Physics

4.5 What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30º with the direction of a uniform magnetic field of 0.15 T?

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4.5 Current in the wire, I = 8 A

Magnitude of the uniform magnetic field, B = 0.15 T

Angle between the wire and the magnetic field, $θ$ = 30 $°$

Magnetic force per unit length of the wire is given as

f = BI $\sin ? θ$ = 0.15 $* 8 * \sin ? 30 °$ = 0.6 N/m

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4.4 A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

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4.4 Current in the power line, I = 90 A

Point is located below power line at a distance, r = 1.5 m

Magnitude of the magnetic field at this point is given as

$|\overset{?}{B}|$ = $\frac{μ_{0}}{4 π} \frac{2 I}{r}$ where $μ_{0}$ = Permeability of free space $π * 10^{- 7}$ = 4 Tm $A^{- 1}$

Hence, $|\overset{?}{B}|$ = $\frac{4 π * 10^{- 7}}{4 π} \frac{2 * 90}{1.5}$ = 1.2 $* 10^{- 5}$ T

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4.3 A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.

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4.3 Current in the wire, I = 50 A

The distance of the point from the wire, r = 2.5 m

Magnitude of the magnetic field at this point is given as

$|\overset{?}{B}|$ = $\frac{μ_{0}}{4 π} \frac{2 I}{r}$ where $μ_{0}$ = Permeability of free space = 4 $π * 10^{- 7}$ Tm $A^{- 1}$

Hence, $|\overset{?}{B}|$ = $\frac{4 π * 10^{- 7}}{4 π} \frac{2 * 50}{2.5}$ = 4.0 $* 10^{- 6}$ T

The direction of the current in the wire is vertically downward. Hence, according to Maxwell's right hand rule, the direction of the magnetic field at the given point is vertically upward.

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4.2 A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

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4.2 Current in the wire, I = 35 A

Distance of a point from the wire, r = 20 cm = 0.2 m

Magnitude of the magnetic field at this point is given as

$|\overset{?}{B}|$ = $\frac{μ_{0}}{4 π} \frac{2 I}{r}$ where $μ_{0}$ = Permeability of free space = 4 $π * 10^{- 7}$ Tm $A^{- 1}$

Hence, $|\overset{?}{B}|$ = $\frac{4 π * 10^{- 7}}{4 π} \frac{2 * 35}{0.2}$ = 3.50 $* 10^{- 5}$ T

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4.1 A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

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4.1 Number of turns of the circular coil, n = 100

Radius of each turn, r = 8.0 cm = 0.08 m

Current flowing in the coil, I = 0.4 A

Magnitude of the magnetic field at the centre of the coil is given as

$|\overset{?}{B}|$ = $\frac{μ_{0}}{4 π} \frac{2 π n I}{r}$ where $μ_{0}$ = Permeability of free space = 4 $π * 10^{- 7}$ Tm $A^{- 1}$
Hence, $|\overset{?}{B}|$ = $\frac{4 π * 10^{- 7}}{4 π} \frac{2 π * 100 * 0.4}{0.08}$ = 3.14 $* 10^{- 4}$ T

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3.23 Figure 3.34 shows a 2.0 V potentiometer used for the determination of internal resistance of a 1.5 V cell. The balance point of the cell in open circuit is 76.3 cm. When a resistor of 9.5 Ω is used in the external circuit of the cell, the balance point shifts to 64.8 cm length of the potentiometer wire. Determine the internal resistance of the cell.

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3.23 Internal resistance of the cell = r

Balance point of the cell in open circuit, $l_{1}$ = 76.3 cm

An external resistance R is connected to the circuit with R = 9.5 Ω

New balance point of the circuit, $l_{2}$ = 64.8 cm

Current flowing through circuit = I

The relation connecting resistance and emf is,

r = ( $\frac{l_{1} - l_{2}}{l_{2}}$ )R = $\frac{76.3 - 64.8}{64.8}$ $* 9.5 =$ 1.69 Ω

Therefore, the internal resistance is 1.69 Ω.

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3.22 Figure 3.33 shows a potentiometer with a cell of 2.0 V and internal resistance 0.40 Ω maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents up to a few mA) gives a balance point at 67.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of 600 kΩ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf $?$ and the balance point found similarly, turns out to be at 82.3 cm length of the wire.

(a) What is the value

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(a) What is the value e ?

(b) What purpose does the high resistance of 600 kΩ have?

(c) Is the balance point affected by this high resistance?

(d) Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0V instead of 2.0V?

(e) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?

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3.22 Constant emf of the standard cell, $E_{1}$ = 1.02 V

Balance point on the wire, $l_{1}$ = 67.3 cm

A cell of unknown emf, $?$ , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm.

The relation of connected emf and balance point is, $\frac{E_{1}}{l_{1}}$ = $\frac{?}{l}$

Hence, $?$ = $\frac{l}{l_{1}} * E_{1}$ = $\frac{82.3}{67.3} * 1.02$ = 1.247 V

The purpose of using high resistance of 600 KΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.

The balance point is not affected by the presence of high resistance.

The point is not affected by the internal resistance of the driver cell.

The met

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3.22 Constant emf of the standard cell, $E_{1}$ = 1.02 V

Balance point on the wire, $l_{1}$ = 67.3 cm

A cell of unknown emf, $?$ , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm.

The relation of connected emf and balance point is, $\frac{E_{1}}{l_{1}}$ = $\frac{?}{l}$

Hence, $?$ = $\frac{l}{l_{1}} * E_{1}$ = $\frac{82.3}{67.3} * 1.02$ = 1.247 V

The purpose of using high resistance of 600 KΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.

The balance point is not affected by the presence of high resistance.

The point is not affected by the internal resistance of the driver cell.

The method would not work if the emf of the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V. This is because if the emf of the driver cell of the potentiometer is less than the emf of the other cell, then there would be no balance point on the wire.

The circuit would not work well for determining an extremely small emf. As the circuit would be unstable, the balance point would be close to end A. Hence, there would be a large percentage of error.

The given circuit can be modified if a series resistance is connected with the wire AB. The potential drop across AB is slightly greater than the emf measured. The percentage error would be small.

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3.21 Determine the current drawn from a 12V supply with internal resistance 0.5 Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.

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3.21 Let the equivalent resistance of the given circuit be R'. The equivalent resistance of an infinite network is given by

R' = 2 + $\frac{R'}{(R^{'} + 1)}$ or R' = $\frac{2 R^{'} + 2 + R'}{(R^{'} + 1)}$ or ${R'}^{2}$ + R' = 2R' + 2 + R'

${R'}^{2} - 2 R^{'} - 2 = 0$

R' = $\frac{2 \pm \sqrt{4 + 8}}{2}$ = 1 $\pm \sqrt 3$

Since R' cannot be negative, hence R' = 1+ $\sqrt 3$ = 2.73 Ω

Internal resistance, r = 0.5Ω

Total resistance = 2.73 + 0.5 = 3.23 Ω

Current drawn from the source = $\frac{12}{3.23}$ A= 3.72 A

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3.20 (a) Given n resistors each of resistance R, how will you combine them to get the (i) maximum (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?

(b) Given the resistances of 1 Ω, 2 Ω, 3 Ω, how will be combine them to get an equivalent resistance of (i) (11/3) Ω (ii) (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?

(c) Determine the equivalent resistance of networks shown in Fig. 3.31.

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3.20 Total number of resistors = n

Resistance of each resistor = R

When the resistors are connected in series, effective resistance $R_{e f f}$ is maximum. $R_{e f f}$ = nR

When n resistors are connected in parallel, the effective resistance, $R_{e f f}$ is minimum. $R_{e f f}$ = $\frac{R}{n}$ . The ratio of maximum to minimum resistance = $\frac{n R}{\frac{R}{n}}$ = $n^{2}$

Let us assume, $R_{1}$ = 1 Ω, $R_{2}$ = 2 Ω, $R_{3}$ = 3 Ω

Required equivalent resistance, R = $\frac{11}{3}$ Ω

From the circuit the equivalent resistance is given by

R = $\frac{2 * 1}{2 + 1}$ + 3 = $\frac{2}{3}$ + 3 = $\frac{11}{3}$ Ω

Required equivalent resistance, R = $\frac{11}{5}$ Ω

From the circuit, the equivalen

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3.20 Total number of resistors = n

Resistance of each resistor = R

When the resistors are connected in series, effective resistance $R_{e f f}$ is maximum. $R_{e f f}$ = nR

Let us assume, $R_{1}$ = 1 Ω, $R_{2}$ = 2 Ω, $R_{3}$ = 3 Ω

Required equivalent resistance, R = $\frac{11}{3}$ Ω

From the circuit the equivalent resistance is given by

R = $\frac{2 * 1}{2 + 1}$ + 3 = $\frac{2}{3}$ + 3 = $\frac{11}{3}$ Ω

Required equivalent resistance, R = $\frac{11}{5}$ Ω

From the circuit, the equivalent resistance is given by

R = $\frac{2 * 3}{2 + 3}$ + 1 = $\frac{6}{5}$ + 1 = $\frac{11}{5}$ Ω

Required equivalent resistance, R = 6 Ω

From the circuit, the equivalent resistance is given by

R = 1 + 2+ 3 = 6 Ω

Required equivalent resistance, R = $\frac{6}{11}$ Ω

From the circuit, the equivalent resistance is given by

R = $\frac{1 * 2 * 3}{1 * 2 + 2 * 3 + 3 * 1}$ = $\frac{6}{11}$ Ω

(a)

It can be observed from the given circuit that in the first small loop, two resistors of resistance 1 Ω each are connected in the series. Hence the equivalent resistance is = 1 + 1= 2 $Ω$ . At the bottom part 2 resistors of 2 Ω each are connected in a series and thus make equivalent resistance of 2 + 2 = 4 Ω. Thus the circuit can be redrawn as 2 Ω and 4 Ω resistances are connected in parallel. Hence the equivalent resistance of each loop is R = $\frac{2 * 4}{2 + 4}$ = $\frac{4}{3}$ Ω. All these 4 loop resistors are connected in series. Thus the total equivalent resistance is $\frac{4}{3}$ Ω $* 4 = \frac{16}{3}$ Ω

Here all 5 resistors are connected in series. So the equivalent resistance is

5 $* R = 5 R$

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Q.3.19 Choose the correct alternative:

(a) Alloys of metals usually have (greater/less) resistivity than that of their constituent metals.

(b) Alloys usually have much (lower/higher) temperature coefficients of resistance than pure metals.

(c) The resistivity of the alloy manganin is nearly independent of/ increases rapidly with increase of temperature.

(d) The resistivity of a typical insulator (e.g., amber) is greater than that of a metal by a factor of the order of (10²²/10²³).

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