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.
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 less