HC Verma Solutions for Class 12 Physics Chapter 36 – Permanent Magnets

Page No 275:

Question 1:

Can we have a single north pole, or a single south pole?

Answer:

No, we cannot have a single north or south pole. Magnetic poles are always found in pairs. They are equal in strength and opposite in nature. Even if we break a magnet into a number of pieces, each piece will become a magnet with equal and opposite poles.

Question 2:

Do two distinct poles actually exist at two nearby points in a magnetic dipole?

Answer:

No, two distinct poles cannot exist at two nearby points in a magnet, as a magnet contains only two distinct poles located at its ends.

Question 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 bare magnet different from that of the middle  region?

Answer:

No, the material making up the middle region of a magnet is the same as that of the material making up its end. When an iron needle is taken closer to one of the ends of a magnet, the pole of the magnet induces a pole of opposite polarity on the needle, making the needle a magnet itself and thereby making it attracted to that pole.
But if we bring the needle closer to the centre of the magnet, then both the poles of the magnet will induce opposite polarity on the needle. As a result, the needle will not get attracted towards the centre of the magnet.

Question 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 north pole and south pole.

Answer:

The direction of the magnetic field is the same in both cases, that is, inside a solenoid and inside a bar magnet. In a solenoid, magnetic field lines are directed from one end to the other internally and externally, so they are in the equivalent combination of north and south poles (as shown in figure).

Question 5:

Sketch the magnetic field lines for a current-carrying circular loop near its centre. Replace the loop by an equivalent magnetic dipole and sketch the magnetic field lines near the centre of the dipole. Identify the difference.

Answer:

The difference between the two configurations is that in the current-carrying loop, the magnetic field lines pass through the centre and are perpendicular to its axis; whereas in the equivalent magnetic dipole, the magnetic field lines do not pass through the centre.

         

 

Question 6:

The force on a north pole,

F→=mB→, parallel to the field

B→. Does it contradict our earlier knowledge that a magnetic field can exert forces only perpendicular to itself?

Answer:

Yes, it seems to contradict with our earlier knowledge that a magnetic field can exert forces only perpendicular to itself.

F→=mB→Here,

B→= Magnetic field
m = Magnetic charge
For a positive magnetic charge, force is along the magnetic field.
For a negative magnetic charge, force is opposite to the magnetic field.
Thus, it contradicts the notion that a magnetic field can exert forces only perpendicular to itself.F

Question 7:

Two bar magnets are placed close to each other with their opposite poles facing each other. In 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 kinetic energy of a system?

Answer:

Yes, it contradicts our earlier knowledge that magnetic forces cannot do any work and hence cannot increase the kinetic energy of the system. When opposite poles are facing each other, an attractive force acts between them so the magnets are pulled towards each other. As the two magnets come close to each other so the force between them increases and hence, the kinetic energy also increases.

Question 8:

Magnetic scalar potential is defined as

Ur→2-Ur→1=-∫r→1r→2B→.dl→Apply this equation to a closed curve enclosing a long straight wire. The RHS of the above equation is then −μ0 i by Ampere’s law. We see that

Ur→2≠Ur→1even when

r→2=r→1. Can we have a magnetic scalar potential in this case?

Answer:

No, we cannot have a magnetic scalar potential here.

Ampere’s law is a method of calculating magnetic field due to current distribution. On the other hand, magnetic scalar potential requires a magnetic field due to pole strength m.
Potential at a distance r is given by
μ0m4πrAs there is no current distribution, no magnetic field due to poles or the pole strength is present. That is why we cannot have a magnetic scalar potential in this case.

Question 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, Earth’s magnetic field is vertical at the poles. A freely suspended magnet becomes vertical at the poles, with its north pole pointing towards Earth’s north pole, which is magnetic south.
The value of the angle of the dip here is 90°.

Page No 276:

Question 10:

Can the dip at a place be (a) zero (b) 90°?

Answer:

(a) Yes, the dip can be zero at the equator of Earth.

(b) Yes, the dip can be 90°â€‹ at the poles of Earth.

Question 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 to tan θ. The procedure works well at Bhuwaneshwar. Will the procedure work if the instrument is taken to Nepal? If there is same error, can it be corrected by correcting the manual or the instrument will have to be taken back to the factory?

Answer:

Yes, the procedure will work if the instrument is taken to Nepal, as the current at a place can be calculated by multiplying the reduction factor K with tan

θ of that place. In our case, we will take the value of tan

θ of Nepal, as tan

θmay vary from place to place. tan

θat any place is determined from the mathematical formula

BBH, where B is the external magnetic field and BH is the horizontal component of Earth’s magnetic field. Thus, we need not take the manual or the instrument back to the factory for correction.

Question 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) end-on position

Points lying on the axis of a magnet are called end-on points. In our case, the point on the axis of the loop (on replacing the circular loop with an equivalent magnetic dipole) lies on the axis of the magnetic dipole or on the end-on position.
If P was the point on the axis of the loop, then it is clear from the figure that P lies on the end-on position of the equivalent magnetic dipole.

Question 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) broadside-on position

The position of the points lying on the equator of a magnetic dipole is called the broadside-on position. In our case, the point on the loop (after replacement of the circular loop with an equivalent magnetic dipole) lies on the equatorial position of the equivalent magnetic dipole. Hence, the point lies on the broadside-on position.
If P was the point on the loop, then it is clear from the figure that point P lies on the broadside-on position of the equivalent magnetic dipole.

Question 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) the product md is fixed

When we replace a circular current-carrying loop with a magnetic dipole to resemble field lines of the circular loop, the pole strength m and the distance between the poles are not fixed.
But the magnetic dipole moment of both systems is always fixed. It is the product of the magnetic moment and the distance between the poles. In other words, md is fixed.
A current loop of area A and current I can be replaced with a magnetic dipole of dipole moment md.
i.e. md = IA

Question 4:

Let r be the distance of a point on the axis of a bar magnet from its centre. The magnetic field at such a point is proportional to

(a)

1r(b)

1r2(c)

1r3(d) none of these

Answer:

(d) None of these

Magnetic field B due to a bar magnet of magnetic moment M at distance r of the point on the axis of the magnet from its centre is given by

B=μo4π2Mrr2-l22
Here, 2l  is the length of the magnet.
So, from the above formula, it can be easily seen that

B∝rr2-l22.

Question 5:

Let r be the distance of a point on the axis of a magnetic dipole from its centre. The magnetic field at such a point is proportional to

(a)

1r(b)

1r2(c)

1r3(d) none of these

Answer:

(c)

1r3Magnetic field B due to a bar magnet of magnetic moment M at distance r of the point on the axis from its centre is given by

B=μ02Mr4πr2-l22Here, 2l  is the length of the magnet.
When the distance of the point where the magnetic field has to be calculated is greater than the length of the magnet, i.e

r >>l, the bar magnet acts like a magnetic dipole whose magnetic field is
B

∝ 1r3Now, l in the denominator can be neglected.
So, the correct option is (c).

Question 6:

Two short magnets of equal dipole moments M are fastened perpendicularly at their centre (figure 36-Q1). The magnitude of the magnetic field at a distance d from the centre on the bisector of the right angle is

(a)

μ04πMd3(b)

μ04π2 Md3(c)

μ04π22Md3(d)

μ04π2Md3Figure

Answer:

(c)

μ04π22Md3
Magnetic field (B1) due to the short dipole A of dipole moment M at an axial point  is given by,

B→1=μ04π2Md3  …1Magnetic field (B2) due to the short dipole B of dipole moment M at an axial point is given by,

B→2=μ04π2Md3   …2Resultant magnetic field (B) will be,
B =

B12+B22B =

μ04π22Md3

Question 7:

Magnetic meridian is

(a) a point
(b) a line along north-south
(c) a horizontal plane
(d) a vertical plane

Answer:

(d) a vertical plane

Magnetic meridian at a place is not a line but a vertical plane passing through the axis of a freely suspended magnet.

Question 8:

A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It

(a) will stay in north-south direction only
(b) will stay in east-west direction only
(c) will become rigid showing no movement
(d) will stay in any position

Answer:

(d) will stay in any position

When taken to a geomagnetic pole, a compass needle that is allowed to move in a horizontal plane will try to suspend itself vertically to the horizontal plane containing the compass. In other words, the horizontal plane containing the compass will restrict the compass to suspend itself in vertical direction; hence, the compass will stay in any position.
However, a freely suspended magnet will become vertical at poles, with its north pole pointing towards Earth at its north pole (which is magnetic south).

Question 9:

A dip circle is taken to geomagnetic equator. The needle is allowed to move in a vertical plane perpendicular to the magnetic meridian. The needle will stay

(a) in horizontal direction only
(b) in vertical direction only
(c) in any direction except vertical and horizontal
(d) in the direction it is released

Answer:

(d) in the direction it is released

At the geomagnetic equator, the needle tries to suspend itself in horizontal direction. But here the needle is restricted to move only in the vertical plane perpendicular to the magnetic meridian. Hence, the needle will stay in the direction it is released.

Question 10:

Which of the following four graphs may best represent the current-deflection relation in a tangent galvanometer?

Figure

Answer:

(c) curve

Since i

∝tan

θ, the only graph that represents this correlation is curve c.

Question 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) remain unchanged

For a tangent galvanometer, deflection is given by

θ = tan-1ik
Here, k is the constant called reduction factor.
From the above formula, we can say that deflection is independent of the number of turns.
Hence, on doubling the number of turns, deflection remains the same.

Question 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) a moving coil galvanometer

The current and deflection dependence of a moving coil galvanometer is given by

i=knABθ ⇒i∝θTherefore, if we double the current, the deflection also gets doubled.
However, in a tangent galvanometer,

i∝tanθ; that is, there is no direct relation between

θand current.
Hence, the correct option is (b).

Question 13:

A very long bar magnet is placed with its north pole coinciding with the centre of a circular loop carrying as 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) very nearly 2

πaiB perpendicular to the plane of the wire

In this case, the north pole of the magnet is coinciding with the centre of the circular loop carrying electric current i. So, the magnetic field lines almost lie on the plane of the ring and the force due to the field lines is perpendicular to the field lines and to the plane of the circular ring.
Let idl be the current element, B be the magnetic field and dF be the force on the current element idl.
Now
dF = Bidl

⇒F=∫02πaBidl

⇒F=2πaiBThus, the force acting on the wire is 2

πaiB and it is perpendicular to the plane of the wire.

Page No 277:

Question 1:

Pick the correct options.

(a) Magnetic field is produced by electric charges only
(b) Magnetic poles are only mathematical assumptions having no real existence
(b) 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) Magnetic field is produced by electric charges only.
(b) Magnetic poles are only mathematical assumptions having no real existence.

Justification of (a) and (b):

Investigators and experimenters have failed to find any sign of magnetic monopoles. So, we can assume that magnetic monopoles are only a mathematical assumption.
A magnetic field is produced by the motion of an electric charge only. In paramagnets or ferromagnets, the motion of an electron (charge) and the alignment of domains (bunch of charges with particular alignment) create paramagnetism and ferromagnetism, respectively.
Therefore, the only cause behind the magnetic field is the motion of an electric charge.

Denial of (c):

The north pole is equivalent to an anticlockwise current and the south pole is equivalent to a clockwise current.

Denial of (d):

A bar magnet is not equivalent to a long, straight current because the distribution and orientation of magnetic field lines do not resemble each other.

Question 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 a diameter of the loop.
(b) The line SN should be perpendicular to the plane of the loop
(c) The south pole should be slow the loop
(d) The north pole should be below the loop

Answer:

(b) The line SN should be perpendicular to the plane of the loop
(d) The north pole should be below the loop.

A horizontal circular loop carrying current in clockwise direction acts like the south pole of a magnet. Hence, the south pole of the magnet coincides with the loop.

Now, when the loop carrying current in clockwise direction is viewed from above, it looks like the magnetic lines of force are entering the loop thus it acts like south pole of a magnet. And if we view from below the loop then it appears that magnetic lines of force are leaving the loop.  Hence, the north pole should be below the loop.

Question 3:

Consider a magnetic dipole kept in the north to south direction. Let P1, P2, Q1, Q2 be four points at the same distance from the dipole towards 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) P1 and P2
(b) Q1 and Q2

We know that magnetic field lines are directed from the north pole to the south pole. From the given figure, we can say that the direction of magnetic field

→Bis the same only at points P1 and P2 and at points Q1 and Q2.

Question 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) P1 and Q1
(d) P2 and Q2

We know that magnetic field lines are directed from the north pole to the south pole. From the given figure, we can say that the direction of the magnetic field

→Bis opposite at points P1 and Q1 and at points P2 and Q2.

Question 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 magnetometer if the earth’s horizontal field is not known

Answer:

(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

Denial of (a):

Tangent galvanometer is an instrument used to measure electric current; it cannot be used to the measure magnetic moment of a bar magnet.

Justification of (b) and (c):

Deflection magnetometer is used to measure

MBH of a permanent bar magnet.
Similarly, oscillation magnetometer is used to measure M BH of a bar magnet. So, if earth’s horizontal field, BH, is known, then the magnetic moment of a bar magnet, M, can be measured.

Justification of (d):

Using deflection and oscillation magnetometers, we can calculate

MBHand M BH, respectively. Therefore, if we multiply the result obtained from both the instruments, then BH cancels out as

MBH × M BH​ = M2. Thus, the value of BH is not required.
Therefore, we can use both deflection and oscillation magnetometers if the earth’s horizontal field is not known.

Question 1:

A long bar magnet has a pole strength of 10 Am. 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:

Given:
Pole strength of the bar magnet, m = 10 Am
Distance of the point from the north pole of the bar magnet, = 5 cm = 0.05 m
We know,
The magnetic field due to magnetic charge

Bis given by

B=μ04π mr2   =10-7×105×10-22=10-625×10-4   =10-225=4×10-4 T

Question 2:

Two long bare magnets are placed with their axes coinciding in such a way that the north pole of the first magnet is 2.0 cm from the south pole of the second. If both the magnets have a pole strength of 10 Am, find the force exerted by one magnet of the other.

Answer:

Given:
Pole strength = m1 = m2 = 10 Am
Distance between the north pole of the first magnet and the south pole of the second magnet, r = 2 cm = 0.02 m
We know,
Force

Fexerted by two magnetic poles on each other is given by

F=μ04πm1m2r2  =4π×10-7×1024π×4×10-4  =2.5×10-2 N

Question 3:

A uniform magnetic field of 0.20 × 10−3 T exists in the space. Find the change in the magnetic scalar potential as one moves through 50 cm along the field.

Answer:

Given:
Magnetic field in the space, B = 0.20 × 10−3 T
Distance moved,

∆r= 50 cm
We know,

B =-dVdl ⇒dV = ∫r1r2 -B.⇀dl
Since the magnetic field is uniform, it can come out of the integration sign.

⇒ ∆V = -B.(∆r)Here,

∆Vis the change in the potential.
∴ Change in the potential = −0.2 × 10−3 × 0.5
= −0.1 × 10−3 T-m
Here, the negative sign shows that the potential decreases.

Question 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.

Figure

Answer:

Given:
Perpendicular distance, dx = 10 sin30â‚’ cm = 0.05 m,
Change in the potential, dV = 0.1 × 10−4 T-m
We know that the relation between the potential and the field is given by

B=-dVdx⇒B = -0.1×10-4 T-m5×10-2 m ⇒B = -2×10-4 TB is perpendicular to the equipotential surface. Here, it is at angle of 120° with the positive x-axis.

Question 5:

The magnetic field at a point, 10 cm away from a magnetic dipole, is found to be 2.0 × 10−4 T. Find the magnetic moment of the dipole if the point is (a) in end-on position of the dipole and (b) in broadside-on position of the dipole.

Answer:

Given:
Magnetic field strength, B = 2 × 10−4 T
Distance of the point from the dipole, d = 10 cm = 0.1 m

(a) If the point is at the end-on position:
The magnetic field

Bon the axial point of the dipole is given by

B=μ04π2Md3Here, M is the magnetic moment of the dipole that we need to find out.∴2×10-4=10-7×2M10-13⇒M=2×10-4×10-310-7×2⇒M=1 A-m2(b) If the point is at broadside-on position (equatorial position):
The magnetic field

Bis given by

B=μ04πMd3⇒2×10-4=10-7×M10-13⇒M=2 A-m2

Question 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 centre of the dipole makes an angle of

tan-1 2with the magnetic axis.

Answer:

Given:
Angle made by observation point P with the axis of the dipole,

θ=

tan-12

⇒tan θ=2 ⇒2=tan2 θ⇒tan θ= cot θ⇒tan θ2=cot θ     ….1We know,tan θ2=tan α          ….2
On comparing (1) and (2), we get
tan α = cot θ

⇒tan α = tan (90 − θ)

⇒α = 90 − θ

θ + α = 90°
Hence, the magnetic field due to the dipole is perpendicular to the magnetic axis.

Question 7:

A bar magnet has a length of 8 cm. The magnetic field at a point at a distance 3 cm from the centre in the broadside-on position is found to be 4 × 10−6 T. Find the pole strength of the magnet.

Answer:

Given:
Length of the magnet, 2l = 8 cm = 8

×10

-2m
Distance of the observation point from the centre of the dipole, d = 3 cm
Magnetic field in the broadside-on position, B = 4 × 10−6 T
The magnetic field due to the dipole on the equatorial point

Bis given by

B=μ0m2l4πd2+l23/2Here, m is the pole strength of the magnet.
On substituting the respective values, we get

4×10-6 = 10-7m×8×10-29×10-4+16×10-43/2⇒ 4×10-6 = m×8×10-9253/2×10-43/2⇒m = 4×10-6×125×10-68×10-9⇒m= 6.25×10-2 A-mThus, the pole strength of the magnet is 6.25

×10

-2A-m.

Question 8:

A magnetic dipole of magnetic moment 1.44 A m2 is placed horizontally with the north pole pointing towards north. Find the position of the neutral point if the horizontal component of the earth’s magnetic field is 18 μT.

Answer:

Given:
Magnetic moment of the magnetic dipole, M = 1.44 A-m2
​Horizontal component of Earth’s magnetic field, BH = 18 μT
We know that for a magnetic dipole with pole pointing to the north, the neutral point always lies in the broadside-on position.
Let d be the perpendicular distance of the neutral point from mid point of the magnet.
The magnetic field due to the dipole at the broadside-on position (B) is given by

B→  = μ0M4πd3This magnetic field strength should be equal to the horizontal component of Earth’s magnetic field at that point, that is, BH due to Earth.
Thus,

B→  = μ0M4πd3⇒B=10-7×1.44d3⇒18×10-6=10-7×1.44d3⇒d3=8×10-3⇒d=2×10-1=20 cm

Question 9:

A magnetic dipole of magnetic moment 0.72 A m2 is placed horizontally with the north pole pointing towards south. Find the position of the neutral point if the horizontal component of the earth’s magnetic field is 18 μT.

Answer:

Given:
Magnetic moment of the magnetic dipole, M = 0.72 Am2
​Horizontal component of Earth’s magnetic field, BH​ = 18 μT
​Let d be the distance of the neutral point from the south of the dipole.
When the magnet is such that its north pole faces the geographic south of Earth, the neutral point lies along the axial line of the magnet.
Thus, the magnetic field on the axial point of the dipole (B) is given by

B=μ04π2Md3This magnetic field strength should be equal to the horizontal component of Earth’s magnetic field.
Thus,

10-7×2×0.72d3=18×10-6⇒d3=2×0.72×10-718×10-6⇒d= 8×10-910-61/3⇒d=2×10-1 m=20 cm

Question 10:

A magnetic dipole of magnetic moment

0.722 A m2is placed horizontally with the north pole pointing towards east. Find the position of the neutral point if the horizontal component of the earth’s magnetic field is 18 μT.

Answer:

Given:
Magnetic moment of magnetic dipole, M

=0.722 A-m2Horizontal component of the earth’s magnetic field, BH = 18 μT
Let d be the distance of neutral point from the dipole.
Magnetic field due to the bar magnet

Bon the equatorial line of the dipole is given by,

B=μ04π Md3⇒4π×10-74π×0.722d3=18×10-6⇒d3=0.72×1.414×10-718×10-6⇒d3=0.005656⇒d=0.0056563⇒d≈0.2 m=20 cm

Question 11:

The magnetic moment of the assumed dipole at the earth’s centre is 8.0 × 1022 A m2. Calculate the magnetic field B at the geomagnetic poles of the earth. Radius of the earth is 6400 km.

Answer:

Given:
Magnetic moment of dipole at Earth’s centre, M = 8.0 × 1022 Am2
Radius of Earth, d = 6400 km
The geomagnetic pole is at the end of the position (axial) of Earth.
Thus, the magnetic field at the axial point of the dipole

B is given by

B=μ04π2Md3⇒B= 10-7×2×8×1022643×1015⇒B= 6×10-5  T=60 μT

Question 12:

If the earth’s magnetic field has a magnitude 3.4 × 10−5 T at the magnetic equator of the earth, what would be its value at the earth’s geomagnetic poles?

Answer:

Given:
Magnetic field at the magnetic equator, B = 3.4 × 10−5 T
Let M be the magnetic moment of Earth’s magnetic dipole and R be the distance of the observation point from the centre of Earth’s magnetic dipole.
As the point on the magnetic equator is on the equatorial position of Earth’s magnet, the magnetic field at the equatorial point

B is given by

B=μ04πMR3⇒μ04π×MR3=3.4×10-5⇒M=3.4×10-5×R3×4π4π×10-7⇒M=3.4×102 R3As the magnetic field on Earth’s geomagnetic poles lies on the axial point of the magnetic dipole, the magnetic field at the axial point

B1is given by

B→1=μ04π×2MR3⇒B1=10-7×2×3.4×102 ×R3R3⇒B1=6.8×10-5 T

Question 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:

Given:
Horizontal component of Earth’s magnetic field, B = 26 μT â€‹
Angle of dip, δ = 60°
The horizontal component of Earth’s magnetic field

BH is given by
BH = Bcosδ
Here,
B = Total magnetic field of Earth
On substituting the respective values, we get

26×10-6=B×12⇒B=52×10-6=52 μTThe vertical component of Earth’s magnetic field

By is given by

By=Bsinδ⇒By=52×10-6×32⇒By=44.98 μT=45 μT

Question 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 23with the horizontal, what would be the dip at that place?

Answer:

Given:
Angle made by the magnetic meridian with the plane of rotation of the needle,

θ = 60°Angle made by the needle with the horizontal,

δ1 = tan-1(23)If

δ is the angle of dip, then

tanδ1=tanδcosθ⇒tanδ=tanδ1cosθ⇒tanδ=tan tan-123cos60°⇒tanδ=23×12=13⇒δ=30°

Page No 278:

Question 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:

Given:
Apparent dip shown by the needle of the dip circle, δ1 = 45°â€‹
Dip shown by the needle of the dip circle on rotating the circle, ​δ= 53° â€‹
True dip

δ is given by
Cot2 δ = Cot2 δ1 + Cot2 δ2

⇒Cot2 δ = Cot2 45° + Cot2 53°

⇒Cot2 δ = 1.56

⇒Cot2 δ = 1.56

δ = 38.6° ≈ 39°

Question 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 = 3.6 × 10−5 T and radius of the coil is 10 cm, find the number of turns in the coil.

Answer:

Given:
Horizontal component of Earth’s magnetic field, BH = 3.6 × 10−5 T
Deflection shown by the tangent galvanometer, θ = 45°
Current through the galvanometer, I = 10 mA = 10−2 A
Radius of the coil, r = 10 cm = 0.1 m
Number of turns in the coil, n = ?
We know,

BHtan θ=μ0 In2r⇒n=BHtan θ×2rμ0I⇒n=3.6×10-5×2×1×10-14π×10-7×10-2⇒n=0.57332×103=573

Question 17:

A moving-coil galvanometer has a 50-turn coil of size 2 cm × 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:

Given:
Number of turns in the coil, n = 50
Area of the cross section of the coil, A = 2 cm × 2 cm = 2 × 2 × 10−4 m2
Current flowing through the coil, I = 20 × 10−3 A
Magnetic field strength due to the presence of the poles, B = 0.5 T
The torque experienced by the coil placed in an external magnetic field

τis given by

τ=nIA→×B→⇒τ=nIABsin 90°⇒τ=50 ×20×10-3×4×10-4×0.5⇒τ=2×10-4 N-m

Question 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:

Given:
Deflection in the magnetometer in the given position when placed in the magnetic field of a short magnet, θ = 37°
Separation between the magnet and the needle, d = 10 cm = 0.1 m
Let M be the magnetic moment of the magnet and BH be Earth’s horizontal magnetic field.
According to the magnetometer theory,

MBH=4πμ0d2-l222dtan θ        For the short magnet, MBH=4πμ0×d42dtan θ⇒MBH=4π4π×10-7×0.132×tan 37°⇒MBH=0.5×0.75×1×104⇒MBH=3.75×103 A-m2/T

Question 19:

The magnetometer of the previous problem is used with the same magnet in Tan-B position. Where should the magnet be placed to produce a 37° deflection of the needle?

Answer:

Given:
Deflection in the magnetometer in the given position when placed in the magnetic field of a short magnet, θ = 37°
Separation between the magnet and the needle, d = 10 cm = 0.1 m
Let M be the magnetic moment of the magnet and BH be Earth’s horizontal magnetic field.
According to the magnetometer theory,

MBH=4πμ0d2-l222dtan θ        For the short magnet,MBH=4πμ0×d42dtan θ⇒MBH=4π4π×10-7×0.132×tan 37°⇒MBH=0.5×0.75×1×104⇒MBH=3.75×103 A-m2/T            = 3.75 × 103 A-m2/T
Deflection in the magnetometer θ = 37°
From the magnetometer theory in Tan-B position, we have

MBH=4πμ0d2+l23/2 tan θSince for the short magnet l << d, we can neglect l w.r.t. d.
Now,

MBH=4πμ0d3 tanθ⇒3.75×103=110-7×d3×0.75⇒d3=3.75×103×10-70.75 =5×10-4⇒d=5×10-43⇒d=0.079 m=7.9 cmMagnet will be at 7.9 cm from the centre.

Question 20:

A deflection magnetometer is placed with its arms in north-south direction. How and where should a short magnet having M/BH = 40 A m2 T−1 be placed so that the needle can stay in any position?

Answer:

Given:
Ratio,

MBH = 40 Am2/TSince the magnet is short, l can be neglected.
So, using the formula for

MBHfrom the magnetometer theory and substituting all values, we get

MBH=4πμ0d32=40⇒d3=40×10-7×2⇒d3=8×10-6⇒d=2×10-2 m=2 cmThus, the magnet should be placed in such a way that its north pole points towards the south and it is 2 cm away from the needle.

Question 21:

A bar magnet takes π/10 second the complete one oscillation in an oscillation magnetometer. The moment of inertia of the magnet about the axis of rotation is 1.2 × 10−4 kg m2 and the earth’s horizontal magnetic field is 30 μT. Find the magnetic moment of the magnet.

Answer:

Given:
Time taken by the bar magnet to complete one oscillation, T =

π10 sMoment of inertia of the magnet about the axis of rotation, I = 1.2 × 10−4 kgm2
Horizontal component of Earth’s magnetic field, BH = 30 μT
Time period of oscillating magnetometer

Tis given by

T=2πIMBHHere,
M = Magnetic moment of the magnet
On substituting the respective values, we get

π10=2π1.2×10-4M×30×10-6⇒1202=1.2×10-4M×30×10-6⇒M=1.2×10-4×40030×10-6⇒M=16×102 A-m2⇒M=1600 A-m2

Question 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:

Given:
Number of oscillations per second made by the combination of bar magnets with like poles,

ν1= 10 s

-1
Number of oscillations per second made by the combination of bar magnets with unlike poles,

ν2= 2 s

-1
The frequency of oscillations in the magnetometer

νis given by

υ=12π MBHIWhen like poles are tied together, the effective magnetic moment is
M = M1M2
When unlike poles are tied together, the effective magnetic moment is
M = M1 +

M2
As the frequency of oscillations is directly proportional to the magnetic moment,

υ1υ2 = M1-M 2M1+M2⇒1022=M1-M2M1+M2⇒251=M1-M2M1+M2⇒25+125-1=M1-M2+M1+M2M1-M2-M1-M2⇒2624=2M1-2M2⇒M1M2=-2624=-1312Hence, the ratio of the effective magnetic moment is

-1312.

Question 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:

Here,
Horizontal component of Earth’s magnetic field, BH = 24 × 10−6 T
Time period of oscillation, T1 = 0.1 s
Downward current in the vertical wire, I = 18 A
Distance of wire from the magnet, r = 20 cm = 0.2 m
In the absence of the wire net magnetic field, BH = 24 × 10−6 T
When a current-carrying wire is placed near the magnet, the effective magnetic field gets changed.
Now the net magnetic field can be obtained by subtracting the magnetic field due to the wire from Earth’s magnetic field.
B = BHBwire
Thus, the magnetic field due to the current-carrying wire

Bwireis given by
B =

μ0I2πrThe net magnetic field

Bis given by
B=24×10-6-μ0 I2πrB=24×10-6-2×10-7×180.2B=24-10×10-6B=14×10-6Time period of the coil

Tis given by

T=2πIMBHLet T1 and T2 be the time periods of the coil in the absence of the wire and in the presence the wire respectively.
As time period is inversely proportional to magnetic field,

T1T2=BBH⇒0.1T2=14×10-624×10-6⇒0.1T22=1424⇒T22=0.01×1424⇒T2=0.076 s

Question 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 this combination. Neglect any induced magnetism.

Answer:

Given:
Frequency of oscillations of the bar magnet in the oscillation magnetometer,

ν= 40 min

-1
Time period for which the bar magnet is placed in the oscillation magnetometer, T1 =

140 minThe time period of oscillations of the bar magnet

Tis given by

T=2πIMBHHere,
I = Moment of inertia
M = Magnetic moment of the bar magnet
BH = Horizontal component of the magnetic field
Now, let T2 be the time period for which the second demagnetised magnet is placed over the magnet.
As the second magnet is demagnetised, the combination will have the same values of M and BH as those for the single magnet. However, variation will be there in the value of I on placing the second demagnetised magnet.

∴T1T2=I1I2⇒140T2=12⇒11600 T22=12⇒T22=1800⇒T2=0.03536 minFor 1 oscillation,
Time taken = 0.03536 min
For 40 oscillations,
Time taken = 0.03536 × 40
= 1.414 min

=2 min

Question 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 north and (b) towards south.

Answer:

Here,
Frequency of oscillations,

ν1 = 40 oscillations/min
Earth’s horizontal magnetic field, BH = 25 μT
Magnetic moment of the second magnet, M = 1.6 A-m2
Distance at which another short magnet is placed, d = 20 cm = 0.2 m

(a) For the north pole of the short magnet facing the north, frequency

v1is given by

v1=12π MBHIHere,
M = Magnetic moment of the magnet
I = Moment of inertia
BH = Horizontal component of the magnetic field
Now, let B be the magnetic field due to the short magnet.
When the north pole of the second magnet faces the north pole of the first magnet, the effective magnetic field

Beffis given by
Beffective = BH

B
The new frequency of oscillations

v2on placing the second magnet is given by

v2=12π MBH-B1The magnetic field produced by the short magnet

Bis given by

B=μ04πmd3⇒B =10-7×1.68×10-3=20 μTSince the frequency is proportional to the magnetic field,

v1v2=BHBH-B⇒40v2=255⇒40v2=5⇒v2=405=17.88        =18 oscillations/min(b) For the north pole facing the south,

v1=12πMBHI⇒v2=12πMB+BH1⇒v1v2=BHB+BH⇒40v2=2545⇒v2=4025/45=54 oscillations/min

HC Verma Solutions for Class 11 Physics – Part 1

HC Verma Solutions for Class 12 Physics – Part 2