Moving Charges and Magnetism

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New answer posted•a year 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?

7 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

4.28 Resistance of the galvanometer coil, G = 15 Ω

Current for which galvanometer shows full deflection, $I_{g}$ = 4 mA = 4 $\times 10^{- 3}$ 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

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New answer posted•a year 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?

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

4.27 Resistance of the galvanometer coil, G = 12 Ω

Current for which there is full scale deflection, $I_{g}$ = 3 mA = 3 $\times 10^{- 3}$ 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 = $\frac{V}{I_{g}}$ -

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New answer posted•a year 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.

7 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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 $\times 10^{- 3}$ kg $\times 10^{- 3}$

Current flowing through the wire, I = 6 A

Acc

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New answer posted•a year 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.)

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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.

Cross-sectional ar

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New answer posted•a year 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?

24 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

4.24 Magnetic field strength, B = 3000G = 3000 $\times 10^{- 4}$ 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 $\times b$ = 10 $\times 5$ = 50 ${c m}^{2}$ = 50 $\times 10^{- 4}$ $m^{2}$

Current in the loop, I = 12 A

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

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New question posted•a year ago

Moving Charges and Magnetism Class 12th

<p title="Click to copy mathml"><strong>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?</strong></p> <div> <div> </div> </div>

15 Views|Posted a year ago

Asked by Payal Gupta

New answer posted•a year 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?

11 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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 mag

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New answer posted•a year 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?

7 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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 = $\frac{μ_{0} I^{2}}{2 π r}$ , where $μ_{0}$ = Permeability of free space = 4 $π \times 10^{- 7}$ T m $A^{- 1}$

Hence F = $\frac{4 π \times 10^{- 7} \times 300^{2}}{2 π \times 0.015}$ N/m = 1.2 N/m

Since the direction of the curr

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New answer posted•a year 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.

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

4.21 Length of the rod. l= 0.45 m

Mass suspended by the wire, m = 60 g = 60 $\times 10^{- 3}$ 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 = $\frac{m g}{I l}$ = $\frac{60 \times 10^{- 3} \times 9.8}{5 \times 0.45}$ = 0.26 T

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

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New answer posted•a year 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?

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

4.20 Magnetic field, B = 0.75 T

Accelerating voltage, V = 15 kV = 15 $\times 10^{- 3}$ V

Electrostatic field, E = 9.0 $\times 10^{5}$ 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

$\frac{1}{2}$ m $v^{2}$ = eV or $\frac{e}{m}$ = $\frac{v^{2}}{2 V}$ ………….(1)

Since the particle remains un-deflected by

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