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Moving Charges and Magnetism

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8 months ago

Moving Charges and Magnetism Class 12th

4.28 A galvanometer coil has a resistance of 15 Ω and the meter shows full scale deflection for a current of 4 mA. How will you convert the meter into an ammeter of range 0 to 6 A?
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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

Moving Charges and Magnetism Class 12th

4.27 A galvanometer coil has a resistance of 12 Ω and the meter shows full scale deflection for a current of 3 mA. How will you convert the meter into a voltmeter of range 0 to 18 V?
0 Follower 4 Views

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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

New answer posted

8 months ago

Moving Charges and Magnetism Class 12th

4.26 A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire? g = 9.8 m s–2.
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Payal Gupta

Contributor-Level 10

4.26 Length of the solenoid, L = 60 cm = 0.6 m

Radius of the solenoid, r = 4.0 cm = 0.04 m

It is given that there are 3 layers of windings of 300 turns each

Hence, total number of turns, n = 900

Length, l = 2 cm = 0.02 m

Mass of the wire, m = 2.5 g = 2.5 * 10 - 3 kg * 10 - 3

Current flowing through the wire, I = 6 A

Acceleration due to gravity, g = 9.8 m/ s 2

We know, magnetic field produced inside the solenoid, B = μ 0 n I L

where μ 0 = Permeability of free space = 4 π * 10 - 7 T m A - 1

Magnetic force is given by the relation

F = Bil = μ 0 n I L * i l

Also the force on the wire is equal to the weight of the wire, F = mg

mg = μ 0 n I i l L

I = m g L μ 0 n i I = 2.5 * 10 - 3 * 9.8 * 0.6 4 π * 10 - 7 * 900 * 0.02 * 6 = 108 A

Hence, the current flowing through the solenoid is 108 A

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New answer posted

8 months ago

Moving Charges and Magnetism Class 12th

4.25 A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the

(a) total torque on the coil,

(b) total force on the coil,

(c) average force on each electron in the coil due to the magnetic field?

(The coil is made of copper wire of cross-sectional area 10–5 m2, and the free electron density in copper is given to be about 1029 m–3.)
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Payal Gupta

Contributor-Level 10

4.25 Number of turns on the circular coil, n =20

Radius of the coil, r = 10 cm = 0.1 m

Magnetic field strength, B = 0.10 T

Current in the coil, I = 5.0 A

The total torque on the coil is zero because the field is uniform.

The total torque on the coil is zero because the field is uniform.

Cross-sectional area of the copper coil, A = 10 - 5 m 2

Number of free electrons per cubic meter of copper, N = 10 29 / m 3

Charge on the electron, e = 1.6 * 10 19 C

Magnetic force, F = Be v d where v d is the drift velocity of electrons

v d = I N e A μ 0

H e n c e F = B e I N e A = 0.1 * 5 10 29 * 10 - 5 = 5 * 10 - 25 N

The average force on each electron is 5 * 10 - 25 N

New answer posted

8 months ago

Moving Charges and Magnetism Class 12th

4.24 A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?
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Payal Gupta

Contributor-Level 10

4.24 Magnetic field strength, B = 3000G = 3000 * 10 - 4 T = 0.3 T

Length of the rectangular loop, l = 10 cm

Width of the rectangular loop, b = 5 cm

Area of the loop, A = l * b = 10 * 5 = 50 c m 2 = 50 * 10 - 4 m 2

Current in the loop, I = 12 A

Assume that the anti-clockwise direction of the current is positive and vice versa.

Torque, τ ? = A ? * B ?

From the given figure, it can be observed that A is normal to the y-z plane and B is directed along z-axis.

τ = 12*( 50 * 10 - 4 ) i ? * 0.3 k ? = - 1.8 * 10 - 2 j ? Nm

The Torque 1.8 * 10 - 2 is Nm along the negative y-direction.

The force on the loop is zero because the angle between A & B is zero.

This case is similar to case (a). The answer is same as case (a)

Torque, τ ? = I A ? * B ?

F

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New question posted

8 months ago

Moving Charges and Magnetism Class 12th

4.24 A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?

 
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New answer posted

8 months ago

Moving Charges and Magnetism Class 12th

4.23 A uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,

(a) the wire intersects the axis,

(b) the wire is turned from N-S to northeast-northwest direction,

(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?
0 Follower 2 Views

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Payal Gupta

Contributor-Level 10

4.23 Magnetic field strength, B = 1.5 T

Radius of the cylindrical region, r = 10 cm = 0.1 m

Current in the wire passing through the cylindrical region, I = 7 A

If the wire intersect the axis, then the length of the wire is the diameter of the cylindrical region, then l = 2r = 0.2 m

Angle between the magnetic field, θ = 90 °

Magnetic force acting on the wire is given by the relation,

F = BIl sin ? θ = 1.5 * 7 * 0.2 * sin ? 90 ° = 2.1 N

Hence, a force of 2.1 N acts on the wire in a vertically downward direction.

If the wire is turned from N-S to NE-NW direction, new length of the wire can be given as

l 1 = l sin ? θ

Angle between magnetic field and current θ = 45 °

Force on the wire,

F = BI l 1 sin ? θ =

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New answer posted

8 months ago

Moving Charges and Magnetism Class 12th

4.22 The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?
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Payal Gupta

Contributor-Level 10

4.22 Current in both the wires, I = 300 A

Distance between the wires, r = 1.5 cm = 0.015 m

Length of the two wires, l = 70 cm = 0.7 m

Now, force between the two wires is given by the relation:

F = μ 0 I 2 2 π r , where μ 0 = Permeability of free space = 4 π * 10 - 7 T m A - 1

Hence F = 4 π * 10 - 7 * 300 2 2 π * 0.015 N/m = 1.2 N/m

Since the direction of the current in the wires is opposite, a repulsive force exists between them.

New answer posted

8 months ago

Moving Charges and Magnetism Class 12th

4.21 A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.

(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?

(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before?

(Ignore the mass of the wires.) g = 9.8 m s–2.
0 Follower 2 Views

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Payal Gupta

Contributor-Level 10

4.21 Length of the rod. l= 0.45 m

Mass suspended by the wire, m = 60 g = 60 * 10 - 3 kg

Acceleration due to gravity, g = 9.8 m/ s 2

Current, I = 5 A

To achieve zero tension, the magnetic field = weight of the wire

BIl = mg or

B = m g I l = 60 * 10 - 3 * 9.8 5 * 0.45 = 0.26 T

The magnetic field should be set up such that it gives an upward magnetic force.

If the direction of current is reversed, then the magnetic force will act downwards and total tension in the wire will be

mg + BIl = 60 * 10 - 3 * 9.8 + 0.26 * 5 * 0.45 = 1.173 T

New answer posted

8 months ago

Moving Charges and Magnetism Class 12th

4.20 A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 * 10–5 V m–1, make a simple guess as to what the beam contains. Why is the answer not unique?
0 Follower 2 Views

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Payal Gupta

Contributor-Level 10

4.20 Magnetic field, B = 0.75 T

Accelerating voltage, V = 15 kV = 15 * 10 - 3 V

Electrostatic field, E = 9.0 * 10 5 V/m

Let the mass of electron = m, Charge of the electron = e, Velocity of the electron = v

Then kinetic energy of the electron = eV

1 2 m v 2 = eV or e m = v 2 2 V ………….(1)

Since the particle remains un-deflected by electric and magnetic field, we can infer that the electric field is balancing the magnetic field.

Hence eE = evB or v = E B ………(2)

Combining equation (1) and (2), we get

e m = 1 2 ( E B ) 2 V = E 2 2 V B 2 = ( 9.0 * 10 5 ) 2 2 * 15000 * 0.75 2 = 48 * 10 6 C/kg

This value of specific charge e/m is equal to the value of deuteron or deuterium ions. This is not a unique answer. Other possible answers ar

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