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PHXII06:ELECTROMAGNETIC INDUCTION

358571 An air cored solenoid with length $20 c m$ , area of cross section $20 c m^{2}$ and number of turns 400 carries a current $2 A .$ The current is suddenly switched off within $10^{- 3} s$ . The average back emf induced across the ends of the open switch in the circuit is (ignore the variation in magnetic field near the ends of the solenoid)

1

2 V

2

4 V

3

3 V

4

5 V

Explanation:

Here, $l = 20 c m = 20 \times 10^{- 2} m$ ,
$A = 20 c m^{2} = 20 \times 10^{- 4} m^{2}$
$N = 400, I_{1} = 2 A, I_{2} = 0, d t = 10^{- 3} s$
$As ε = \frac{d ϕ}{d t} = \frac{d (B A N)}{d t}$
$= \frac{μ_{0} N d I A N}{l d t}$
$= \frac{μ_{0} N (I_{1} - I_{2}) A N}{l d t}$
$= \frac{4 π \times 10^{- 7} \times {(400)}^{2} \times 2 \times 20 \times 10^{- 4}}{20 \times 10^{- 2} \times 10^{- 3}} = 4 V$

PHXII06:ELECTROMAGNETIC INDUCTION

358572 A coil of $N = 50$ turns carries a current $I = 1 A$ and creates a magnetic flux $ϕ = 10^{- 5} T m^{- 2}$ . The value of its inductance $L$ will be

1

10^{- 4} H

2

10^{- 3} H

3

10^{- 5} H

4

2 \times 10^{- 4} H

Explanation:

$ϕ = L i \Rightarrow L = \frac{ϕ}{i} = 10^{- 5} H$

PHXII06:ELECTROMAGNETIC INDUCTION

358573 The current in a coil of inductance $0.2 H$ changes from $5 A$ to $2 A$ in $0.5 \sec$ . The magnitude of the average induced emf in the coil is

1

30 V

2

0.3 V

3

0.6 V

4

1.2 V

Explanation:

$Given | \frac{Δ I}{Δ t} | = \frac{5 - 2}{0.5} = \frac{3}{0.5} = 6 A s^{- 1}$
and $L = 0.2 H$
$∴$ Induced emf, $| ε | = L | \frac{Δ l}{Δ t} | = 0.2 \times 6 = 1.2 V$

KCET - 2020

PHXII06:ELECTROMAGNETIC INDUCTION

358574 A coil of wire of a certain radius has 600 turns per unit length and a self-inductance of $108 m H .$ The self-inductance of a $2^{n d}$ similar coil of 500 turns per unit length will be

1

75 m H

2

74 m H

3

77 m H

4

76 m H

Explanation:

$\frac{L_{B}}{L_{A}} = {(\frac{n_{B}}{n_{A}})}^{2} \Rightarrow L_{B} = {(\frac{500}{600})}^{2} \times 108 = 75 m H$

PHXII06:ELECTROMAGNETIC INDUCTION

358575 The number of turns, cross-sectional area, and lengths for four solenoids are given in the following table.
supporting img

The solenoid with maximum self inductance is :

1 1

2 2

3 3

4 4

Explanation:

$L = μ_{0} n^{2} π r^{2} l$
$A = π r^{2}$
$n = \frac{N}{l} \Rightarrow L = μ_{0} \frac{N^{2}}{l} A$
By putting the given values, it can be seen that it is maximum for solenoid no. 4.

PHXII06:ELECTROMAGNETIC INDUCTION

358571 An air cored solenoid with length $20 c m$ , area of cross section $20 c m^{2}$ and number of turns 400 carries a current $2 A .$ The current is suddenly switched off within $10^{- 3} s$ . The average back emf induced across the ends of the open switch in the circuit is (ignore the variation in magnetic field near the ends of the solenoid)

1

2 V

2

4 V

3

3 V

4

5 V

Explanation:

Here, $l = 20 c m = 20 \times 10^{- 2} m$ ,
$A = 20 c m^{2} = 20 \times 10^{- 4} m^{2}$
$N = 400, I_{1} = 2 A, I_{2} = 0, d t = 10^{- 3} s$
$As ε = \frac{d ϕ}{d t} = \frac{d (B A N)}{d t}$
$= \frac{μ_{0} N d I A N}{l d t}$
$= \frac{μ_{0} N (I_{1} - I_{2}) A N}{l d t}$
$= \frac{4 π \times 10^{- 7} \times {(400)}^{2} \times 2 \times 20 \times 10^{- 4}}{20 \times 10^{- 2} \times 10^{- 3}} = 4 V$

PHXII06:ELECTROMAGNETIC INDUCTION

358572 A coil of $N = 50$ turns carries a current $I = 1 A$ and creates a magnetic flux $ϕ = 10^{- 5} T m^{- 2}$ . The value of its inductance $L$ will be

1

10^{- 4} H

2

10^{- 3} H

3

10^{- 5} H

4

2 \times 10^{- 4} H

Explanation:

$ϕ = L i \Rightarrow L = \frac{ϕ}{i} = 10^{- 5} H$

PHXII06:ELECTROMAGNETIC INDUCTION

358573 The current in a coil of inductance $0.2 H$ changes from $5 A$ to $2 A$ in $0.5 \sec$ . The magnitude of the average induced emf in the coil is

1

30 V

2

0.3 V

3

0.6 V

4

1.2 V

Explanation:

$Given | \frac{Δ I}{Δ t} | = \frac{5 - 2}{0.5} = \frac{3}{0.5} = 6 A s^{- 1}$
and $L = 0.2 H$
$∴$ Induced emf, $| ε | = L | \frac{Δ l}{Δ t} | = 0.2 \times 6 = 1.2 V$

KCET - 2020

PHXII06:ELECTROMAGNETIC INDUCTION

358574 A coil of wire of a certain radius has 600 turns per unit length and a self-inductance of $108 m H .$ The self-inductance of a $2^{n d}$ similar coil of 500 turns per unit length will be

1

75 m H

2

74 m H

3

77 m H

4

76 m H

Explanation:

$\frac{L_{B}}{L_{A}} = {(\frac{n_{B}}{n_{A}})}^{2} \Rightarrow L_{B} = {(\frac{500}{600})}^{2} \times 108 = 75 m H$

PHXII06:ELECTROMAGNETIC INDUCTION

358575 The number of turns, cross-sectional area, and lengths for four solenoids are given in the following table.
supporting img

The solenoid with maximum self inductance is :

1 1

2 2

3 3

4 4

Explanation:

$L = μ_{0} n^{2} π r^{2} l$
$A = π r^{2}$
$n = \frac{N}{l} \Rightarrow L = μ_{0} \frac{N^{2}}{l} A$
By putting the given values, it can be seen that it is maximum for solenoid no. 4.

PHXII06:ELECTROMAGNETIC INDUCTION

358571 An air cored solenoid with length $20 c m$ , area of cross section $20 c m^{2}$ and number of turns 400 carries a current $2 A .$ The current is suddenly switched off within $10^{- 3} s$ . The average back emf induced across the ends of the open switch in the circuit is (ignore the variation in magnetic field near the ends of the solenoid)

1

2 V

2

4 V

3

3 V

4

5 V

Explanation:

Here, $l = 20 c m = 20 \times 10^{- 2} m$ ,
$A = 20 c m^{2} = 20 \times 10^{- 4} m^{2}$
$N = 400, I_{1} = 2 A, I_{2} = 0, d t = 10^{- 3} s$
$As ε = \frac{d ϕ}{d t} = \frac{d (B A N)}{d t}$
$= \frac{μ_{0} N d I A N}{l d t}$
$= \frac{μ_{0} N (I_{1} - I_{2}) A N}{l d t}$
$= \frac{4 π \times 10^{- 7} \times {(400)}^{2} \times 2 \times 20 \times 10^{- 4}}{20 \times 10^{- 2} \times 10^{- 3}} = 4 V$

PHXII06:ELECTROMAGNETIC INDUCTION

358572 A coil of $N = 50$ turns carries a current $I = 1 A$ and creates a magnetic flux $ϕ = 10^{- 5} T m^{- 2}$ . The value of its inductance $L$ will be

1

10^{- 4} H

2

10^{- 3} H

3

10^{- 5} H

4

2 \times 10^{- 4} H

Explanation:

$ϕ = L i \Rightarrow L = \frac{ϕ}{i} = 10^{- 5} H$

PHXII06:ELECTROMAGNETIC INDUCTION

358573 The current in a coil of inductance $0.2 H$ changes from $5 A$ to $2 A$ in $0.5 \sec$ . The magnitude of the average induced emf in the coil is

1

30 V

2

0.3 V

3

0.6 V

4

1.2 V

Explanation:

$Given | \frac{Δ I}{Δ t} | = \frac{5 - 2}{0.5} = \frac{3}{0.5} = 6 A s^{- 1}$
and $L = 0.2 H$
$∴$ Induced emf, $| ε | = L | \frac{Δ l}{Δ t} | = 0.2 \times 6 = 1.2 V$

KCET - 2020

PHXII06:ELECTROMAGNETIC INDUCTION

358574 A coil of wire of a certain radius has 600 turns per unit length and a self-inductance of $108 m H .$ The self-inductance of a $2^{n d}$ similar coil of 500 turns per unit length will be

1

75 m H

2

74 m H

3

77 m H

4

76 m H

Explanation:

$\frac{L_{B}}{L_{A}} = {(\frac{n_{B}}{n_{A}})}^{2} \Rightarrow L_{B} = {(\frac{500}{600})}^{2} \times 108 = 75 m H$

PHXII06:ELECTROMAGNETIC INDUCTION

358575 The number of turns, cross-sectional area, and lengths for four solenoids are given in the following table.
supporting img

The solenoid with maximum self inductance is :

1 1

2 2

3 3

4 4

Explanation:

$L = μ_{0} n^{2} π r^{2} l$
$A = π r^{2}$
$n = \frac{N}{l} \Rightarrow L = μ_{0} \frac{N^{2}}{l} A$
By putting the given values, it can be seen that it is maximum for solenoid no. 4.

PHXII06:ELECTROMAGNETIC INDUCTION

358571 An air cored solenoid with length $20 c m$ , area of cross section $20 c m^{2}$ and number of turns 400 carries a current $2 A .$ The current is suddenly switched off within $10^{- 3} s$ . The average back emf induced across the ends of the open switch in the circuit is (ignore the variation in magnetic field near the ends of the solenoid)

1

2 V

2

4 V

3

3 V

4

5 V

Explanation:

Here, $l = 20 c m = 20 \times 10^{- 2} m$ ,
$A = 20 c m^{2} = 20 \times 10^{- 4} m^{2}$
$N = 400, I_{1} = 2 A, I_{2} = 0, d t = 10^{- 3} s$
$As ε = \frac{d ϕ}{d t} = \frac{d (B A N)}{d t}$
$= \frac{μ_{0} N d I A N}{l d t}$
$= \frac{μ_{0} N (I_{1} - I_{2}) A N}{l d t}$
$= \frac{4 π \times 10^{- 7} \times {(400)}^{2} \times 2 \times 20 \times 10^{- 4}}{20 \times 10^{- 2} \times 10^{- 3}} = 4 V$

PHXII06:ELECTROMAGNETIC INDUCTION

358572 A coil of $N = 50$ turns carries a current $I = 1 A$ and creates a magnetic flux $ϕ = 10^{- 5} T m^{- 2}$ . The value of its inductance $L$ will be

1

10^{- 4} H

2

10^{- 3} H

3

10^{- 5} H

4

2 \times 10^{- 4} H

Explanation:

$ϕ = L i \Rightarrow L = \frac{ϕ}{i} = 10^{- 5} H$

PHXII06:ELECTROMAGNETIC INDUCTION

358573 The current in a coil of inductance $0.2 H$ changes from $5 A$ to $2 A$ in $0.5 \sec$ . The magnitude of the average induced emf in the coil is

1

30 V

2

0.3 V

3

0.6 V

4

1.2 V

Explanation:

$Given | \frac{Δ I}{Δ t} | = \frac{5 - 2}{0.5} = \frac{3}{0.5} = 6 A s^{- 1}$
and $L = 0.2 H$
$∴$ Induced emf, $| ε | = L | \frac{Δ l}{Δ t} | = 0.2 \times 6 = 1.2 V$

KCET - 2020

PHXII06:ELECTROMAGNETIC INDUCTION

358574 A coil of wire of a certain radius has 600 turns per unit length and a self-inductance of $108 m H .$ The self-inductance of a $2^{n d}$ similar coil of 500 turns per unit length will be

1

75 m H

2

74 m H

3

77 m H

4

76 m H

Explanation:

$\frac{L_{B}}{L_{A}} = {(\frac{n_{B}}{n_{A}})}^{2} \Rightarrow L_{B} = {(\frac{500}{600})}^{2} \times 108 = 75 m H$

PHXII06:ELECTROMAGNETIC INDUCTION

358575 The number of turns, cross-sectional area, and lengths for four solenoids are given in the following table.
supporting img

The solenoid with maximum self inductance is :

1 1

2 2

3 3

4 4

Explanation:

$L = μ_{0} n^{2} π r^{2} l$
$A = π r^{2}$
$n = \frac{N}{l} \Rightarrow L = μ_{0} \frac{N^{2}}{l} A$
By putting the given values, it can be seen that it is maximum for solenoid no. 4.

PHXII06:ELECTROMAGNETIC INDUCTION

358571 An air cored solenoid with length $20 c m$ , area of cross section $20 c m^{2}$ and number of turns 400 carries a current $2 A .$ The current is suddenly switched off within $10^{- 3} s$ . The average back emf induced across the ends of the open switch in the circuit is (ignore the variation in magnetic field near the ends of the solenoid)

1

2 V

2

4 V

3

3 V

4

5 V

Explanation:

Here, $l = 20 c m = 20 \times 10^{- 2} m$ ,
$A = 20 c m^{2} = 20 \times 10^{- 4} m^{2}$
$N = 400, I_{1} = 2 A, I_{2} = 0, d t = 10^{- 3} s$
$As ε = \frac{d ϕ}{d t} = \frac{d (B A N)}{d t}$
$= \frac{μ_{0} N d I A N}{l d t}$
$= \frac{μ_{0} N (I_{1} - I_{2}) A N}{l d t}$
$= \frac{4 π \times 10^{- 7} \times {(400)}^{2} \times 2 \times 20 \times 10^{- 4}}{20 \times 10^{- 2} \times 10^{- 3}} = 4 V$

PHXII06:ELECTROMAGNETIC INDUCTION

358572 A coil of $N = 50$ turns carries a current $I = 1 A$ and creates a magnetic flux $ϕ = 10^{- 5} T m^{- 2}$ . The value of its inductance $L$ will be

1

10^{- 4} H

2

10^{- 3} H

3

10^{- 5} H

4

2 \times 10^{- 4} H

Explanation:

$ϕ = L i \Rightarrow L = \frac{ϕ}{i} = 10^{- 5} H$

PHXII06:ELECTROMAGNETIC INDUCTION

358573 The current in a coil of inductance $0.2 H$ changes from $5 A$ to $2 A$ in $0.5 \sec$ . The magnitude of the average induced emf in the coil is

1

30 V

2

0.3 V

3

0.6 V

4

1.2 V

Explanation:

$Given | \frac{Δ I}{Δ t} | = \frac{5 - 2}{0.5} = \frac{3}{0.5} = 6 A s^{- 1}$
and $L = 0.2 H$
$∴$ Induced emf, $| ε | = L | \frac{Δ l}{Δ t} | = 0.2 \times 6 = 1.2 V$

KCET - 2020

PHXII06:ELECTROMAGNETIC INDUCTION

358574 A coil of wire of a certain radius has 600 turns per unit length and a self-inductance of $108 m H .$ The self-inductance of a $2^{n d}$ similar coil of 500 turns per unit length will be

1

75 m H

2

74 m H

3

77 m H

4

76 m H

Explanation:

$\frac{L_{B}}{L_{A}} = {(\frac{n_{B}}{n_{A}})}^{2} \Rightarrow L_{B} = {(\frac{500}{600})}^{2} \times 108 = 75 m H$

PHXII06:ELECTROMAGNETIC INDUCTION

358575 The number of turns, cross-sectional area, and lengths for four solenoids are given in the following table.
supporting img

The solenoid with maximum self inductance is :

1 1

2 2

3 3

4 4

Explanation:

$L = μ_{0} n^{2} π r^{2} l$
$A = π r^{2}$
$n = \frac{N}{l} \Rightarrow L = μ_{0} \frac{N^{2}}{l} A$
By putting the given values, it can be seen that it is maximum for solenoid no. 4.

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