Current in the power line, I = 90 A Point is located below the power line at distance, r = 1.5 m Hence, magnetic field at that point is given by the relation, |B|= (μ0/4π) 2l/r Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹ |B|= (4π x 10⁻⁷ )/4π x (2x 90 )/1.5 = 41.2 x 10⁻⁵ T The curreRead more
Current in the power line, I = 90 A
Point is located below the power line at distance, r = 1.5 m
Hence, magnetic field at that point is given by the relation,
|B|= (μ0/4π) 2l/r
Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹
|B|= (4π x 10⁻⁷ )/4π x (2x 90 )/1.5 = 41.2 x 10⁻⁵ T
The current is flowing from East to West. The point is below the power line. Hence, according to Maxwell’s right hand thumb rule, the direction of the magnetic field is towards the South.
Current in the wire, I = 50 A A point is 2.5 m away from the East of the wire. Therefore magnitude of the distance of the point from the wire, r = 2.5 m. Magnitude of the magnetic field at that point is given by the relation, |B|= (μ0/4π) 2l/r Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm ARead more
Current in the wire, I = 50 A
A point is 2.5 m away from the East of the wire.
Therefore magnitude of the distance of the point from the wire, r = 2.5 m.
Magnitude of the magnetic field at that point is given by the relation, |B|= (μ0/4π) 2l/r
Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹
|B|= (4π x 10⁻⁷ )/4π x (2x 50 )/2.5 = 4 x 10⁻⁶ T
The point is located normal to the wire length at a distance of 2.5 m. The direction of the current in the wire is vertically downward. Hence, according to the Maxwell’s right hand thumb rule, the direction of the magnetic field at the given point is vertically upward.
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: |B|= (μ0/4π) 2l/r Where, μ0 = Permeability of free space = 4π x 10-7 T m A-1 |B|= (4π x 10-7 )/4π x ( 2 x 35 )/0.2 = 3.5 x 10⁻⁵ T Hence, the magnitude of thRead more
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: |B|= (μ0/4π) 2l/r
Where, μ0 = Permeability of free space = 4π x 10-7 T m A-1
|B|= (4π x 10-7 )/4π x ( 2 x 35 )/0.2 = 3.5 x 10⁻⁵ T
Hence, the magnitude of the magnetic field at a point 20 cm from the wire is 3.5 x 10–5 T.
Number of turns on 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 by the relation, |B|= (μ0/4π) 2πnl/r Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹ So, |B|= (4Read more
Number of turns on 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 by the relation,
|B|= (μ0/4π) 2πnl/r
Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹
So,
|B|= (4π x 10⁻⁷)/4π x ( 2π x 100 x 0.4 )/r =3.14 x 10⁻⁴ T
Hence, the magnitude of the magnetic field is 3.14 x 10⁻⁴ T.
Internal resistance of the cell = r Balance point of the cell in open circuit, l₁ = 76.3 cm An external resistance (R) is connected to the circuit with R = 9.5 Ω New balance point of the circuit, l₂ = 64.8 cm Current flowing through the circuit = I The relation connecting resistance and emf is, r =Read more
Internal resistance of the cell = r
Balance point of the cell in open circuit, l₁ = 76.3 cm
An external resistance (R) is connected to the circuit with R = 9.5 Ω
New balance point of the circuit, l₂ = 64.8 cm
Current flowing through the circuit = I
The relation connecting resistance and emf is,
r = (l₁ -l₂ )/l₂ x R
= (76.3 -64.8)64.8 x 9.5 =1.68 Ω
Therefore, the internal resistance of the cell is 1.68Ω.
Resistance of the standard resistor, R = 10.0 Ω Balance point for this resistance, l₁ = 58.3 cm Current in the potentiometer wire = i Hence, potential drop across R, E₁ = iR Resistance of the unknown resistor = X Balance point for this resistor, l₂ = 68.5 cm Hence, potential drop across X, E2 = iX TRead more
Resistance of the standard resistor, R = 10.0 Ω
Balance point for this resistance, l₁ = 58.3 cm
Current in the potentiometer wire = i
Hence, potential drop across R, E₁ = iR
Resistance of the unknown resistor = X
Balance point for this resistor, l₂ = 68.5 cm
Hence, potential drop across X, E2 = iX
The relation connecting emf and balance point is,
E₁/E₂ =l₁/l₂
=> iR/iX =l₁/l₂
X= (l₁/l₂ ) x R
= 68.5/58.3 x 10 = 11.749 Ω
Therefore, the value of the unknown resistance, X, is 11.75Ω.
If we fail to find a balance point with the given cell of emf, ε , then the potential drop across R and X must be reduced by putting a resistance in series with it. Only if the potential drop across R or X is smaller than the potential drop across the potentiometer wire AB, a balance point is obtained.
Ans (a). Constant emf of the given standard cell, E₁ = 1.02 V Balance point on the wire, l₁ = 67.3 cm A cell of unknown emf, ε , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm The relation connecting emf and balance point is, E₁/l₁ = ε /l ε = (l/l₁ ) x E₁ = (82.3 /Read more
Ans (a).
Constant emf of the given standard cell, E₁ = 1.02 V
Balance point on the wire, l₁ = 67.3 cm
A cell of unknown emf, ε , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm
The relation connecting emf and balance point is,
E₁/l₁ = ε /l
ε = (l/l₁ ) x E₁ = (82.3 /67.3 ) x 1.02 = 1.247 V
The value of unknown emf is 1.247 V.
Ans (b).
The purpose of using the high resistance of 600 kΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.
Ans (c).
The balance point is not affected by the presence of high resistance.
Ans (d).
The point is not affected by the internal resistance of the driver cell.
Ans (e).
The method would not work if 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.
Ans (f).
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.
The resistance of each resistor connected in the given circuit, R = 1 Ω Equivalent resistance of the given circuit = R' The network is infinite. Hence, equivalent resistance is given by the relation, R' = 2 + R'/(R+1) (R')² -2 R' -2 = 0 R' = [2 ± √ (4+8)]/2 = [2 ± √12]/2 = 1 ± √3 Negative value of RRead more
The resistance of each resistor connected in the given circuit,
R = 1 Ω
Equivalent resistance of the given circuit = R’
The network is infinite.
Hence, equivalent resistance is given by the relation,
R’ = 2 + R’/(R+1)
(R’)² -2 R’ -2 = 0
R’ = [2 ± √ (4+8)]/2
= [2 ± √12]/2 = 1 ± √3
Negative value of R’ cannot be accepted.
Hence, equivalent resistance,
R’=(1 + √3)= 1 + 1.73 = 2.73 Ω
Internal resistance of the circuit, r = 0.5 Ω
Hence, total resistance of the given circuit
= 2.73 + 0.5 = 3.23 Ω
Supply voltage, V = 12 V
According to Ohm’s Law, current drawn from the source is given by the ratio, 12/3.23= 3.72 A
Ans (a). Alloys of metals usually have greater resistivity than that of their constituent metals. Ans (b). Alloys usually have lower temperature coefficients of resistance than pure metals. Ans (c). The resistivity of the alloy, manganin, is nearly independent of increase of temperature. Ans (d). ThRead more
Ans (a).
Alloys of metals usually have greater resistivity than that of their constituent metals.
Ans (b).
Alloys usually have lower temperature coefficients of resistance than pure metals.
Ans (c).
The resistivity of the alloy, manganin, is nearly independent of increase of temperature.
Ans (d).
The resistivity of a typical insulator is greater than that of a metal by a factor of the order of 1022.
Ans (a). When a steady current flows in a metallic conductor of non-uniform cross-section, the current flowing through the conductor is constant. Current density, electric field, and drift speed are inversely proportional to the area of cross-section. Therefore, they are not constant. Ans (b). No, ORead more
Ans (a).
When a steady current flows in a metallic conductor of non-uniform cross-section, the current flowing through the conductor is constant. Current density, electric field, and drift speed are inversely proportional to the area of cross-section. Therefore, they are not constant.
Ans (b).
No, Ohm’s law is not universally applicable for all conducting elements. Vacuum diode semi-conductor is a non-ohmic conductor. Ohm’s law is not valid for it.
Ans (c).
According to Ohm’s law, the relation for the potential is V = IR Voltage (V) is directly proportional to current (I).
R is the internal resistance of the source. I =V/R
If V is low, then R must be very low, so that high current can be drawn from the source.
Ans (d).
In order to prohibit the current from exceeding the safety limit, a high tension supply must have a very large internal resistance. If the internal resistance is not large, then the current drawn can exceed the safety limits in case of a short circuit.
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?
Current in the power line, I = 90 A Point is located below the power line at distance, r = 1.5 m Hence, magnetic field at that point is given by the relation, |B|= (μ0/4π) 2l/r Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹ |B|= (4π x 10⁻⁷ )/4π x (2x 90 )/1.5 = 41.2 x 10⁻⁵ T The curreRead more
Current in the power line, I = 90 A
Point is located below the power line at distance, r = 1.5 m
Hence, magnetic field at that point is given by the relation,
|B|= (μ0/4π) 2l/r
Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹
|B|= (4π x 10⁻⁷ )/4π x (2x 90 )/1.5 = 41.2 x 10⁻⁵ T
The current is flowing from East to West. The point is below the power line. Hence, according to Maxwell’s right hand thumb rule, the direction of the magnetic field is towards the South.
See lessA 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.
Current in the wire, I = 50 A A point is 2.5 m away from the East of the wire. Therefore magnitude of the distance of the point from the wire, r = 2.5 m. Magnitude of the magnetic field at that point is given by the relation, |B|= (μ0/4π) 2l/r Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm ARead more
Current in the wire, I = 50 A
A point is 2.5 m away from the East of the wire.
Therefore magnitude of the distance of the point from the wire, r = 2.5 m.
Magnitude of the magnetic field at that point is given by the relation,
|B|= (μ0/4π) 2l/r
Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹
|B|= (4π x 10⁻⁷ )/4π x (2x 50 )/2.5 = 4 x 10⁻⁶ T
The point is located normal to the wire length at a distance of 2.5 m. The direction of the current in the wire is vertically downward. Hence, according to the Maxwell’s right hand thumb rule, the direction of the magnetic field at the given point is vertically upward.
See lessA 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?
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: |B|= (μ0/4π) 2l/r Where, μ0 = Permeability of free space = 4π x 10-7 T m A-1 |B|= (4π x 10-7 )/4π x ( 2 x 35 )/0.2 = 3.5 x 10⁻⁵ T Hence, the magnitude of thRead more
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:
|B|= (μ0/4π) 2l/r
Where, μ0 = Permeability of free space = 4π x 10-7 T m A-1
|B|= (4π x 10-7 )/4π x ( 2 x 35 )/0.2 = 3.5 x 10⁻⁵ T
Hence, the magnitude of the magnetic field at a point 20 cm from the wire is 3.5 x 10–5 T.
See lessA 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?
Number of turns on 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 by the relation, |B|= (μ0/4π) 2πnl/r Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹ So, |B|= (4Read more
Number of turns on 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 by the relation,
|B|= (μ0/4π) 2πnl/r
Where , μ0 = Permeability of free space = 4π x 10⁻⁷ Tm A⁻¹
So,
|B|= (4π x 10⁻⁷)/4π x ( 2π x 100 x 0.4 )/r =3.14 x 10⁻⁴ T
Hence, the magnitude of the magnetic field is 3.14 x 10⁻⁴ T.
See lessFigure 3.35 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.
Internal resistance of the cell = r Balance point of the cell in open circuit, l₁ = 76.3 cm An external resistance (R) is connected to the circuit with R = 9.5 Ω New balance point of the circuit, l₂ = 64.8 cm Current flowing through the circuit = I The relation connecting resistance and emf is, r =Read more
Internal resistance of the cell = r
Balance point of the cell in open circuit, l₁ = 76.3 cm
An external resistance (R) is connected to the circuit with R = 9.5 Ω
New balance point of the circuit, l₂ = 64.8 cm
Current flowing through the circuit = I
The relation connecting resistance and emf is,
r = (l₁ -l₂ )/l₂ x R
= (76.3 -64.8)64.8 x 9.5 =1.68 Ω
Therefore, the internal resistance of the cell is 1.68Ω.
See lessFigure 3.34 shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor R = 10.0 Ω is found to be 58.3 cm, while that with the unknown resistance X is 68.5 cm. Determine the value of X. What might you do if you failed to find a balance point with the given cell of emf ε ?
Resistance of the standard resistor, R = 10.0 Ω Balance point for this resistance, l₁ = 58.3 cm Current in the potentiometer wire = i Hence, potential drop across R, E₁ = iR Resistance of the unknown resistor = X Balance point for this resistor, l₂ = 68.5 cm Hence, potential drop across X, E2 = iX TRead more
Resistance of the standard resistor, R = 10.0 Ω
Balance point for this resistance, l₁ = 58.3 cm
Current in the potentiometer wire = i
Hence, potential drop across R, E₁ = iR
Resistance of the unknown resistor = X
Balance point for this resistor, l₂ = 68.5 cm
Hence, potential drop across X, E2 = iX
The relation connecting emf and balance point is,
E₁/E₂ =l₁/l₂
=> iR/iX =l₁/l₂
X= (l₁/l₂ ) x R
= 68.5/58.3 x 10 = 11.749 Ω
Therefore, the value of the unknown resistance, X, is 11.75Ω.
If we fail to find a balance point with the given cell of emf, ε , then the potential drop across R and X must be reduced by putting a resistance in series with it. Only if the potential drop across R or X is smaller than the potential drop across the potentiometer wire AB, a balance point is obtained.
See lessFigure 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 upto 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 ε and the balance point found similarly, turns out to be at 82.3 cm length of the wire.(a) What is the value ε? (b) What purpose does the high resistance of 600 kΩ have?(c) Is the balance point affected by this high resistance? (d) Is the balance point affected by the internal resistance of the driver cell? (e) 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? (f ) 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?
Ans (a). Constant emf of the given standard cell, E₁ = 1.02 V Balance point on the wire, l₁ = 67.3 cm A cell of unknown emf, ε , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm The relation connecting emf and balance point is, E₁/l₁ = ε /l ε = (l/l₁ ) x E₁ = (82.3 /Read more
Ans (a).
Constant emf of the given standard cell, E₁ = 1.02 V
Balance point on the wire, l₁ = 67.3 cm
A cell of unknown emf, ε , replaced the standard cell. Therefore, new balance point on the wire, l = 82.3 cm
The relation connecting emf and balance point is,
E₁/l₁ = ε /l
ε = (l/l₁ ) x E₁ = (82.3 /67.3 ) x 1.02 = 1.247 V
The value of unknown emf is 1.247 V.
Ans (b).
The purpose of using the high resistance of 600 kΩ is to reduce the current through the galvanometer when the movable contact is far from the balance point.
Ans (c).
The balance point is not affected by the presence of high resistance.
Ans (d).
The point is not affected by the internal resistance of the driver cell.
Ans (e).
The method would not work if 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.
Ans (f).
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.
See lessDetermine the current drawn from a 12V supply with internal resistance 0.5Ω by the infinite network shown in Fig. 3.32. Each resistor has 1Ω resistance.
The resistance of each resistor connected in the given circuit, R = 1 Ω Equivalent resistance of the given circuit = R' The network is infinite. Hence, equivalent resistance is given by the relation, R' = 2 + R'/(R+1) (R')² -2 R' -2 = 0 R' = [2 ± √ (4+8)]/2 = [2 ± √12]/2 = 1 ± √3 Negative value of RRead more
The resistance of each resistor connected in the given circuit,
R = 1 Ω
Equivalent resistance of the given circuit = R’
The network is infinite.
Hence, equivalent resistance is given by the relation,
R’ = 2 + R’/(R+1)
(R’)² -2 R’ -2 = 0
R’ = [2 ± √ (4+8)]/2
= [2 ± √12]/2 = 1 ± √3
Negative value of R’ cannot be accepted.
Hence, equivalent resistance,
R’=(1 + √3)= 1 + 1.73 = 2.73 Ω
Internal resistance of the circuit, r = 0.5 Ω
Hence, total resistance of the given circuit
= 2.73 + 0.5 = 3.23 Ω
Supply voltage, V = 12 V
According to Ohm’s Law, current drawn from the source is given by the ratio, 12/3.23= 3.72 A
See lessChoose 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 (10²²/10³ ).
Ans (a). Alloys of metals usually have greater resistivity than that of their constituent metals. Ans (b). Alloys usually have lower temperature coefficients of resistance than pure metals. Ans (c). The resistivity of the alloy, manganin, is nearly independent of increase of temperature. Ans (d). ThRead more
Ans (a).
Alloys of metals usually have greater resistivity than that of their constituent metals.
Ans (b).
Alloys usually have lower temperature coefficients of resistance than pure metals.
Ans (c).
The resistivity of the alloy, manganin, is nearly independent of increase of temperature.
Ans (d).
See lessThe resistivity of a typical insulator is greater than that of a metal by a factor of the order of 1022.
Answer the following questions: (a) A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor: current, current density, electric field, drift speed? (b) Is Ohm’s law universally applicable for all conducting elements? If not, give examples of elements which do not obey Ohm’s law. (c) A low voltage supply from which one needs high currents must have very low internal resistance. Why? (d) A high tension (HT) supply of, say, 6 kV must have a very large internal resistance. Why?
Ans (a). When a steady current flows in a metallic conductor of non-uniform cross-section, the current flowing through the conductor is constant. Current density, electric field, and drift speed are inversely proportional to the area of cross-section. Therefore, they are not constant. Ans (b). No, ORead more
Ans (a).
When a steady current flows in a metallic conductor of non-uniform cross-section, the current flowing through the conductor is constant. Current density, electric field, and drift speed are inversely proportional to the area of cross-section. Therefore, they are not constant.
Ans (b).
No, Ohm’s law is not universally applicable for all conducting elements. Vacuum diode semi-conductor is a non-ohmic conductor. Ohm’s law is not valid for it.
Ans (c).
According to Ohm’s law, the relation for the potential is V = IR Voltage (V) is directly proportional to current (I).
R is the internal resistance of the source. I =V/R
If V is low, then R must be very low, so that high current can be drawn from the source.
Ans (d).
In order to prohibit the current from exceeding the safety limit, a high tension supply must have a very large internal resistance. If the internal resistance is not large, then the current drawn can exceed the safety limits in case of a short circuit.
See less