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Current Electricity

151723 A horizontal wire carries $160 A$ current below which another wire of linear density $10 g / m$ carrying a current is kept at $4 c m$ distance. If the wire is kept below hangs in air. What is the current in this wire when the direction of current in both the wires is same?
$(g = 10 m / s^{2}$ and $μ_{0} = 4 π \times 10^{- 7}$ )

1

125 A

2

140 A

3

110 A

4

100 A

Explanation:

A Given that,
Current, $(i_{1}) = 160 A, \frac{m}{l} = 10 g / m = 10^{- 2} kg / m$
$g = 10 m / s^{2}, μ_{o} = 4 π \times 10^{- 7}$
$d = 4 cm = 0.04 m$

For wire second be in air at equilibrium,
$F = mg$
$\frac{F}{l} = \frac{m}{l} g = \frac{μ_{0} i_{1} i_{2}}{2 π d}$
$10^{- 2} \times 10 = \frac{4 π \times 10^{- 7} \times 160 i_{2}}{2 π \times 0.04}$
$i_{2} = \frac{10^{- 1} \times 2 π \times 0.04}{160 \times 4 π \times 10^{- 7}}$
$i_{2} = 125 A$

TS EAMCET 18.07.2022

Current Electricity

151724 Find the current in the three resistors as shown in the following figure?

1

i_{1} = 0, i_{2} = \frac{4 V}{R}, i_{3} = \frac{2 v}{R}

2

i_{1} = 0, i_{2} = 0, i_{3} = 0

3

i_{1} = 0, i_{2} = \frac{2 V}{R}, i_{3} = \frac{4 v}{R}

4

i_{1} = 0, i_{2} = \frac{2 V}{R}, i_{3} = \frac{2 v}{R}

Explanation:

B
(i) For first loop,
Using KVL,
$2 + i_{1} R - 2 = 0$
$i_{1} = 0$
(ii) For second loop,
By using KVL
$2 + i_{2} R - 2 - i_{1} R = 0$
$i_{2} R - i_{1} R = 0$
$i_{2} = i_{1} = 0$
(iii) For third loop,
By using KVL,
$2 + i_{3} R - 2 - i_{2} R = 0$
$i_{3} = i_{2} = 0$
$i_{3} = 0$

TS EAMCET 18.07.2022

Current Electricity

151725 The resistivity of a metal is $1 \times 10^{- 8} Ω m$ . If it contains $9 \times 10^{28}$ electrons per $m^{3}$ then the relaxation time of electrons inside the metal is nearly.
$(electron mass = 9.1 \times 10^{- 31} kg)$

1

4 \times 10^{- 14} s

2

7 \times 10^{- 14} s

3

1.0 \times 10^{- 14} s

4

9 \times 10^{- 14} s

Explanation:

A Given,
No. of free electron $n = 9 \times 10^{28}$ electron per $m^{3}$
Resistivity of metal $ρ = 1 \times 10^{- 8} Ω m$
Mass of electron $m = 9.1 \times 10^{- 31} kg$
Relaxation time (successive collision) $τ =$ ?
The resistivity of material of a conductor is given by -
$ρ = \frac{m}{{ne}^{2} τ}$
$τ = \frac{9.1 \times 10^{- 31}}{9 \times 10^{28} \times {(1.6 \times 10^{- 19})}^{2} \times 1 \times 10^{- 8}}$
$τ = 0.394 \times 10^{- 13}$
$τ = 3.94 \times 10^{- 14}$
$τ = 4 \times 10^{- 14} s$

TS EAMCET 19.07.2022

Current Electricity

151726 Find the mobility of electron in a wire, if its average collision time is $9.1 \times 10^{- 15} s$ .
(charge of electron $= 1.6 \times 10^{- 19} C$ and mass of electron $= 9.1 \times 10^{- 31} kg$ )

1

9.1 \times 10^{- 3} m^{2} / V - s

2

1.6 \times 10^{- 3} m^{2} / V - s

3

1.75 \times 10^{- 3} m^{2} / V - s

4

1 \times 10^{- 3} m^{2} / V - s

Explanation:

B Given,
$τ = 9.1 \times 10^{- 15} s, e = 1.6 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} kg$
Mobility ' $μ$ ' of charge carrier is defined as magnitude of drift velocity per unit electric field is -
$μ = \frac{| {\vec{v}}_{d} |}{E} = \frac{e τ}{m}$
$μ = \frac{1.6 \times 10^{- 19} \times 9.1 \times 10^{- 15}}{9.1 \times 10^{- 31}}$
$μ = 1.6 \times 10^{- 3} m^{2} / V - s$

TS EAMCET 20.07.2022

Current Electricity

151723 A horizontal wire carries $160 A$ current below which another wire of linear density $10 g / m$ carrying a current is kept at $4 c m$ distance. If the wire is kept below hangs in air. What is the current in this wire when the direction of current in both the wires is same?
$(g = 10 m / s^{2}$ and $μ_{0} = 4 π \times 10^{- 7}$ )

1

125 A

2

140 A

3

110 A

4

100 A

Explanation:

A Given that,
Current, $(i_{1}) = 160 A, \frac{m}{l} = 10 g / m = 10^{- 2} kg / m$
$g = 10 m / s^{2}, μ_{o} = 4 π \times 10^{- 7}$
$d = 4 cm = 0.04 m$

For wire second be in air at equilibrium,
$F = mg$
$\frac{F}{l} = \frac{m}{l} g = \frac{μ_{0} i_{1} i_{2}}{2 π d}$
$10^{- 2} \times 10 = \frac{4 π \times 10^{- 7} \times 160 i_{2}}{2 π \times 0.04}$
$i_{2} = \frac{10^{- 1} \times 2 π \times 0.04}{160 \times 4 π \times 10^{- 7}}$
$i_{2} = 125 A$

TS EAMCET 18.07.2022

Current Electricity

151724 Find the current in the three resistors as shown in the following figure?

1

i_{1} = 0, i_{2} = \frac{4 V}{R}, i_{3} = \frac{2 v}{R}

2

i_{1} = 0, i_{2} = 0, i_{3} = 0

3

i_{1} = 0, i_{2} = \frac{2 V}{R}, i_{3} = \frac{4 v}{R}

4

i_{1} = 0, i_{2} = \frac{2 V}{R}, i_{3} = \frac{2 v}{R}

Explanation:

B
(i) For first loop,
Using KVL,
$2 + i_{1} R - 2 = 0$
$i_{1} = 0$
(ii) For second loop,
By using KVL
$2 + i_{2} R - 2 - i_{1} R = 0$
$i_{2} R - i_{1} R = 0$
$i_{2} = i_{1} = 0$
(iii) For third loop,
By using KVL,
$2 + i_{3} R - 2 - i_{2} R = 0$
$i_{3} = i_{2} = 0$
$i_{3} = 0$

TS EAMCET 18.07.2022

Current Electricity

151725 The resistivity of a metal is $1 \times 10^{- 8} Ω m$ . If it contains $9 \times 10^{28}$ electrons per $m^{3}$ then the relaxation time of electrons inside the metal is nearly.
$(electron mass = 9.1 \times 10^{- 31} kg)$

1

4 \times 10^{- 14} s

2

7 \times 10^{- 14} s

3

1.0 \times 10^{- 14} s

4

9 \times 10^{- 14} s

Explanation:

A Given,
No. of free electron $n = 9 \times 10^{28}$ electron per $m^{3}$
Resistivity of metal $ρ = 1 \times 10^{- 8} Ω m$
Mass of electron $m = 9.1 \times 10^{- 31} kg$
Relaxation time (successive collision) $τ =$ ?
The resistivity of material of a conductor is given by -
$ρ = \frac{m}{{ne}^{2} τ}$
$τ = \frac{9.1 \times 10^{- 31}}{9 \times 10^{28} \times {(1.6 \times 10^{- 19})}^{2} \times 1 \times 10^{- 8}}$
$τ = 0.394 \times 10^{- 13}$
$τ = 3.94 \times 10^{- 14}$
$τ = 4 \times 10^{- 14} s$

TS EAMCET 19.07.2022

Current Electricity

151726 Find the mobility of electron in a wire, if its average collision time is $9.1 \times 10^{- 15} s$ .
(charge of electron $= 1.6 \times 10^{- 19} C$ and mass of electron $= 9.1 \times 10^{- 31} kg$ )

1

9.1 \times 10^{- 3} m^{2} / V - s

2

1.6 \times 10^{- 3} m^{2} / V - s

3

1.75 \times 10^{- 3} m^{2} / V - s

4

1 \times 10^{- 3} m^{2} / V - s

Explanation:

B Given,
$τ = 9.1 \times 10^{- 15} s, e = 1.6 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} kg$
Mobility ' $μ$ ' of charge carrier is defined as magnitude of drift velocity per unit electric field is -
$μ = \frac{| {\vec{v}}_{d} |}{E} = \frac{e τ}{m}$
$μ = \frac{1.6 \times 10^{- 19} \times 9.1 \times 10^{- 15}}{9.1 \times 10^{- 31}}$
$μ = 1.6 \times 10^{- 3} m^{2} / V - s$

TS EAMCET 20.07.2022

Current Electricity

151723 A horizontal wire carries $160 A$ current below which another wire of linear density $10 g / m$ carrying a current is kept at $4 c m$ distance. If the wire is kept below hangs in air. What is the current in this wire when the direction of current in both the wires is same?
$(g = 10 m / s^{2}$ and $μ_{0} = 4 π \times 10^{- 7}$ )

1

125 A

2

140 A

3

110 A

4

100 A

Explanation:

A Given that,
Current, $(i_{1}) = 160 A, \frac{m}{l} = 10 g / m = 10^{- 2} kg / m$
$g = 10 m / s^{2}, μ_{o} = 4 π \times 10^{- 7}$
$d = 4 cm = 0.04 m$

For wire second be in air at equilibrium,
$F = mg$
$\frac{F}{l} = \frac{m}{l} g = \frac{μ_{0} i_{1} i_{2}}{2 π d}$
$10^{- 2} \times 10 = \frac{4 π \times 10^{- 7} \times 160 i_{2}}{2 π \times 0.04}$
$i_{2} = \frac{10^{- 1} \times 2 π \times 0.04}{160 \times 4 π \times 10^{- 7}}$
$i_{2} = 125 A$

TS EAMCET 18.07.2022

Current Electricity

151724 Find the current in the three resistors as shown in the following figure?

1

i_{1} = 0, i_{2} = \frac{4 V}{R}, i_{3} = \frac{2 v}{R}

2

i_{1} = 0, i_{2} = 0, i_{3} = 0

3

i_{1} = 0, i_{2} = \frac{2 V}{R}, i_{3} = \frac{4 v}{R}

4

i_{1} = 0, i_{2} = \frac{2 V}{R}, i_{3} = \frac{2 v}{R}

Explanation:

B
(i) For first loop,
Using KVL,
$2 + i_{1} R - 2 = 0$
$i_{1} = 0$
(ii) For second loop,
By using KVL
$2 + i_{2} R - 2 - i_{1} R = 0$
$i_{2} R - i_{1} R = 0$
$i_{2} = i_{1} = 0$
(iii) For third loop,
By using KVL,
$2 + i_{3} R - 2 - i_{2} R = 0$
$i_{3} = i_{2} = 0$
$i_{3} = 0$

TS EAMCET 18.07.2022

Current Electricity

151725 The resistivity of a metal is $1 \times 10^{- 8} Ω m$ . If it contains $9 \times 10^{28}$ electrons per $m^{3}$ then the relaxation time of electrons inside the metal is nearly.
$(electron mass = 9.1 \times 10^{- 31} kg)$

1

4 \times 10^{- 14} s

2

7 \times 10^{- 14} s

3

1.0 \times 10^{- 14} s

4

9 \times 10^{- 14} s

Explanation:

A Given,
No. of free electron $n = 9 \times 10^{28}$ electron per $m^{3}$
Resistivity of metal $ρ = 1 \times 10^{- 8} Ω m$
Mass of electron $m = 9.1 \times 10^{- 31} kg$
Relaxation time (successive collision) $τ =$ ?
The resistivity of material of a conductor is given by -
$ρ = \frac{m}{{ne}^{2} τ}$
$τ = \frac{9.1 \times 10^{- 31}}{9 \times 10^{28} \times {(1.6 \times 10^{- 19})}^{2} \times 1 \times 10^{- 8}}$
$τ = 0.394 \times 10^{- 13}$
$τ = 3.94 \times 10^{- 14}$
$τ = 4 \times 10^{- 14} s$

TS EAMCET 19.07.2022

Current Electricity

151726 Find the mobility of electron in a wire, if its average collision time is $9.1 \times 10^{- 15} s$ .
(charge of electron $= 1.6 \times 10^{- 19} C$ and mass of electron $= 9.1 \times 10^{- 31} kg$ )

1

9.1 \times 10^{- 3} m^{2} / V - s

2

1.6 \times 10^{- 3} m^{2} / V - s

3

1.75 \times 10^{- 3} m^{2} / V - s

4

1 \times 10^{- 3} m^{2} / V - s

Explanation:

B Given,
$τ = 9.1 \times 10^{- 15} s, e = 1.6 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} kg$
Mobility ' $μ$ ' of charge carrier is defined as magnitude of drift velocity per unit electric field is -
$μ = \frac{| {\vec{v}}_{d} |}{E} = \frac{e τ}{m}$
$μ = \frac{1.6 \times 10^{- 19} \times 9.1 \times 10^{- 15}}{9.1 \times 10^{- 31}}$
$μ = 1.6 \times 10^{- 3} m^{2} / V - s$

TS EAMCET 20.07.2022

Current Electricity

151723 A horizontal wire carries $160 A$ current below which another wire of linear density $10 g / m$ carrying a current is kept at $4 c m$ distance. If the wire is kept below hangs in air. What is the current in this wire when the direction of current in both the wires is same?
$(g = 10 m / s^{2}$ and $μ_{0} = 4 π \times 10^{- 7}$ )

1

125 A

2

140 A

3

110 A

4

100 A

Explanation:

A Given that,
Current, $(i_{1}) = 160 A, \frac{m}{l} = 10 g / m = 10^{- 2} kg / m$
$g = 10 m / s^{2}, μ_{o} = 4 π \times 10^{- 7}$
$d = 4 cm = 0.04 m$

For wire second be in air at equilibrium,
$F = mg$
$\frac{F}{l} = \frac{m}{l} g = \frac{μ_{0} i_{1} i_{2}}{2 π d}$
$10^{- 2} \times 10 = \frac{4 π \times 10^{- 7} \times 160 i_{2}}{2 π \times 0.04}$
$i_{2} = \frac{10^{- 1} \times 2 π \times 0.04}{160 \times 4 π \times 10^{- 7}}$
$i_{2} = 125 A$

TS EAMCET 18.07.2022

Current Electricity

151724 Find the current in the three resistors as shown in the following figure?

1

i_{1} = 0, i_{2} = \frac{4 V}{R}, i_{3} = \frac{2 v}{R}

2

i_{1} = 0, i_{2} = 0, i_{3} = 0

3

i_{1} = 0, i_{2} = \frac{2 V}{R}, i_{3} = \frac{4 v}{R}

4

i_{1} = 0, i_{2} = \frac{2 V}{R}, i_{3} = \frac{2 v}{R}

Explanation:

B
(i) For first loop,
Using KVL,
$2 + i_{1} R - 2 = 0$
$i_{1} = 0$
(ii) For second loop,
By using KVL
$2 + i_{2} R - 2 - i_{1} R = 0$
$i_{2} R - i_{1} R = 0$
$i_{2} = i_{1} = 0$
(iii) For third loop,
By using KVL,
$2 + i_{3} R - 2 - i_{2} R = 0$
$i_{3} = i_{2} = 0$
$i_{3} = 0$

TS EAMCET 18.07.2022

Current Electricity

151725 The resistivity of a metal is $1 \times 10^{- 8} Ω m$ . If it contains $9 \times 10^{28}$ electrons per $m^{3}$ then the relaxation time of electrons inside the metal is nearly.
$(electron mass = 9.1 \times 10^{- 31} kg)$

1

4 \times 10^{- 14} s

2

7 \times 10^{- 14} s

3

1.0 \times 10^{- 14} s

4

9 \times 10^{- 14} s

Explanation:

A Given,
No. of free electron $n = 9 \times 10^{28}$ electron per $m^{3}$
Resistivity of metal $ρ = 1 \times 10^{- 8} Ω m$
Mass of electron $m = 9.1 \times 10^{- 31} kg$
Relaxation time (successive collision) $τ =$ ?
The resistivity of material of a conductor is given by -
$ρ = \frac{m}{{ne}^{2} τ}$
$τ = \frac{9.1 \times 10^{- 31}}{9 \times 10^{28} \times {(1.6 \times 10^{- 19})}^{2} \times 1 \times 10^{- 8}}$
$τ = 0.394 \times 10^{- 13}$
$τ = 3.94 \times 10^{- 14}$
$τ = 4 \times 10^{- 14} s$

TS EAMCET 19.07.2022

Current Electricity

151726 Find the mobility of electron in a wire, if its average collision time is $9.1 \times 10^{- 15} s$ .
(charge of electron $= 1.6 \times 10^{- 19} C$ and mass of electron $= 9.1 \times 10^{- 31} kg$ )

1

9.1 \times 10^{- 3} m^{2} / V - s

2

1.6 \times 10^{- 3} m^{2} / V - s

3

1.75 \times 10^{- 3} m^{2} / V - s

4

1 \times 10^{- 3} m^{2} / V - s

Explanation:

B Given,
$τ = 9.1 \times 10^{- 15} s, e = 1.6 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} kg$
Mobility ' $μ$ ' of charge carrier is defined as magnitude of drift velocity per unit electric field is -
$μ = \frac{| {\vec{v}}_{d} |}{E} = \frac{e τ}{m}$
$μ = \frac{1.6 \times 10^{- 19} \times 9.1 \times 10^{- 15}}{9.1 \times 10^{- 31}}$
$μ = 1.6 \times 10^{- 3} m^{2} / V - s$

TS EAMCET 20.07.2022

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