QBManager

Electro Magnetic Induction

154762 The mutual inductance between two coils is 0.09 Henry. If the current in the primary coil changes from 0 to $20 A$ in $0.006 s$ , the e.m.f. induced in the secondary coil at that instant is

1

120 V

2

200 V

3

180 V

4

300 V

Explanation:

D Given,
Mutual inductance $M = 0.09 H$
Change in current $| Δ I | = I_{1} - I_{2} = 0 - 20$
$Δ I = 20 A$
Time $Δ t = 0.006 s$
$∵ ε = \frac{M Δ I}{Δ t} = \frac{0.09 \times 20}{0.006}$ $ε = 300 V$

MHT-CET 2020

Electro Magnetic Induction

154763 Two coaxial coils $A$ and $B$ of radii ' $R_{1}$ ' and ' $R_{2}$ ' are placed in the same plane. $(R_{2} > R_{1})$ . If a current is passed through coil $B$ , the coefficient of mutual inductance between the coils is proportional to

1

\frac{R_{1}^{2}}{R_{2}}

2

\frac{R_{2}^{2}}{R_{1}}

3

\frac{1}{R_{1} R_{2}}

4

R_{1} R_{2}

Explanation:

A :
For bigger loop-
$B = \frac{μ_{o} I_{1}}{2 R_{2}}$
So, the flux associated within inner loop
$ϕ_{2} = B. A_{1}$
$ϕ_{2} = \frac{μ_{o} I_{1}}{2 R_{2}} \times π R_{1}^{2}$
$∵ ϕ_{2} = {MI}_{1}$
$∴ M = \frac{ϕ_{2}}{I_{1}} = \frac{μ_{o} I_{1}}{2 R_{2}} \times \frac{π R_{1}^{2}}{I_{1}}$
$M = \frac{μ_{o} π R_{1}^{2}}{2 R_{2}}$
$∴ M \propto \frac{R_{1}^{2}}{R_{2}}$

MHT-CET 2020

Electro Magnetic Induction

154764 A current $I = 10 \sin (50 π t)$ ampere is passed in the first coil which induces a maximum e.m.f. of $5 π$ volt in the second coil. The mutual inductance between the coils is

1

5 mH

2

1 mH

3

0.1 mH

4

10 mH

Explanation:

D Given,
$I = 10 \sin (50 π t)$
Maximum induced emf $= 5 π$
Comparing equation (i) with standered equation
$I = I_{o} \sin (ω t)$
$I_{o} = 10 A$
$ω = 50 π$
Induced emf
$ε = M \frac{d I}{d t}$
$= M \frac{d}{d t} [10 \sin (50 π t)]$
$= 10 M (\cos 50 π t) \times 50 π$
For maximum value of $ε, \cos 50 π t = 1$
$ε_{max} = 10 M \times 50 π t$
$M = \frac{ε_{max}}{10 \times 50 π} = \frac{5 π}{10 \times 50 π} = \frac{1}{100}$
$M = 1 \times 10^{- 2}$
$M = 10 \times 10^{- 3}$
$M = 10 mH$

MHT-CET 2020

Electro Magnetic Induction

154765 A toroidal solenoid with air core has an average radius ' $R$ ', number of turns ' $N$ ' and area of cross-section ' $A$ '. The self-inductance of the solenoid is (Neglect the field variation across-section of the toroid)

1

\frac{μ_{0} N^{2} A}{R}

2

\frac{μ_{0} NA}{2 π R}

3

\frac{μ_{0} NA}{R}

4

\frac{μ_{0} N^{2} A}{2 π R}

Explanation:

D Given,
Average radius of toroidal solenoid $= R$ .
Number of turns $= N$
Area of crosesoction $= A$
Then, self inductance of solenoide,
$= \frac{μ_{o} N^{2} A}{2 π R}$

MHT-CET 2020

Electro Magnetic Induction

154767 Two coils $P$ and $Q$ have mutual inductance ' $M$ ' $H$ . If the current in the primary is $I = I_{0} \sin ω t$ , then the maximum value of e.m.f. induced in coil $Q$ is

1

\frac{M}{I_{0} ω}

2

I_{0} M ω

3

\frac{I_{0}}{M ω}

4

\frac{ω}{I_{0} M}

Explanation:

B Given
$I = I_{o} \sin ω t$
Mutual inductance $= M$
Induced emf in the coil
$ε = M \frac{d I}{d t}$
$= M \frac{d}{d t} (I_{0} \sin ω t)$
$ε = M ω I_{0} \cos ω t$
For maximum value of $ε, \cos ω t = 1$
$∴ ε_{max} = M I_{0} ω$

MHT-CET 2020

Electro Magnetic Induction

154762 The mutual inductance between two coils is 0.09 Henry. If the current in the primary coil changes from 0 to $20 A$ in $0.006 s$ , the e.m.f. induced in the secondary coil at that instant is

1

120 V

2

200 V

3

180 V

4

300 V

Explanation:

D Given,
Mutual inductance $M = 0.09 H$
Change in current $| Δ I | = I_{1} - I_{2} = 0 - 20$
$Δ I = 20 A$
Time $Δ t = 0.006 s$
$∵ ε = \frac{M Δ I}{Δ t} = \frac{0.09 \times 20}{0.006}$ $ε = 300 V$

MHT-CET 2020

Electro Magnetic Induction

154763 Two coaxial coils $A$ and $B$ of radii ' $R_{1}$ ' and ' $R_{2}$ ' are placed in the same plane. $(R_{2} > R_{1})$ . If a current is passed through coil $B$ , the coefficient of mutual inductance between the coils is proportional to

1

\frac{R_{1}^{2}}{R_{2}}

2

\frac{R_{2}^{2}}{R_{1}}

3

\frac{1}{R_{1} R_{2}}

4

R_{1} R_{2}

Explanation:

A :
For bigger loop-
$B = \frac{μ_{o} I_{1}}{2 R_{2}}$
So, the flux associated within inner loop
$ϕ_{2} = B. A_{1}$
$ϕ_{2} = \frac{μ_{o} I_{1}}{2 R_{2}} \times π R_{1}^{2}$
$∵ ϕ_{2} = {MI}_{1}$
$∴ M = \frac{ϕ_{2}}{I_{1}} = \frac{μ_{o} I_{1}}{2 R_{2}} \times \frac{π R_{1}^{2}}{I_{1}}$
$M = \frac{μ_{o} π R_{1}^{2}}{2 R_{2}}$
$∴ M \propto \frac{R_{1}^{2}}{R_{2}}$

MHT-CET 2020

Electro Magnetic Induction

154764 A current $I = 10 \sin (50 π t)$ ampere is passed in the first coil which induces a maximum e.m.f. of $5 π$ volt in the second coil. The mutual inductance between the coils is

1

5 mH

2

1 mH

3

0.1 mH

4

10 mH

Explanation:

D Given,
$I = 10 \sin (50 π t)$
Maximum induced emf $= 5 π$
Comparing equation (i) with standered equation
$I = I_{o} \sin (ω t)$
$I_{o} = 10 A$
$ω = 50 π$
Induced emf
$ε = M \frac{d I}{d t}$
$= M \frac{d}{d t} [10 \sin (50 π t)]$
$= 10 M (\cos 50 π t) \times 50 π$
For maximum value of $ε, \cos 50 π t = 1$
$ε_{max} = 10 M \times 50 π t$
$M = \frac{ε_{max}}{10 \times 50 π} = \frac{5 π}{10 \times 50 π} = \frac{1}{100}$
$M = 1 \times 10^{- 2}$
$M = 10 \times 10^{- 3}$
$M = 10 mH$

MHT-CET 2020

Electro Magnetic Induction

154765 A toroidal solenoid with air core has an average radius ' $R$ ', number of turns ' $N$ ' and area of cross-section ' $A$ '. The self-inductance of the solenoid is (Neglect the field variation across-section of the toroid)

1

\frac{μ_{0} N^{2} A}{R}

2

\frac{μ_{0} NA}{2 π R}

3

\frac{μ_{0} NA}{R}

4

\frac{μ_{0} N^{2} A}{2 π R}

Explanation:

D Given,
Average radius of toroidal solenoid $= R$ .
Number of turns $= N$
Area of crosesoction $= A$
Then, self inductance of solenoide,
$= \frac{μ_{o} N^{2} A}{2 π R}$

MHT-CET 2020

Electro Magnetic Induction

154767 Two coils $P$ and $Q$ have mutual inductance ' $M$ ' $H$ . If the current in the primary is $I = I_{0} \sin ω t$ , then the maximum value of e.m.f. induced in coil $Q$ is

1

\frac{M}{I_{0} ω}

2

I_{0} M ω

3

\frac{I_{0}}{M ω}

4

\frac{ω}{I_{0} M}

Explanation:

B Given
$I = I_{o} \sin ω t$
Mutual inductance $= M$
Induced emf in the coil
$ε = M \frac{d I}{d t}$
$= M \frac{d}{d t} (I_{0} \sin ω t)$
$ε = M ω I_{0} \cos ω t$
For maximum value of $ε, \cos ω t = 1$
$∴ ε_{max} = M I_{0} ω$

MHT-CET 2020

Electro Magnetic Induction

154762 The mutual inductance between two coils is 0.09 Henry. If the current in the primary coil changes from 0 to $20 A$ in $0.006 s$ , the e.m.f. induced in the secondary coil at that instant is

1

120 V

2

200 V

3

180 V

4

300 V

Explanation:

D Given,
Mutual inductance $M = 0.09 H$
Change in current $| Δ I | = I_{1} - I_{2} = 0 - 20$
$Δ I = 20 A$
Time $Δ t = 0.006 s$
$∵ ε = \frac{M Δ I}{Δ t} = \frac{0.09 \times 20}{0.006}$ $ε = 300 V$

MHT-CET 2020

Electro Magnetic Induction

154763 Two coaxial coils $A$ and $B$ of radii ' $R_{1}$ ' and ' $R_{2}$ ' are placed in the same plane. $(R_{2} > R_{1})$ . If a current is passed through coil $B$ , the coefficient of mutual inductance between the coils is proportional to

1

\frac{R_{1}^{2}}{R_{2}}

2

\frac{R_{2}^{2}}{R_{1}}

3

\frac{1}{R_{1} R_{2}}

4

R_{1} R_{2}

Explanation:

A :
For bigger loop-
$B = \frac{μ_{o} I_{1}}{2 R_{2}}$
So, the flux associated within inner loop
$ϕ_{2} = B. A_{1}$
$ϕ_{2} = \frac{μ_{o} I_{1}}{2 R_{2}} \times π R_{1}^{2}$
$∵ ϕ_{2} = {MI}_{1}$
$∴ M = \frac{ϕ_{2}}{I_{1}} = \frac{μ_{o} I_{1}}{2 R_{2}} \times \frac{π R_{1}^{2}}{I_{1}}$
$M = \frac{μ_{o} π R_{1}^{2}}{2 R_{2}}$
$∴ M \propto \frac{R_{1}^{2}}{R_{2}}$

MHT-CET 2020

Electro Magnetic Induction

154764 A current $I = 10 \sin (50 π t)$ ampere is passed in the first coil which induces a maximum e.m.f. of $5 π$ volt in the second coil. The mutual inductance between the coils is

1

5 mH

2

1 mH

3

0.1 mH

4

10 mH

Explanation:

D Given,
$I = 10 \sin (50 π t)$
Maximum induced emf $= 5 π$
Comparing equation (i) with standered equation
$I = I_{o} \sin (ω t)$
$I_{o} = 10 A$
$ω = 50 π$
Induced emf
$ε = M \frac{d I}{d t}$
$= M \frac{d}{d t} [10 \sin (50 π t)]$
$= 10 M (\cos 50 π t) \times 50 π$
For maximum value of $ε, \cos 50 π t = 1$
$ε_{max} = 10 M \times 50 π t$
$M = \frac{ε_{max}}{10 \times 50 π} = \frac{5 π}{10 \times 50 π} = \frac{1}{100}$
$M = 1 \times 10^{- 2}$
$M = 10 \times 10^{- 3}$
$M = 10 mH$

MHT-CET 2020

Electro Magnetic Induction

154765 A toroidal solenoid with air core has an average radius ' $R$ ', number of turns ' $N$ ' and area of cross-section ' $A$ '. The self-inductance of the solenoid is (Neglect the field variation across-section of the toroid)

1

\frac{μ_{0} N^{2} A}{R}

2

\frac{μ_{0} NA}{2 π R}

3

\frac{μ_{0} NA}{R}

4

\frac{μ_{0} N^{2} A}{2 π R}

Explanation:

D Given,
Average radius of toroidal solenoid $= R$ .
Number of turns $= N$
Area of crosesoction $= A$
Then, self inductance of solenoide,
$= \frac{μ_{o} N^{2} A}{2 π R}$

MHT-CET 2020

Electro Magnetic Induction

154767 Two coils $P$ and $Q$ have mutual inductance ' $M$ ' $H$ . If the current in the primary is $I = I_{0} \sin ω t$ , then the maximum value of e.m.f. induced in coil $Q$ is

1

\frac{M}{I_{0} ω}

2

I_{0} M ω

3

\frac{I_{0}}{M ω}

4

\frac{ω}{I_{0} M}

Explanation:

B Given
$I = I_{o} \sin ω t$
Mutual inductance $= M$
Induced emf in the coil
$ε = M \frac{d I}{d t}$
$= M \frac{d}{d t} (I_{0} \sin ω t)$
$ε = M ω I_{0} \cos ω t$
For maximum value of $ε, \cos ω t = 1$
$∴ ε_{max} = M I_{0} ω$

MHT-CET 2020

Electro Magnetic Induction

154762 The mutual inductance between two coils is 0.09 Henry. If the current in the primary coil changes from 0 to $20 A$ in $0.006 s$ , the e.m.f. induced in the secondary coil at that instant is

1

120 V

2

200 V

3

180 V

4

300 V

Explanation:

D Given,
Mutual inductance $M = 0.09 H$
Change in current $| Δ I | = I_{1} - I_{2} = 0 - 20$
$Δ I = 20 A$
Time $Δ t = 0.006 s$
$∵ ε = \frac{M Δ I}{Δ t} = \frac{0.09 \times 20}{0.006}$ $ε = 300 V$

MHT-CET 2020

Electro Magnetic Induction

154763 Two coaxial coils $A$ and $B$ of radii ' $R_{1}$ ' and ' $R_{2}$ ' are placed in the same plane. $(R_{2} > R_{1})$ . If a current is passed through coil $B$ , the coefficient of mutual inductance between the coils is proportional to

1

\frac{R_{1}^{2}}{R_{2}}

2

\frac{R_{2}^{2}}{R_{1}}

3

\frac{1}{R_{1} R_{2}}

4

R_{1} R_{2}

Explanation:

A :
For bigger loop-
$B = \frac{μ_{o} I_{1}}{2 R_{2}}$
So, the flux associated within inner loop
$ϕ_{2} = B. A_{1}$
$ϕ_{2} = \frac{μ_{o} I_{1}}{2 R_{2}} \times π R_{1}^{2}$
$∵ ϕ_{2} = {MI}_{1}$
$∴ M = \frac{ϕ_{2}}{I_{1}} = \frac{μ_{o} I_{1}}{2 R_{2}} \times \frac{π R_{1}^{2}}{I_{1}}$
$M = \frac{μ_{o} π R_{1}^{2}}{2 R_{2}}$
$∴ M \propto \frac{R_{1}^{2}}{R_{2}}$

MHT-CET 2020

Electro Magnetic Induction

154764 A current $I = 10 \sin (50 π t)$ ampere is passed in the first coil which induces a maximum e.m.f. of $5 π$ volt in the second coil. The mutual inductance between the coils is

1

5 mH

2

1 mH

3

0.1 mH

4

10 mH

Explanation:

D Given,
$I = 10 \sin (50 π t)$
Maximum induced emf $= 5 π$
Comparing equation (i) with standered equation
$I = I_{o} \sin (ω t)$
$I_{o} = 10 A$
$ω = 50 π$
Induced emf
$ε = M \frac{d I}{d t}$
$= M \frac{d}{d t} [10 \sin (50 π t)]$
$= 10 M (\cos 50 π t) \times 50 π$
For maximum value of $ε, \cos 50 π t = 1$
$ε_{max} = 10 M \times 50 π t$
$M = \frac{ε_{max}}{10 \times 50 π} = \frac{5 π}{10 \times 50 π} = \frac{1}{100}$
$M = 1 \times 10^{- 2}$
$M = 10 \times 10^{- 3}$
$M = 10 mH$

MHT-CET 2020

Electro Magnetic Induction

154765 A toroidal solenoid with air core has an average radius ' $R$ ', number of turns ' $N$ ' and area of cross-section ' $A$ '. The self-inductance of the solenoid is (Neglect the field variation across-section of the toroid)

1

\frac{μ_{0} N^{2} A}{R}

2

\frac{μ_{0} NA}{2 π R}

3

\frac{μ_{0} NA}{R}

4

\frac{μ_{0} N^{2} A}{2 π R}

Explanation:

D Given,
Average radius of toroidal solenoid $= R$ .
Number of turns $= N$
Area of crosesoction $= A$
Then, self inductance of solenoide,
$= \frac{μ_{o} N^{2} A}{2 π R}$

MHT-CET 2020

Electro Magnetic Induction

154767 Two coils $P$ and $Q$ have mutual inductance ' $M$ ' $H$ . If the current in the primary is $I = I_{0} \sin ω t$ , then the maximum value of e.m.f. induced in coil $Q$ is

1

\frac{M}{I_{0} ω}

2

I_{0} M ω

3

\frac{I_{0}}{M ω}

4

\frac{ω}{I_{0} M}

Explanation:

B Given
$I = I_{o} \sin ω t$
Mutual inductance $= M$
Induced emf in the coil
$ε = M \frac{d I}{d t}$
$= M \frac{d}{d t} (I_{0} \sin ω t)$
$ε = M ω I_{0} \cos ω t$
For maximum value of $ε, \cos ω t = 1$
$∴ ε_{max} = M I_{0} ω$

MHT-CET 2020

Electro Magnetic Induction

154762 The mutual inductance between two coils is 0.09 Henry. If the current in the primary coil changes from 0 to $20 A$ in $0.006 s$ , the e.m.f. induced in the secondary coil at that instant is

1

120 V

2

200 V

3

180 V

4

300 V

Explanation:

D Given,
Mutual inductance $M = 0.09 H$
Change in current $| Δ I | = I_{1} - I_{2} = 0 - 20$
$Δ I = 20 A$
Time $Δ t = 0.006 s$
$∵ ε = \frac{M Δ I}{Δ t} = \frac{0.09 \times 20}{0.006}$ $ε = 300 V$

MHT-CET 2020

Electro Magnetic Induction

154763 Two coaxial coils $A$ and $B$ of radii ' $R_{1}$ ' and ' $R_{2}$ ' are placed in the same plane. $(R_{2} > R_{1})$ . If a current is passed through coil $B$ , the coefficient of mutual inductance between the coils is proportional to

1

\frac{R_{1}^{2}}{R_{2}}

2

\frac{R_{2}^{2}}{R_{1}}

3

\frac{1}{R_{1} R_{2}}

4

R_{1} R_{2}

Explanation:

A :
For bigger loop-
$B = \frac{μ_{o} I_{1}}{2 R_{2}}$
So, the flux associated within inner loop
$ϕ_{2} = B. A_{1}$
$ϕ_{2} = \frac{μ_{o} I_{1}}{2 R_{2}} \times π R_{1}^{2}$
$∵ ϕ_{2} = {MI}_{1}$
$∴ M = \frac{ϕ_{2}}{I_{1}} = \frac{μ_{o} I_{1}}{2 R_{2}} \times \frac{π R_{1}^{2}}{I_{1}}$
$M = \frac{μ_{o} π R_{1}^{2}}{2 R_{2}}$
$∴ M \propto \frac{R_{1}^{2}}{R_{2}}$

MHT-CET 2020

Electro Magnetic Induction

154764 A current $I = 10 \sin (50 π t)$ ampere is passed in the first coil which induces a maximum e.m.f. of $5 π$ volt in the second coil. The mutual inductance between the coils is

1

5 mH

2

1 mH

3

0.1 mH

4

10 mH

Explanation:

D Given,
$I = 10 \sin (50 π t)$
Maximum induced emf $= 5 π$
Comparing equation (i) with standered equation
$I = I_{o} \sin (ω t)$
$I_{o} = 10 A$
$ω = 50 π$
Induced emf
$ε = M \frac{d I}{d t}$
$= M \frac{d}{d t} [10 \sin (50 π t)]$
$= 10 M (\cos 50 π t) \times 50 π$
For maximum value of $ε, \cos 50 π t = 1$
$ε_{max} = 10 M \times 50 π t$
$M = \frac{ε_{max}}{10 \times 50 π} = \frac{5 π}{10 \times 50 π} = \frac{1}{100}$
$M = 1 \times 10^{- 2}$
$M = 10 \times 10^{- 3}$
$M = 10 mH$

MHT-CET 2020

Electro Magnetic Induction

154765 A toroidal solenoid with air core has an average radius ' $R$ ', number of turns ' $N$ ' and area of cross-section ' $A$ '. The self-inductance of the solenoid is (Neglect the field variation across-section of the toroid)

1

\frac{μ_{0} N^{2} A}{R}

2

\frac{μ_{0} NA}{2 π R}

3

\frac{μ_{0} NA}{R}

4

\frac{μ_{0} N^{2} A}{2 π R}

Explanation:

D Given,
Average radius of toroidal solenoid $= R$ .
Number of turns $= N$
Area of crosesoction $= A$
Then, self inductance of solenoide,
$= \frac{μ_{o} N^{2} A}{2 π R}$

MHT-CET 2020

Electro Magnetic Induction

154767 Two coils $P$ and $Q$ have mutual inductance ' $M$ ' $H$ . If the current in the primary is $I = I_{0} \sin ω t$ , then the maximum value of e.m.f. induced in coil $Q$ is

1

\frac{M}{I_{0} ω}

2

I_{0} M ω

3

\frac{I_{0}}{M ω}

4

\frac{ω}{I_{0} M}

Explanation:

B Given
$I = I_{o} \sin ω t$
Mutual inductance $= M$
Induced emf in the coil
$ε = M \frac{d I}{d t}$
$= M \frac{d}{d t} (I_{0} \sin ω t)$
$ε = M ω I_{0} \cos ω t$
For maximum value of $ε, \cos ω t = 1$
$∴ ε_{max} = M I_{0} ω$

MHT-CET 2020