ANSWER: Time period, T = π/10 s. The moment of inertia of the magnet, I =1.2x10⁻⁴ kg-m². BH =30µT. If the magnetic moment of the magnet be M, then,
T =2π√(I/MBH)
→M =4π²I/BHT²
=4π²*1.2x10⁻⁴/(30x10⁻⁶*π²/100)
=1.6x10³ A-m²
=1600 A-m².
22. The combination of two bar magnets makes 10 oscillations per second in an oscillation magnetometer when like poles are tied together and 2 oscillations per second when unlike poles are tied together. Find the ratio of the magnetic moments of the magnets. Neglect any induced magnetism.
ANSWER: Let the magnetic moments of the magnets be M and M'. When like poles are tied together the net magnetic moment =M +M'. When unlike poles are tied together, the net magnetic moment =M -M'.
In the first case time period, T =1/10 s =0.10 s. In the second case time period, T' =1/2 s =0.50s.
The relation between magnetic moment, time period, the moment of inertia of the magnet, and BH is given as
T =2π√(I/MBH).
→M =4π²I/BHT²
Hence in the first case,
M+M' =4π²I/BHT² ---------- (i)
Where I is the combined moment of inertia of the magnets.
In the second case,
M-M' =4π²I/BHT'² ----------- (ii)
Dividing (i) by (ii), we get
(M+M')/(M-M') =T'²/T² =(0.5/0.1)² =25/1
From componendo and dividendo rule
(M+M'+M-M')/(M+M'-M+M')=(25+1)/(25-1)
→2M/2M' =26/24
→M/M' =13:12.
23. A short magnet oscillates in an oscillation magnetometer with a time period of 0.10 s where the earth's horizontal magnetic field is 24 µT. A downward current of 18 A is established in a vertical wire placed 20 cm east of the magnet. Find the new time period.
ANSWER: Since the current in the wire is downwards, the circular magnetic field around it will be clockwise when viewed from above. Also, since the magnet is short, the magnetic field, B' due to the current will be taken towards the north.
B' =µₒi/(2πd)
=(µₒ/2π)*18/0.2
=180µₒ/4π T
=180*10⁻⁷ T
=18 µT.  |
| Diagram for Q-23 |
Now the total magnetic field in the north direction B" =B'+BH
=18 +24 T =42 µT.
The time period of the oscillation without the current,
T =2π√(I/MBH) and the time period with the current,
T' =2π√(I/MB").
Now, T'/T =√(BH/B")
→T' =T√(24/42)
=0.10*0.76 s =0.076 s.
So the new time period is 0.076 s.
24. A bar magnet makes 40 oscillations per minute in an oscillation magnetometer. An identical magnet is demagnetized completely and is placed over the magnet in the magnetometer. Find the time taken for 40 oscillations by the combination. Neglect any induced magnetism.
ANSWER: The time period in the first case, T =1/40 minutes =1.5 s. If the moment of inertia of the magnet is I, then in the second case the combined moment of inertia =2I. Let the magnetic moment of the magnet and the horizontal component of the E arth's magnetic field be M and BH, respectively.
T =2π√(I/MBH) ----- --- (i)
If the new time period is T', then,
T' =2π√(2I/MBH) ----- --- (ii)
Dividing (ii) by (i),
T'/T =√2
→T' =√2T =1.5√2 s
It is the time for one oscillation, hence the time taken to perform 40 oscillations =40*1.5√2 s =60√2 s =√2 minutes.
25. A short magnet makes 40 oscillations per minute when used in an oscillation magnetometer at a place where the earth's horizontal magnetic field is 25 µT. Another short magnet of magnetic moment 1.6 A-m² is placed 20 cm east of the oscillating magnet, Find the new frequency of oscillation if the magnet has its north pole (a) towards the north and (b) towards the south.
ANSWER: The oscillating magnet is at the broadside-on position of the short magnet placed 20 cm east of it. The magnetic field due to the second magnet at the position of the first magnet,
B' =(µₒ/4π)M/d³
=10⁻⁷*1.6/0.2³ T
=20x10⁻⁶ T
=20 µT.
The time period of oscillation before placing the magnet, T =60/40 s =1.5 s.
So, T =2π√(I/MBH)
(a) When the second magnet has its north pole towards the north, the direction of B' is towards the south i.e. opposite to BH. The net horizontal field at the first magnet, B" =BH-B' =25-20 µT
→B" =5 µT.
So the time period now,
T' =2π√(I/MB")
Hence,
T'/T =√(BH/B") ---- ----- (i)
→T' =1.5√(25/5) =3.35 s
Oscillation per minute =60/3.35 ≈18.
(b) When the second magnet has its north pole towards the south the direction of B' at the position of the first magnet is towards the north i.e. same as BH. Hence the net horizontal magnetic field at the oscillating magnet,
B" =BH+B' =25+20 =45 µT.
The new time period T' is given from (i) as,
T' =1.5√(25/45) =1.12 s
Hence oscillation per minute =60/1.12 =54.
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