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New answer posted

8 months ago

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Payal Gupta

Contributor-Level 10

5.8 (a) Number of turns, n = 2000

Area of cross-section, A = 1.6 * 10 - 4 m 2

Current, I = 4.0 A

The magnetic moment along the axis of the solenoid is calculated as

M = nAI = 2000 * 1.6 * 10 - 4 * 4 = 1.28 A m 2

(b) Magnetic field, B = 7.5 * 10 - 2 T

Angle between magnetic field and the axis of the solenoid, θ = 30 °

Torque τ = M B sin ? θ

= 1.28 * 7.5 * 10 - 2 * sin ? 30 °

= 4.8 * 10 - 2 Nm

New answer posted

8 months ago

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P
Payal Gupta

Contributor-Level 10

5.7 Magnetic moment, M = 1.5 J/T

Magnetic field strength, B = 0.22 T

Initial angle between the axis and the magnetic field, θ 1 = 0 ° and the final angle, θ 2 = 90 °

Work required to make the magnetic moment normal to the direction of the magnetic field is given as:

W = -MB( cos θ 2 - cos θ 1 ) = - 1.5 * 0.22 ( cos ? 90 ° - cos 0 ° ) = 0.33 J

Initial angle between the axis and the magnetic field, θ 1 = 0 ° and the final angle, θ 2 = 180

Work required to make the magnetic moment normal to the direction of the magnetic field is given as:

W = -MB( cos θ 2 - cos θ 1 ) = - 1.5 * 0.22 ( cos ? 180 ° - cos 0 ° ) = 0.33 J

For case

Torque τ = M B sin ? θ , here θ = θ 2 = 90 °

= 1.5 * 0.22 sin ? 90 ° = 0.33 J

Torque τ = M B sin ? θ , here θ = θ 2 = 180 °

= 1.5 * 0.22 sin ? 180 ° =

...more

New answer posted

8 months ago

0 Follower 9 Views

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Payal Gupta

Contributor-Level 10

5.6 Magnetic field strength, B = 0.25 T

Magnetic moment, M = 0.6 J/T

The angle θ , between the axis of the solenoid and the direction of the applied field is 30 ° . Therefore, the torque acting on the solenoid is given as

τ = M B s i n θ = 0.6 * 0.25 * sin ? 30 ° = 7.5 * 10 - 2 J

New answer posted

8 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

5.5 Number of turns, n = 800

Area of the cross-section, A = 2.5 * 10 - 4 m 2

Current flowing, I = 3.0 A

A current carrying solenoid behaves like a bar magnet because a magnetic field develops along its axis (along the length). °

The magnetic moment associated is calculated as M = nIA = 800 * 3 * 2.5 * 10 - 4 J/T= 0.6 J/T

New answer posted

8 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

5.4 Moment of the bar magnet, M = 0.32 J/T

Magnetic field, B = 0.15 T

The bar magnet is aligned along the magnetic field. This system is considered as being in stable equilibrium. Hence, the angle θ , between the bar magnet and the magnetic field is 0 ° .

Potential energy of the system = -MBcos θ = - 0.32 * 0.15 * cos ? 0 ° = -4.8 * 10 - 2 J

When the bar magnet is oriented 180 ° to the magnetic field, it becomes unstable equilibrium.

Potential energy = - MBcos θ = - 0.32 * 0.15 * cos ? 180 ° = 4.8 * 10 - 2 J

New answer posted

8 months ago

0 Follower 5 Views

P
Payal Gupta

Contributor-Level 10

5.3 Magnetic field strength, B = 0.25 T

Torque on the bar magnet, τ = 4.5 * 10 - 2 J

Angle between the bar magnet and the external magnetic field, θ = 30 °

From the relation T = MB sin ? θ , where M = Magnetic moment, we get

M = τ B sin ? θ = 4.5 * 10 - 2 0.25 s i n 30 ° = 0.36 J/T

Hence the magnetic moment is 0.36 J/T

New answer posted

8 months ago

5.2 Answer the following questions:

(a) The earth's magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably?

(b) The earth's core is known to contain iron. Yet geologists do not regard this as a source of the earth's magnetism. Why?

(c) The charged currents in the outer conducting regions of the earth's core are thought to be responsible for earth's magnetism. What might be the 'battery' (i.e., the source of energy) to sustain these currents?

(d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion y

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Payal Gupta

Contributor-Level 10

5.2 (a) Earth's magnetic field changes with time. It takes a few hundred years to change by an appreciable amount. The variation in earth's magnetic field with the time cannot be neglected.

(b) Earth's core contains molten iron. This form of iron is not ferromagnetic. Hence this is not considered as a source of earth's magnetism.

(c) The radioactivity in earth's interior is the source of energy that sustains the currents in the outer conducting regions of earth's core. These charged currents are considered to be responsible for earth's magnetism.

(d) The change of earth's magnetic field got weakly recorded in rocks during their solidif

...more

New answer posted

8 months ago

5.1 Answer the following questions regarding earth's magnetism:

(a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth's magnetic field.

(b) The angle of dip at a location in southern India is about 18°. Would you expect a greater or smaller dip angle in Britain?

(c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground?

(d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole?

(e) The ear

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Payal Gupta

Contributor-Level 10

5.1 (a) Earth's magnetic field can be specified by three following independent quantities

Magnetic declination

Angle of dip

Horizontal component of earth's magnetic field.

(b) The angle of dip at a point depends on how far the point is located with respect to North pole or South pole. The angle of dip will be more in Britain than Southern India as Britain is closer to Magnetic North pole than South India to the Magnetic South pole.

(c) It is a hypothesis that a huge bar magnet is embedded deep in Earth's ground with its north pole near magnetic south pole of earth and south pole is near magnetic north pole of earth. Magnetic field lines

...more

New answer posted

8 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

4.28 Resistance of the galvanometer coil, G = 15 Ω

Current for which galvanometer shows full deflection, I g = 4 mA = 4 * 10 - 3 A

Range of ammeter has to be converted from 0 to 6 A, hence I = 6 A

A shunt resistor S is to be connected in parallel with the galvanometer to convert it to an ammeter. The value of S is given as

S = 10 mΩ

Hence, a shunt resistor of 10 mΩ is to be connected to galvanometer to convert it to an ammeter.

 

New answer posted

8 months ago

0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

4.27 Resistance of the galvanometer coil, G = 12 Ω

Current for which there is full scale deflection, I g = 3 mA = 3 * 10 - 3 A

Range of voltmeter = 0, to be converted to 18 V, hence V = 18 V

Let there be a resistor R connected in series with the galvanometer to convert it into a voltmeter. R is given as

R = V I g - G = 18 3 * 10 - 3 - 12 = 5988 Ω

Hence the required value of resistor is 5988 Ω

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