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Electro Magnetic Induction

154870 A rectangular coil of 100 turns and size $0.1 m \times$ $0.05 m$ is placed perpendicular to a magnetic field of $0.1 T$ . The induced emf when the field drops to $0.05 T$ in $0.05 s$ is-

1

0.5 V

2

1.0 V

3

1.5 V

4

2.0 V

Explanation:

A Given that,
Number of turns $N = 100$ turns
Area $A = 0.1 \times 0.05 m$
Primary magnetic field $B_{1} = 0.1 T$
Secondary magnetic field $B_{2} = 0.05 T$
Time $Δ T = 0.05 \sec$
$∵ emf ε = \frac{Nd ϕ}{dt}$
$∴$ emf $ε = N \frac{d (BA)}{dt} = 100 \times 0.1 \times 0.05 \times \frac{(0.1 - 0.05)}{0.05}$
$emf ε = 0.5 V$

BCECE-2009

Electro Magnetic Induction

154871 A rod of length $1.0 m$ is rotated in a plane perpendicular to a uniform magnetic field of induction $0.25 T$ with a frequency of $12 rev / s$ . The induced emf across the ends of the rod is-

1

18.89 V

2

3 V

3

15 V

4

9.42 V

Explanation:

D Given that,
Length of $rod l = 1.0 m$
Magnetic field, $B = 0.25 T$
Frequency, $ω = 12 rev / \sec$
$emf ε = BA ω$
emf $ε = 0.25 \times π (1)^{2} \times 12$
emf $ε = 9.42 V$

AP EAMCET (22.09.2020) Shift-II

Electro Magnetic Induction

154872 The inductance of a coil is $L = 10 H$ and resistance $R = 5 Ω$ . If applied voltage of battery is $10 V$ and it switches off in 1 millisecond, find induced emf of inductor.

1

2 \times 10^{4} V

2

1.2 \times 10^{4} V

3

2 \times 10^{- 4} V

4 None of these

Explanation:

A Given that,
Self inductance $L = 10 H$ ,
Resistance $R = 5 Ω$
Voltage of battery $V = 10 V$
Time $Δ t = 1 \times 10^{- 3} \sec$
Induced current, $I = \frac{V}{R} = \frac{10}{5} = 2 A$
$∵$ emf $ε = - L \frac{dI}{dt}$
$∴$ emf $ε = 10 \times \frac{2}{10^{- 3}}$
$| ε | = 2 \times 10^{4} V$

BCECE-2008

Electro Magnetic Induction

154873 The current in a coil decreases from $1 A$ to $0.2 A$ in 10s. The coefficient of self-inductance, if induced of self-inductance, if induced emf is $0.4 V$ , is :

1

5 H

2

3 H

3

4 H

4

2 H

Explanation:

A Given that,
Primary current $I_{1} = 1 A$ ,
Secondary current $I_{2} = 0.2 A$
Time $Δ t = 10 \sec$ ,
emf $ε = 0.4 V$
$L = \frac{emf}{Δ I / Δ t}$
$L = \frac{0.4}{(1 - 0.2) / 10} = \frac{4}{0.8} = 5 H$
$L = 5 H$

BCECE-2004

Electro Magnetic Induction

154870 A rectangular coil of 100 turns and size $0.1 m \times$ $0.05 m$ is placed perpendicular to a magnetic field of $0.1 T$ . The induced emf when the field drops to $0.05 T$ in $0.05 s$ is-

1

0.5 V

2

1.0 V

3

1.5 V

4

2.0 V

Explanation:

A Given that,
Number of turns $N = 100$ turns
Area $A = 0.1 \times 0.05 m$
Primary magnetic field $B_{1} = 0.1 T$
Secondary magnetic field $B_{2} = 0.05 T$
Time $Δ T = 0.05 \sec$
$∵ emf ε = \frac{Nd ϕ}{dt}$
$∴$ emf $ε = N \frac{d (BA)}{dt} = 100 \times 0.1 \times 0.05 \times \frac{(0.1 - 0.05)}{0.05}$
$emf ε = 0.5 V$

BCECE-2009

Electro Magnetic Induction

154871 A rod of length $1.0 m$ is rotated in a plane perpendicular to a uniform magnetic field of induction $0.25 T$ with a frequency of $12 rev / s$ . The induced emf across the ends of the rod is-

1

18.89 V

2

3 V

3

15 V

4

9.42 V

Explanation:

D Given that,
Length of $rod l = 1.0 m$
Magnetic field, $B = 0.25 T$
Frequency, $ω = 12 rev / \sec$
$emf ε = BA ω$
emf $ε = 0.25 \times π (1)^{2} \times 12$
emf $ε = 9.42 V$

AP EAMCET (22.09.2020) Shift-II

Electro Magnetic Induction

154872 The inductance of a coil is $L = 10 H$ and resistance $R = 5 Ω$ . If applied voltage of battery is $10 V$ and it switches off in 1 millisecond, find induced emf of inductor.

1

2 \times 10^{4} V

2

1.2 \times 10^{4} V

3

2 \times 10^{- 4} V

4 None of these

Explanation:

A Given that,
Self inductance $L = 10 H$ ,
Resistance $R = 5 Ω$
Voltage of battery $V = 10 V$
Time $Δ t = 1 \times 10^{- 3} \sec$
Induced current, $I = \frac{V}{R} = \frac{10}{5} = 2 A$
$∵$ emf $ε = - L \frac{dI}{dt}$
$∴$ emf $ε = 10 \times \frac{2}{10^{- 3}}$
$| ε | = 2 \times 10^{4} V$

BCECE-2008

Electro Magnetic Induction

154873 The current in a coil decreases from $1 A$ to $0.2 A$ in 10s. The coefficient of self-inductance, if induced of self-inductance, if induced emf is $0.4 V$ , is :

1

5 H

2

3 H

3

4 H

4

2 H

Explanation:

A Given that,
Primary current $I_{1} = 1 A$ ,
Secondary current $I_{2} = 0.2 A$
Time $Δ t = 10 \sec$ ,
emf $ε = 0.4 V$
$L = \frac{emf}{Δ I / Δ t}$
$L = \frac{0.4}{(1 - 0.2) / 10} = \frac{4}{0.8} = 5 H$
$L = 5 H$

BCECE-2004

Electro Magnetic Induction

154870 A rectangular coil of 100 turns and size $0.1 m \times$ $0.05 m$ is placed perpendicular to a magnetic field of $0.1 T$ . The induced emf when the field drops to $0.05 T$ in $0.05 s$ is-

1

0.5 V

2

1.0 V

3

1.5 V

4

2.0 V

Explanation:

A Given that,
Number of turns $N = 100$ turns
Area $A = 0.1 \times 0.05 m$
Primary magnetic field $B_{1} = 0.1 T$
Secondary magnetic field $B_{2} = 0.05 T$
Time $Δ T = 0.05 \sec$
$∵ emf ε = \frac{Nd ϕ}{dt}$
$∴$ emf $ε = N \frac{d (BA)}{dt} = 100 \times 0.1 \times 0.05 \times \frac{(0.1 - 0.05)}{0.05}$
$emf ε = 0.5 V$

BCECE-2009

Electro Magnetic Induction

154871 A rod of length $1.0 m$ is rotated in a plane perpendicular to a uniform magnetic field of induction $0.25 T$ with a frequency of $12 rev / s$ . The induced emf across the ends of the rod is-

1

18.89 V

2

3 V

3

15 V

4

9.42 V

Explanation:

D Given that,
Length of $rod l = 1.0 m$
Magnetic field, $B = 0.25 T$
Frequency, $ω = 12 rev / \sec$
$emf ε = BA ω$
emf $ε = 0.25 \times π (1)^{2} \times 12$
emf $ε = 9.42 V$

AP EAMCET (22.09.2020) Shift-II

Electro Magnetic Induction

154872 The inductance of a coil is $L = 10 H$ and resistance $R = 5 Ω$ . If applied voltage of battery is $10 V$ and it switches off in 1 millisecond, find induced emf of inductor.

1

2 \times 10^{4} V

2

1.2 \times 10^{4} V

3

2 \times 10^{- 4} V

4 None of these

Explanation:

A Given that,
Self inductance $L = 10 H$ ,
Resistance $R = 5 Ω$
Voltage of battery $V = 10 V$
Time $Δ t = 1 \times 10^{- 3} \sec$
Induced current, $I = \frac{V}{R} = \frac{10}{5} = 2 A$
$∵$ emf $ε = - L \frac{dI}{dt}$
$∴$ emf $ε = 10 \times \frac{2}{10^{- 3}}$
$| ε | = 2 \times 10^{4} V$

BCECE-2008

Electro Magnetic Induction

154873 The current in a coil decreases from $1 A$ to $0.2 A$ in 10s. The coefficient of self-inductance, if induced of self-inductance, if induced emf is $0.4 V$ , is :

1

5 H

2

3 H

3

4 H

4

2 H

Explanation:

A Given that,
Primary current $I_{1} = 1 A$ ,
Secondary current $I_{2} = 0.2 A$
Time $Δ t = 10 \sec$ ,
emf $ε = 0.4 V$
$L = \frac{emf}{Δ I / Δ t}$
$L = \frac{0.4}{(1 - 0.2) / 10} = \frac{4}{0.8} = 5 H$
$L = 5 H$

BCECE-2004

Electro Magnetic Induction

154870 A rectangular coil of 100 turns and size $0.1 m \times$ $0.05 m$ is placed perpendicular to a magnetic field of $0.1 T$ . The induced emf when the field drops to $0.05 T$ in $0.05 s$ is-

1

0.5 V

2

1.0 V

3

1.5 V

4

2.0 V

Explanation:

A Given that,
Number of turns $N = 100$ turns
Area $A = 0.1 \times 0.05 m$
Primary magnetic field $B_{1} = 0.1 T$
Secondary magnetic field $B_{2} = 0.05 T$
Time $Δ T = 0.05 \sec$
$∵ emf ε = \frac{Nd ϕ}{dt}$
$∴$ emf $ε = N \frac{d (BA)}{dt} = 100 \times 0.1 \times 0.05 \times \frac{(0.1 - 0.05)}{0.05}$
$emf ε = 0.5 V$

BCECE-2009

Electro Magnetic Induction

154871 A rod of length $1.0 m$ is rotated in a plane perpendicular to a uniform magnetic field of induction $0.25 T$ with a frequency of $12 rev / s$ . The induced emf across the ends of the rod is-

1

18.89 V

2

3 V

3

15 V

4

9.42 V

Explanation:

D Given that,
Length of $rod l = 1.0 m$
Magnetic field, $B = 0.25 T$
Frequency, $ω = 12 rev / \sec$
$emf ε = BA ω$
emf $ε = 0.25 \times π (1)^{2} \times 12$
emf $ε = 9.42 V$

AP EAMCET (22.09.2020) Shift-II

Electro Magnetic Induction

154872 The inductance of a coil is $L = 10 H$ and resistance $R = 5 Ω$ . If applied voltage of battery is $10 V$ and it switches off in 1 millisecond, find induced emf of inductor.

1

2 \times 10^{4} V

2

1.2 \times 10^{4} V

3

2 \times 10^{- 4} V

4 None of these

Explanation:

A Given that,
Self inductance $L = 10 H$ ,
Resistance $R = 5 Ω$
Voltage of battery $V = 10 V$
Time $Δ t = 1 \times 10^{- 3} \sec$
Induced current, $I = \frac{V}{R} = \frac{10}{5} = 2 A$
$∵$ emf $ε = - L \frac{dI}{dt}$
$∴$ emf $ε = 10 \times \frac{2}{10^{- 3}}$
$| ε | = 2 \times 10^{4} V$

BCECE-2008

Electro Magnetic Induction

154873 The current in a coil decreases from $1 A$ to $0.2 A$ in 10s. The coefficient of self-inductance, if induced of self-inductance, if induced emf is $0.4 V$ , is :

1

5 H

2

3 H

3

4 H

4

2 H

Explanation:

A Given that,
Primary current $I_{1} = 1 A$ ,
Secondary current $I_{2} = 0.2 A$
Time $Δ t = 10 \sec$ ,
emf $ε = 0.4 V$
$L = \frac{emf}{Δ I / Δ t}$
$L = \frac{0.4}{(1 - 0.2) / 10} = \frac{4}{0.8} = 5 H$
$L = 5 H$

BCECE-2004

Question Bank Series

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Question Bank Series

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