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Electro Magnetic Induction

154834 The equivalent inductance of two inductor is $2.4 H$ when connected in parallel and $10 H$ when connected in series. The difference between the two inductance is

1

2 H

2

3 H

3

4 H

4

5 H

Explanation:

A Given,
$L_{parallel} = 2.4 H$
$L_{series} = 10 H$
We know that, when two inducator connected in series-
$L_{series} = L_{1} + L_{2}$
$10 = L_{1} + L_{2}$
$L_{1} = 10 - L_{2}$
When inductor connected in parallel-
$L_{parallel} = \frac{L_{1} \times L_{2}}{(L_{1} + L_{2})}$
$2.4 = \frac{(10 - L_{2}) L_{2}}{10}$
$L_{2}^{2} - 10 L_{2} + 24 = 0$
$(L_{2} - 6) (L_{2} - 4) = 0$
$L_{2} = 6, 4 .$
Putting the value in equation (i), we get-
$L_{1} + L_{2} = 10$
$6 + L_{2} = 10$
$L_{2} = 4 H$
$L_{1} + L_{2} = 10$
$4 + L_{2} = 10$
$L_{2} = 6 H$
The two inductors is $6 H$ and $4 H$ .
Hence, the difference between two inductors is-
$= 6 - 4 = 2 H$

CG PET- 2004

Electro Magnetic Induction

154835 The inductance of coil is $60 μ H$ . A current in this coil increases from 1.0 $A$ to $1.5 A$ in $0.1 s$ . The magnitude of the induced emf is

1

60 \times 10^{- 6} V

2

300 \times 10^{- 4} V

3

30 \times 10^{- 4} V

4

3 \times 10^{- 4} V

Explanation:

D Given,
Inductance of coil $(L) = 60 μ H = 60 \times 10^{- 6} H$
Current $(dI) = 1.5 - 1.0 = 0.5 A$
Time, $(dt) = 0.1 s$
We know that,
Induced emf $(ε) = L \cdot \frac{dI}{dt}$
$ε = 60 \times 10^{- 6} \times \frac{0.5}{0.1}$
$ε = 300 \times 10^{- 6} V$
$ε = 3 \times 10^{- 4} V$

CG PET- 2004

Electro Magnetic Induction

154836 A loop of area $0.1 m^{2}$ rotates with a speed of 60 $rev / s$ with the axis of rotation perpendicular to a magnetic field $B = 0.4 T$ . If there are 100 turns in the loop, the maximum voltage induced in the loop is

1

15.07 V

2

150.7 V

3

1507 V

4

240 V

Explanation:

C Given,
Number of turns $= 100$
Area $(A) = 0.1 m^{2}$
Speed $(N) = 60 rev / s$ Magnetic field $(B) = 0.4 T$
We know that,
Induced emf in the coil -
$ε = NBA ω \sin ω t$
For maximum induced voltage -
$\sin ω t = 1$
So,
$ε_{0} = N B A ω$
$ε_{0} = 100 \times 0.4 \times 0.1 \times (2 π \times 60)$
$ε_{0} = 480 π$
$ε_{0} = 1507.2 V = 1507 V$

CG PET- 2004

Electro Magnetic Induction

154839 Two solenoids of same cross-sectional area have their lengths and number of turns in ratio of 1:2. The ratio of self-inductance of two solenoids is

1

1 : 1

2

1 : 2

3

2 : 1

4

1 : 4

Explanation:

B Given,
The ratio of number of turns $= 1 : 2$
We know that,
$Self inductance (L) = \frac{μ_{0} N^{2} IA}{l}$
$L \propto \frac{N^{2}}{l}$
So, $\frac{L_{1}}{L_{2}} = \frac{μ_{\circ} N_{1}^{2} IA / l_{1}}{μ_{\circ} N_{2}^{2} IA / l_{2}}$
$\frac{L_{1}}{L_{2}} = {(\frac{N_{1}}{N_{2}})}^{2} \times (\frac{l_{2}}{l_{1}})$
$\frac{L_{1}}{L_{2}} = {(\frac{1}{2})}^{2} \times (\frac{2}{1}) (∵ \frac{N_{1}}{N_{2}} & \frac{l_{1}}{l_{2}} = \frac{1}{2})$
$\frac{L_{1}}{L_{2}} = \frac{1}{2}$

BITSAT-2010

Electro Magnetic Induction

154834 The equivalent inductance of two inductor is $2.4 H$ when connected in parallel and $10 H$ when connected in series. The difference between the two inductance is

1

2 H

2

3 H

3

4 H

4

5 H

Explanation:

A Given,
$L_{parallel} = 2.4 H$
$L_{series} = 10 H$
We know that, when two inducator connected in series-
$L_{series} = L_{1} + L_{2}$
$10 = L_{1} + L_{2}$
$L_{1} = 10 - L_{2}$
When inductor connected in parallel-
$L_{parallel} = \frac{L_{1} \times L_{2}}{(L_{1} + L_{2})}$
$2.4 = \frac{(10 - L_{2}) L_{2}}{10}$
$L_{2}^{2} - 10 L_{2} + 24 = 0$
$(L_{2} - 6) (L_{2} - 4) = 0$
$L_{2} = 6, 4 .$
Putting the value in equation (i), we get-
$L_{1} + L_{2} = 10$
$6 + L_{2} = 10$
$L_{2} = 4 H$
$L_{1} + L_{2} = 10$
$4 + L_{2} = 10$
$L_{2} = 6 H$
The two inductors is $6 H$ and $4 H$ .
Hence, the difference between two inductors is-
$= 6 - 4 = 2 H$

CG PET- 2004

Electro Magnetic Induction

154835 The inductance of coil is $60 μ H$ . A current in this coil increases from 1.0 $A$ to $1.5 A$ in $0.1 s$ . The magnitude of the induced emf is

1

60 \times 10^{- 6} V

2

300 \times 10^{- 4} V

3

30 \times 10^{- 4} V

4

3 \times 10^{- 4} V

Explanation:

D Given,
Inductance of coil $(L) = 60 μ H = 60 \times 10^{- 6} H$
Current $(dI) = 1.5 - 1.0 = 0.5 A$
Time, $(dt) = 0.1 s$
We know that,
Induced emf $(ε) = L \cdot \frac{dI}{dt}$
$ε = 60 \times 10^{- 6} \times \frac{0.5}{0.1}$
$ε = 300 \times 10^{- 6} V$
$ε = 3 \times 10^{- 4} V$

CG PET- 2004

Electro Magnetic Induction

154836 A loop of area $0.1 m^{2}$ rotates with a speed of 60 $rev / s$ with the axis of rotation perpendicular to a magnetic field $B = 0.4 T$ . If there are 100 turns in the loop, the maximum voltage induced in the loop is

1

15.07 V

2

150.7 V

3

1507 V

4

240 V

Explanation:

C Given,
Number of turns $= 100$
Area $(A) = 0.1 m^{2}$
Speed $(N) = 60 rev / s$ Magnetic field $(B) = 0.4 T$
We know that,
Induced emf in the coil -
$ε = NBA ω \sin ω t$
For maximum induced voltage -
$\sin ω t = 1$
So,
$ε_{0} = N B A ω$
$ε_{0} = 100 \times 0.4 \times 0.1 \times (2 π \times 60)$
$ε_{0} = 480 π$
$ε_{0} = 1507.2 V = 1507 V$

CG PET- 2004

Electro Magnetic Induction

154839 Two solenoids of same cross-sectional area have their lengths and number of turns in ratio of 1:2. The ratio of self-inductance of two solenoids is

1

1 : 1

2

1 : 2

3

2 : 1

4

1 : 4

Explanation:

B Given,
The ratio of number of turns $= 1 : 2$
We know that,
$Self inductance (L) = \frac{μ_{0} N^{2} IA}{l}$
$L \propto \frac{N^{2}}{l}$
So, $\frac{L_{1}}{L_{2}} = \frac{μ_{\circ} N_{1}^{2} IA / l_{1}}{μ_{\circ} N_{2}^{2} IA / l_{2}}$
$\frac{L_{1}}{L_{2}} = {(\frac{N_{1}}{N_{2}})}^{2} \times (\frac{l_{2}}{l_{1}})$
$\frac{L_{1}}{L_{2}} = {(\frac{1}{2})}^{2} \times (\frac{2}{1}) (∵ \frac{N_{1}}{N_{2}} & \frac{l_{1}}{l_{2}} = \frac{1}{2})$
$\frac{L_{1}}{L_{2}} = \frac{1}{2}$

BITSAT-2010

Electro Magnetic Induction

154834 The equivalent inductance of two inductor is $2.4 H$ when connected in parallel and $10 H$ when connected in series. The difference between the two inductance is

1

2 H

2

3 H

3

4 H

4

5 H

Explanation:

A Given,
$L_{parallel} = 2.4 H$
$L_{series} = 10 H$
We know that, when two inducator connected in series-
$L_{series} = L_{1} + L_{2}$
$10 = L_{1} + L_{2}$
$L_{1} = 10 - L_{2}$
When inductor connected in parallel-
$L_{parallel} = \frac{L_{1} \times L_{2}}{(L_{1} + L_{2})}$
$2.4 = \frac{(10 - L_{2}) L_{2}}{10}$
$L_{2}^{2} - 10 L_{2} + 24 = 0$
$(L_{2} - 6) (L_{2} - 4) = 0$
$L_{2} = 6, 4 .$
Putting the value in equation (i), we get-
$L_{1} + L_{2} = 10$
$6 + L_{2} = 10$
$L_{2} = 4 H$
$L_{1} + L_{2} = 10$
$4 + L_{2} = 10$
$L_{2} = 6 H$
The two inductors is $6 H$ and $4 H$ .
Hence, the difference between two inductors is-
$= 6 - 4 = 2 H$

CG PET- 2004

Electro Magnetic Induction

154835 The inductance of coil is $60 μ H$ . A current in this coil increases from 1.0 $A$ to $1.5 A$ in $0.1 s$ . The magnitude of the induced emf is

1

60 \times 10^{- 6} V

2

300 \times 10^{- 4} V

3

30 \times 10^{- 4} V

4

3 \times 10^{- 4} V

Explanation:

D Given,
Inductance of coil $(L) = 60 μ H = 60 \times 10^{- 6} H$
Current $(dI) = 1.5 - 1.0 = 0.5 A$
Time, $(dt) = 0.1 s$
We know that,
Induced emf $(ε) = L \cdot \frac{dI}{dt}$
$ε = 60 \times 10^{- 6} \times \frac{0.5}{0.1}$
$ε = 300 \times 10^{- 6} V$
$ε = 3 \times 10^{- 4} V$

CG PET- 2004

Electro Magnetic Induction

154836 A loop of area $0.1 m^{2}$ rotates with a speed of 60 $rev / s$ with the axis of rotation perpendicular to a magnetic field $B = 0.4 T$ . If there are 100 turns in the loop, the maximum voltage induced in the loop is

1

15.07 V

2

150.7 V

3

1507 V

4

240 V

Explanation:

C Given,
Number of turns $= 100$
Area $(A) = 0.1 m^{2}$
Speed $(N) = 60 rev / s$ Magnetic field $(B) = 0.4 T$
We know that,
Induced emf in the coil -
$ε = NBA ω \sin ω t$
For maximum induced voltage -
$\sin ω t = 1$
So,
$ε_{0} = N B A ω$
$ε_{0} = 100 \times 0.4 \times 0.1 \times (2 π \times 60)$
$ε_{0} = 480 π$
$ε_{0} = 1507.2 V = 1507 V$

CG PET- 2004

Electro Magnetic Induction

154839 Two solenoids of same cross-sectional area have their lengths and number of turns in ratio of 1:2. The ratio of self-inductance of two solenoids is

1

1 : 1

2

1 : 2

3

2 : 1

4

1 : 4

Explanation:

B Given,
The ratio of number of turns $= 1 : 2$
We know that,
$Self inductance (L) = \frac{μ_{0} N^{2} IA}{l}$
$L \propto \frac{N^{2}}{l}$
So, $\frac{L_{1}}{L_{2}} = \frac{μ_{\circ} N_{1}^{2} IA / l_{1}}{μ_{\circ} N_{2}^{2} IA / l_{2}}$
$\frac{L_{1}}{L_{2}} = {(\frac{N_{1}}{N_{2}})}^{2} \times (\frac{l_{2}}{l_{1}})$
$\frac{L_{1}}{L_{2}} = {(\frac{1}{2})}^{2} \times (\frac{2}{1}) (∵ \frac{N_{1}}{N_{2}} & \frac{l_{1}}{l_{2}} = \frac{1}{2})$
$\frac{L_{1}}{L_{2}} = \frac{1}{2}$

BITSAT-2010

Electro Magnetic Induction

154834 The equivalent inductance of two inductor is $2.4 H$ when connected in parallel and $10 H$ when connected in series. The difference between the two inductance is

1

2 H

2

3 H

3

4 H

4

5 H

Explanation:

A Given,
$L_{parallel} = 2.4 H$
$L_{series} = 10 H$
We know that, when two inducator connected in series-
$L_{series} = L_{1} + L_{2}$
$10 = L_{1} + L_{2}$
$L_{1} = 10 - L_{2}$
When inductor connected in parallel-
$L_{parallel} = \frac{L_{1} \times L_{2}}{(L_{1} + L_{2})}$
$2.4 = \frac{(10 - L_{2}) L_{2}}{10}$
$L_{2}^{2} - 10 L_{2} + 24 = 0$
$(L_{2} - 6) (L_{2} - 4) = 0$
$L_{2} = 6, 4 .$
Putting the value in equation (i), we get-
$L_{1} + L_{2} = 10$
$6 + L_{2} = 10$
$L_{2} = 4 H$
$L_{1} + L_{2} = 10$
$4 + L_{2} = 10$
$L_{2} = 6 H$
The two inductors is $6 H$ and $4 H$ .
Hence, the difference between two inductors is-
$= 6 - 4 = 2 H$

CG PET- 2004

Electro Magnetic Induction

154835 The inductance of coil is $60 μ H$ . A current in this coil increases from 1.0 $A$ to $1.5 A$ in $0.1 s$ . The magnitude of the induced emf is

1

60 \times 10^{- 6} V

2

300 \times 10^{- 4} V

3

30 \times 10^{- 4} V

4

3 \times 10^{- 4} V

Explanation:

D Given,
Inductance of coil $(L) = 60 μ H = 60 \times 10^{- 6} H$
Current $(dI) = 1.5 - 1.0 = 0.5 A$
Time, $(dt) = 0.1 s$
We know that,
Induced emf $(ε) = L \cdot \frac{dI}{dt}$
$ε = 60 \times 10^{- 6} \times \frac{0.5}{0.1}$
$ε = 300 \times 10^{- 6} V$
$ε = 3 \times 10^{- 4} V$

CG PET- 2004

Electro Magnetic Induction

154836 A loop of area $0.1 m^{2}$ rotates with a speed of 60 $rev / s$ with the axis of rotation perpendicular to a magnetic field $B = 0.4 T$ . If there are 100 turns in the loop, the maximum voltage induced in the loop is

1

15.07 V

2

150.7 V

3

1507 V

4

240 V

Explanation:

C Given,
Number of turns $= 100$
Area $(A) = 0.1 m^{2}$
Speed $(N) = 60 rev / s$ Magnetic field $(B) = 0.4 T$
We know that,
Induced emf in the coil -
$ε = NBA ω \sin ω t$
For maximum induced voltage -
$\sin ω t = 1$
So,
$ε_{0} = N B A ω$
$ε_{0} = 100 \times 0.4 \times 0.1 \times (2 π \times 60)$
$ε_{0} = 480 π$
$ε_{0} = 1507.2 V = 1507 V$

CG PET- 2004

Electro Magnetic Induction

154839 Two solenoids of same cross-sectional area have their lengths and number of turns in ratio of 1:2. The ratio of self-inductance of two solenoids is

1

1 : 1

2

1 : 2

3

2 : 1

4

1 : 4

Explanation:

B Given,
The ratio of number of turns $= 1 : 2$
We know that,
$Self inductance (L) = \frac{μ_{0} N^{2} IA}{l}$
$L \propto \frac{N^{2}}{l}$
So, $\frac{L_{1}}{L_{2}} = \frac{μ_{\circ} N_{1}^{2} IA / l_{1}}{μ_{\circ} N_{2}^{2} IA / l_{2}}$
$\frac{L_{1}}{L_{2}} = {(\frac{N_{1}}{N_{2}})}^{2} \times (\frac{l_{2}}{l_{1}})$
$\frac{L_{1}}{L_{2}} = {(\frac{1}{2})}^{2} \times (\frac{2}{1}) (∵ \frac{N_{1}}{N_{2}} & \frac{l_{1}}{l_{2}} = \frac{1}{2})$
$\frac{L_{1}}{L_{2}} = \frac{1}{2}$

BITSAT-2010