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Electro Magnetic Induction

154773 The energy stored in a $50 mH$ inductor carrying a current of $4 A$ is

1

0.4 J

2

4.0 J

3

0.8 J

4

0.04 J

Explanation:

A Given that,
$L = 50 mH = 50 \times 10^{- 3} H$
$I = 4 A$
Energy stored, $E = \frac{1}{2} {LI}^{2}$
$E = \frac{1}{2} \times 50 \times 10^{- 3} \times (4)^{2}$
$E = 0.4 J$

AP EAMCET (18.09.2020) Shift-I

Electro Magnetic Induction

154774 Two concentric circular coils, one of small radius $r$ and the other of large radius $R$ are placed co-axially with centers coinciding. If the radius $r$ is changed by $2 %$ , then the change in mutual inductance of the arrangement is (Assume, $r << R$ )

1

2 %

2

1.5 %

3

4 %

4

0 %

Explanation:

C Mutual inductance $M = \frac{μ_{0} N_{1} N_{2} A}{l} = \frac{μ_{0} N_{1} N_{2} π r^{2}}{l}$
$N_{1} =$ number of turns in inner coil
$N_{2} =$ number of turns in outer coil
$A =$ Area of coil
$M \propto r^{2}$
Given that, $\frac{Δ r}{r} \times 100 % = 2 %$
So, By the formula of error analysis,
$\frac{Δ M}{M} \times 100 % = 2 (\frac{Δ r}{r} \times 100 %) = 2 (2 %) = 4 %$

TS- EAMCET-14.09.2020

Electro Magnetic Induction

154775 A varying current in a coil changes from $10 A$ to zero in $1.5 s$ . If the average emf induced in the coil is $200 V$ , the self-inductance of the coil is

1

25 H

2

30 H

3

50 H

4

45 H

Explanation:

B Given,
Varying current $(dI) = 10 A$
Total time $(dt) = 1.5 \sec$
Induced emf $(ε) = 200 V$
We know that,
Induced emf $(ε) = L \frac{dI}{dt}$
$L = \frac{ε}{dI / dt} = \frac{200}{10 / 1.5}$
$L = 30$ Henry

TS- EAMCET-10.09.2020

Electro Magnetic Induction

154776 Two coils have mutual inductance of $1.5 H$ . If current in primary coil is suddenly raised to $5 A$ in one mille second, the induced emf in the secondary coil is

1

75 V

2

750 V

3

7500 V

4

75000 V

Explanation:

C Given,
Mutual inductacne $(M) = 1.5 H$
Current $(dI) = 5 A$
Time $(dt) = 1 ms = 1 \times 10^{- 3} \sec$
We know that,
Induced emf $(ε) = M \frac{dI}{dt}$
$ε = 1.5 \times \frac{5}{(10)^{- 3}}$
$ε = 1.5 \times 5 \times 10^{3} = 7500 V$

Manipal UGET -2020

Electro Magnetic Induction

154773 The energy stored in a $50 mH$ inductor carrying a current of $4 A$ is

1

0.4 J

2

4.0 J

3

0.8 J

4

0.04 J

Explanation:

A Given that,
$L = 50 mH = 50 \times 10^{- 3} H$
$I = 4 A$
Energy stored, $E = \frac{1}{2} {LI}^{2}$
$E = \frac{1}{2} \times 50 \times 10^{- 3} \times (4)^{2}$
$E = 0.4 J$

AP EAMCET (18.09.2020) Shift-I

Electro Magnetic Induction

154774 Two concentric circular coils, one of small radius $r$ and the other of large radius $R$ are placed co-axially with centers coinciding. If the radius $r$ is changed by $2 %$ , then the change in mutual inductance of the arrangement is (Assume, $r << R$ )

1

2 %

2

1.5 %

3

4 %

4

0 %

Explanation:

C Mutual inductance $M = \frac{μ_{0} N_{1} N_{2} A}{l} = \frac{μ_{0} N_{1} N_{2} π r^{2}}{l}$
$N_{1} =$ number of turns in inner coil
$N_{2} =$ number of turns in outer coil
$A =$ Area of coil
$M \propto r^{2}$
Given that, $\frac{Δ r}{r} \times 100 % = 2 %$
So, By the formula of error analysis,
$\frac{Δ M}{M} \times 100 % = 2 (\frac{Δ r}{r} \times 100 %) = 2 (2 %) = 4 %$

TS- EAMCET-14.09.2020

Electro Magnetic Induction

154775 A varying current in a coil changes from $10 A$ to zero in $1.5 s$ . If the average emf induced in the coil is $200 V$ , the self-inductance of the coil is

1

25 H

2

30 H

3

50 H

4

45 H

Explanation:

B Given,
Varying current $(dI) = 10 A$
Total time $(dt) = 1.5 \sec$
Induced emf $(ε) = 200 V$
We know that,
Induced emf $(ε) = L \frac{dI}{dt}$
$L = \frac{ε}{dI / dt} = \frac{200}{10 / 1.5}$
$L = 30$ Henry

TS- EAMCET-10.09.2020

Electro Magnetic Induction

154776 Two coils have mutual inductance of $1.5 H$ . If current in primary coil is suddenly raised to $5 A$ in one mille second, the induced emf in the secondary coil is

1

75 V

2

750 V

3

7500 V

4

75000 V

Explanation:

C Given,
Mutual inductacne $(M) = 1.5 H$
Current $(dI) = 5 A$
Time $(dt) = 1 ms = 1 \times 10^{- 3} \sec$
We know that,
Induced emf $(ε) = M \frac{dI}{dt}$
$ε = 1.5 \times \frac{5}{(10)^{- 3}}$
$ε = 1.5 \times 5 \times 10^{3} = 7500 V$

Manipal UGET -2020

Electro Magnetic Induction

154773 The energy stored in a $50 mH$ inductor carrying a current of $4 A$ is

1

0.4 J

2

4.0 J

3

0.8 J

4

0.04 J

Explanation:

A Given that,
$L = 50 mH = 50 \times 10^{- 3} H$
$I = 4 A$
Energy stored, $E = \frac{1}{2} {LI}^{2}$
$E = \frac{1}{2} \times 50 \times 10^{- 3} \times (4)^{2}$
$E = 0.4 J$

AP EAMCET (18.09.2020) Shift-I

Electro Magnetic Induction

154774 Two concentric circular coils, one of small radius $r$ and the other of large radius $R$ are placed co-axially with centers coinciding. If the radius $r$ is changed by $2 %$ , then the change in mutual inductance of the arrangement is (Assume, $r << R$ )

1

2 %

2

1.5 %

3

4 %

4

0 %

Explanation:

C Mutual inductance $M = \frac{μ_{0} N_{1} N_{2} A}{l} = \frac{μ_{0} N_{1} N_{2} π r^{2}}{l}$
$N_{1} =$ number of turns in inner coil
$N_{2} =$ number of turns in outer coil
$A =$ Area of coil
$M \propto r^{2}$
Given that, $\frac{Δ r}{r} \times 100 % = 2 %$
So, By the formula of error analysis,
$\frac{Δ M}{M} \times 100 % = 2 (\frac{Δ r}{r} \times 100 %) = 2 (2 %) = 4 %$

TS- EAMCET-14.09.2020

Electro Magnetic Induction

154775 A varying current in a coil changes from $10 A$ to zero in $1.5 s$ . If the average emf induced in the coil is $200 V$ , the self-inductance of the coil is

1

25 H

2

30 H

3

50 H

4

45 H

Explanation:

B Given,
Varying current $(dI) = 10 A$
Total time $(dt) = 1.5 \sec$
Induced emf $(ε) = 200 V$
We know that,
Induced emf $(ε) = L \frac{dI}{dt}$
$L = \frac{ε}{dI / dt} = \frac{200}{10 / 1.5}$
$L = 30$ Henry

TS- EAMCET-10.09.2020

Electro Magnetic Induction

154776 Two coils have mutual inductance of $1.5 H$ . If current in primary coil is suddenly raised to $5 A$ in one mille second, the induced emf in the secondary coil is

1

75 V

2

750 V

3

7500 V

4

75000 V

Explanation:

C Given,
Mutual inductacne $(M) = 1.5 H$
Current $(dI) = 5 A$
Time $(dt) = 1 ms = 1 \times 10^{- 3} \sec$
We know that,
Induced emf $(ε) = M \frac{dI}{dt}$
$ε = 1.5 \times \frac{5}{(10)^{- 3}}$
$ε = 1.5 \times 5 \times 10^{3} = 7500 V$

Manipal UGET -2020

Electro Magnetic Induction

154773 The energy stored in a $50 mH$ inductor carrying a current of $4 A$ is

1

0.4 J

2

4.0 J

3

0.8 J

4

0.04 J

Explanation:

A Given that,
$L = 50 mH = 50 \times 10^{- 3} H$
$I = 4 A$
Energy stored, $E = \frac{1}{2} {LI}^{2}$
$E = \frac{1}{2} \times 50 \times 10^{- 3} \times (4)^{2}$
$E = 0.4 J$

AP EAMCET (18.09.2020) Shift-I

Electro Magnetic Induction

154774 Two concentric circular coils, one of small radius $r$ and the other of large radius $R$ are placed co-axially with centers coinciding. If the radius $r$ is changed by $2 %$ , then the change in mutual inductance of the arrangement is (Assume, $r << R$ )

1

2 %

2

1.5 %

3

4 %

4

0 %

Explanation:

C Mutual inductance $M = \frac{μ_{0} N_{1} N_{2} A}{l} = \frac{μ_{0} N_{1} N_{2} π r^{2}}{l}$
$N_{1} =$ number of turns in inner coil
$N_{2} =$ number of turns in outer coil
$A =$ Area of coil
$M \propto r^{2}$
Given that, $\frac{Δ r}{r} \times 100 % = 2 %$
So, By the formula of error analysis,
$\frac{Δ M}{M} \times 100 % = 2 (\frac{Δ r}{r} \times 100 %) = 2 (2 %) = 4 %$

TS- EAMCET-14.09.2020

Electro Magnetic Induction

154775 A varying current in a coil changes from $10 A$ to zero in $1.5 s$ . If the average emf induced in the coil is $200 V$ , the self-inductance of the coil is

1

25 H

2

30 H

3

50 H

4

45 H

Explanation:

B Given,
Varying current $(dI) = 10 A$
Total time $(dt) = 1.5 \sec$
Induced emf $(ε) = 200 V$
We know that,
Induced emf $(ε) = L \frac{dI}{dt}$
$L = \frac{ε}{dI / dt} = \frac{200}{10 / 1.5}$
$L = 30$ Henry

TS- EAMCET-10.09.2020

Electro Magnetic Induction

154776 Two coils have mutual inductance of $1.5 H$ . If current in primary coil is suddenly raised to $5 A$ in one mille second, the induced emf in the secondary coil is

1

75 V

2

750 V

3

7500 V

4

75000 V

Explanation:

C Given,
Mutual inductacne $(M) = 1.5 H$
Current $(dI) = 5 A$
Time $(dt) = 1 ms = 1 \times 10^{- 3} \sec$
We know that,
Induced emf $(ε) = M \frac{dI}{dt}$
$ε = 1.5 \times \frac{5}{(10)^{- 3}}$
$ε = 1.5 \times 5 \times 10^{3} = 7500 V$

Manipal UGET -2020

Question Bank Series

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