# H C VERMA Solutions for Class 12-science Physics Chapter 14 - Permanent Magnets

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## Chapter 14 - Permanent Magnets Exercise 277

Question 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 5cm from the north pole of the magnet.

Solution 1

m=10 A-m

r=5cm=0.05m

Magnetic field,

T

Question 2

Two long bar magnets are placed with their axes coinciding in such a way that the north pole of the first magnet is 2.0cm 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.

Solution 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 move through 50cm along the field.

Solution 3

T-m

Question 4

Figure shows some of the equipotential surfaces of the magnetic scalar potential. Find the magnetic field B at a point in the region.

Solution 4

Perpendicular distance between the equi-potential surfaces

Now,

Magnetic field,

Direction of magnetic field is from higher potential to lower potential and perpendicular to equi-potential surface.

So, magnetic field is at  with the positive x-axis.

Question 5

The magnetic field at a point, 10cm away from a magnetic field 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.

Solution 5

T

r=10cm=0.1m

(a) For end-on position

M=1 A-

M=2 A-

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 center of the dipole makes an angle of  with the magnetic axis.

Solution 6

Let at point P, magnetic field is perpendicular to the magnetic axis.

Here,

We know that,

Question 7

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

Solution 7

Length of the magnet, 2l=8cml=4cm

Here, r=3cm

B=T

A-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 18𝜇T.

Solution 8

The neutral point for a magnet with north pole pointing in north will be at broadside on position.

At neutral point

Magnetic field due to magnet=earth's horizontal magnetic field

r=0.2m=20cm

from the midpoint of magnet of plane bisecting the dipole.

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 18𝜇T.

Solution 9

The neutral point for a magnet with north pole pointing in south will be at end on position.

At neutral point

Magnetic field due to magnet=earth's horizontal magnetic field

r=0.2m=20cm

from center of magnet.

Question 10

A magnetic dipole of magnetic moment 0.72A-m2 is 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 18𝜇T.

Solution 10

Let at point P, net magnetic field is zero.

Here,

We know that,

At neutral point,

Magnetic field due to magnet= earth's horizontal field

at an angle of  south of east.

Question 11

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

Solution 11

M=8.0×1022 A-m2

r=6400km

At axial point, magnetic field

Question 12

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

Solution 12

We know that,

Question 13

The magnetic field due to the earth has a horizontal component of 26𝜇Tat a place where the dip is 60°. Find the vertical component and the magnitude of the field.

Solution 13

Now, total magnetic field,

B=

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 the direction making an angle of  with the horizontal, what would be the dip at that place?

Solution 14

Apparent dip

Angle between magnetic meridian and plane of needle

Let true dip be δ

We know that,

## Chapter 14 - Permanent Magnets Exercise 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.

Solution 15

True dip (δ) is given by

Question 16

A tangent galvanometer shows a deflection of 45° when 10mA of current is passed through it. If the horizontal component of the earth's magnetic field is T and radius of the coil is 10cm, find the number of turns in the coil.

Solution 16

T

R=10cm=0.1m

I=10mA

We know that

Question 17

A moving-coil galvanometer has a 50-turn coil of size 2cm×2cm. It is suspended between the magnetic poles producing a magnetic field of 0.5T. Find the torque on the coil due to the magnetic field when a current of 20mA passes through it.

Solution 17

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 10cm from the needle. Find the ratio of the magnetic moment of the magnet to the earth's horizontal magnetic field.

Solution 18

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?

Solution 19

r=0.079m=7.9cm from the center

Question 20

A deflection magnetometer is placed with its arms in north-south direction. How and where should a short magnet having  be placed so that the needle can stay in any position?

Solution 20

For short magnet,

m=2cm

From the needle, north pole pointing towards south.

Question 21

A bar magnet takes  second to complete oscillation in an oscillation magnetometer. The moment od inertia of the magnet about the axis of rotation is  and the earth's horizontal magnetic field is 30T. Find the magnetic moment of the magnet.

Solution 21

M=1600

Question 22

The combination of two bar magnets makes 10 oscillations per second in an oscillation magnetometer when like poles are tied together. Find the second when unlike poles are tied together. Find the ratio of the magnetic moments of the magnets. Neglect any induced magnetism.

Solution 22

Time period is given by

Let magnetic moment of two poles be M1 and M2 and moment of inertia be l1and l2 respectively.

For like poles

For unlike poles

Divide (i) and (ii)

Question 23

A short magnet oscillates in an oscillation magnetometer with a time period of 0.10s where the earth's horizontal magnetic field is 24𝜇T. A downward current of 18A is established in a vertical wire placed 20cm east of the magnet. Find the new time period.

Solution 23

Initially, magnetic field

and time period T=0.1sec

Now, when wire is placed, net magnetic field

We know that,

sec

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.

Solution 24

Initially, time period of one oscillation=min

And moment of inertia=I

Later, moment of inertia=2I

We know that,

min

For 40 oscillations, time period=minutes

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 25𝜇T. Another short magnet of magnetic moment 1.6 A-m2 is placed 20cm 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.

Solution 25

Frequency oscillation

(a) for north facing north,

oscillation/minute

(b) for north pole facing south

oscillation/minute

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