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PHXII06:ELECTROMAGNETIC INDUCTION

358701 The magnetic flux through a circuit of resistance $R$ changes by an amount $Δ ϕ$ in time $Δ t$ . The total quantity of electric charge $Q$ which passes during this time through any point of circuit is given by

1

Q = \frac{Δ ϕ}{Δ t} \times R

2

Q = - \frac{Δ ϕ}{Δ t} + R

3

Q = \frac{Δ ϕ}{Δ t}

4

Q = \frac{Δ ϕ}{R}

Explanation:

The emf induced $ε = \frac{Δ ϕ}{Δ t}$
$i = \frac{Q}{Δ t} = \frac{ε}{R} = \frac{Δ ϕ}{R Δ t} \Rightarrow Q = \frac{Δ ϕ}{R}$

PHXII06:ELECTROMAGNETIC INDUCTION

358702 As shown in the figure, a magnet is moved with a fast speed towards a coil at rest. Due to this induced electromotive force, induced current and induced charge in the coil is $E$ , $I$ and $Q$ respectively. If the speed of the magnet is doubled, the incorrect statements is
supporting img

1

I

increases

2

E

increases

3

Q

increases

4

Q

remains same

Explanation:

Induced charge is given by
$\begin{aligned} i_{ind} = \frac{d q}{d t} = \frac{ε_{ind}}{R} = \frac{1}{R} \frac{d ϕ}{d t} \\ d q = \frac{d ϕ}{R} \\ Δ q = \frac{Δ ϕ}{R} \end{aligned}$
So the induced charge does not depend on speed of the incoming magnet.

NCERT Exempler

PHXII06:ELECTROMAGNETIC INDUCTION

358703 A current $i = 2.5 (1 + 2 t) \times 10^{3} A$ increases at a steady rate in a long straight wire. A small circular loop of radius $10^{- 3} m$ has its plane parallel to the wire and its centre is placed at a distance of 1 $m$ from the wire. If the resistance of the loop is $5 π \times 10^{- 4} Ω$ , find the magnitude of the induced current in the loop.
supporting img

1

1.0 μ A

2

3.0 μ A

3

5.0 μ A

4

2.0 μ A

Explanation:

The field due to the wire at the centre of loop is
$B = \frac{μ_{0} I}{2 π d} = \frac{4 π \times 10^{- 7} \times I}{2 π \times 1} = 2 I \times 10^{- 7} W / m^{2}$
So the flux linked with the loop wire
$ϕ = B A = B \times π r^{2} = 2 I \times 10^{- 7} \times π \times {(10^{- 3})}^{2}$
$= 2 π I \times 10^{- 13} W$
The magnitude of emf induced in the loop due to change of current
$| e | = \frac{d ϕ}{d t} = 2 π \times 10^{- 13} \frac{d I}{d t} V (1)$
$∵$ $I = 2.5 (1 + 2 t) \times 10^{3}$
$∴ \frac{d I}{d t} = 5.0 \times 10^{3} A / s (2)$
From (1) and (2), we get
$e = 2 π \times 10^{- 13} \times 5.0 \times 10^{3} = 10 π \times 10^{- 10} V$
And the induced current in the loop
$i = \frac{e}{R} = \frac{10 π \times 10^{- 10}}{5 π \times 10^{- 4}} = 2.0 \times 10^{- 6} A = 2.0 μ A$
Due to increase in current in the wire the flux linked with the loop will increase, so in accordance with Len's law the direction of current induced in the loop will be inverse of that in wire, i.e., anticlockwise.

PHXII06:ELECTROMAGNETIC INDUCTION

358704 Statement A :
The presence of large magnetic flux through a coil produces current in the coil.
Statement B :
Magnetic flux is essential to maintain an induced current in the coil.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

PHXII06:ELECTROMAGNETIC INDUCTION

358701 The magnetic flux through a circuit of resistance $R$ changes by an amount $Δ ϕ$ in time $Δ t$ . The total quantity of electric charge $Q$ which passes during this time through any point of circuit is given by

1

Q = \frac{Δ ϕ}{Δ t} \times R

2

Q = - \frac{Δ ϕ}{Δ t} + R

3

Q = \frac{Δ ϕ}{Δ t}

4

Q = \frac{Δ ϕ}{R}

Explanation:

The emf induced $ε = \frac{Δ ϕ}{Δ t}$
$i = \frac{Q}{Δ t} = \frac{ε}{R} = \frac{Δ ϕ}{R Δ t} \Rightarrow Q = \frac{Δ ϕ}{R}$

PHXII06:ELECTROMAGNETIC INDUCTION

358702 As shown in the figure, a magnet is moved with a fast speed towards a coil at rest. Due to this induced electromotive force, induced current and induced charge in the coil is $E$ , $I$ and $Q$ respectively. If the speed of the magnet is doubled, the incorrect statements is
supporting img

1

I

increases

2

E

increases

3

Q

increases

4

Q

remains same

Explanation:

Induced charge is given by
$\begin{aligned} i_{ind} = \frac{d q}{d t} = \frac{ε_{ind}}{R} = \frac{1}{R} \frac{d ϕ}{d t} \\ d q = \frac{d ϕ}{R} \\ Δ q = \frac{Δ ϕ}{R} \end{aligned}$
So the induced charge does not depend on speed of the incoming magnet.

NCERT Exempler

PHXII06:ELECTROMAGNETIC INDUCTION

358703 A current $i = 2.5 (1 + 2 t) \times 10^{3} A$ increases at a steady rate in a long straight wire. A small circular loop of radius $10^{- 3} m$ has its plane parallel to the wire and its centre is placed at a distance of 1 $m$ from the wire. If the resistance of the loop is $5 π \times 10^{- 4} Ω$ , find the magnitude of the induced current in the loop.
supporting img

1

1.0 μ A

2

3.0 μ A

3

5.0 μ A

4

2.0 μ A

Explanation:

The field due to the wire at the centre of loop is
$B = \frac{μ_{0} I}{2 π d} = \frac{4 π \times 10^{- 7} \times I}{2 π \times 1} = 2 I \times 10^{- 7} W / m^{2}$
So the flux linked with the loop wire
$ϕ = B A = B \times π r^{2} = 2 I \times 10^{- 7} \times π \times {(10^{- 3})}^{2}$
$= 2 π I \times 10^{- 13} W$
The magnitude of emf induced in the loop due to change of current
$| e | = \frac{d ϕ}{d t} = 2 π \times 10^{- 13} \frac{d I}{d t} V (1)$
$∵$ $I = 2.5 (1 + 2 t) \times 10^{3}$
$∴ \frac{d I}{d t} = 5.0 \times 10^{3} A / s (2)$
From (1) and (2), we get
$e = 2 π \times 10^{- 13} \times 5.0 \times 10^{3} = 10 π \times 10^{- 10} V$
And the induced current in the loop
$i = \frac{e}{R} = \frac{10 π \times 10^{- 10}}{5 π \times 10^{- 4}} = 2.0 \times 10^{- 6} A = 2.0 μ A$
Due to increase in current in the wire the flux linked with the loop will increase, so in accordance with Len's law the direction of current induced in the loop will be inverse of that in wire, i.e., anticlockwise.

PHXII06:ELECTROMAGNETIC INDUCTION

358704 Statement A :
The presence of large magnetic flux through a coil produces current in the coil.
Statement B :
Magnetic flux is essential to maintain an induced current in the coil.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

PHXII06:ELECTROMAGNETIC INDUCTION

358701 The magnetic flux through a circuit of resistance $R$ changes by an amount $Δ ϕ$ in time $Δ t$ . The total quantity of electric charge $Q$ which passes during this time through any point of circuit is given by

1

Q = \frac{Δ ϕ}{Δ t} \times R

2

Q = - \frac{Δ ϕ}{Δ t} + R

3

Q = \frac{Δ ϕ}{Δ t}

4

Q = \frac{Δ ϕ}{R}

Explanation:

The emf induced $ε = \frac{Δ ϕ}{Δ t}$
$i = \frac{Q}{Δ t} = \frac{ε}{R} = \frac{Δ ϕ}{R Δ t} \Rightarrow Q = \frac{Δ ϕ}{R}$

PHXII06:ELECTROMAGNETIC INDUCTION

358702 As shown in the figure, a magnet is moved with a fast speed towards a coil at rest. Due to this induced electromotive force, induced current and induced charge in the coil is $E$ , $I$ and $Q$ respectively. If the speed of the magnet is doubled, the incorrect statements is
supporting img

1

I

increases

2

E

increases

3

Q

increases

4

Q

remains same

Explanation:

Induced charge is given by
$\begin{aligned} i_{ind} = \frac{d q}{d t} = \frac{ε_{ind}}{R} = \frac{1}{R} \frac{d ϕ}{d t} \\ d q = \frac{d ϕ}{R} \\ Δ q = \frac{Δ ϕ}{R} \end{aligned}$
So the induced charge does not depend on speed of the incoming magnet.

NCERT Exempler

PHXII06:ELECTROMAGNETIC INDUCTION

358703 A current $i = 2.5 (1 + 2 t) \times 10^{3} A$ increases at a steady rate in a long straight wire. A small circular loop of radius $10^{- 3} m$ has its plane parallel to the wire and its centre is placed at a distance of 1 $m$ from the wire. If the resistance of the loop is $5 π \times 10^{- 4} Ω$ , find the magnitude of the induced current in the loop.
supporting img

1

1.0 μ A

2

3.0 μ A

3

5.0 μ A

4

2.0 μ A

Explanation:

The field due to the wire at the centre of loop is
$B = \frac{μ_{0} I}{2 π d} = \frac{4 π \times 10^{- 7} \times I}{2 π \times 1} = 2 I \times 10^{- 7} W / m^{2}$
So the flux linked with the loop wire
$ϕ = B A = B \times π r^{2} = 2 I \times 10^{- 7} \times π \times {(10^{- 3})}^{2}$
$= 2 π I \times 10^{- 13} W$
The magnitude of emf induced in the loop due to change of current
$| e | = \frac{d ϕ}{d t} = 2 π \times 10^{- 13} \frac{d I}{d t} V (1)$
$∵$ $I = 2.5 (1 + 2 t) \times 10^{3}$
$∴ \frac{d I}{d t} = 5.0 \times 10^{3} A / s (2)$
From (1) and (2), we get
$e = 2 π \times 10^{- 13} \times 5.0 \times 10^{3} = 10 π \times 10^{- 10} V$
And the induced current in the loop
$i = \frac{e}{R} = \frac{10 π \times 10^{- 10}}{5 π \times 10^{- 4}} = 2.0 \times 10^{- 6} A = 2.0 μ A$
Due to increase in current in the wire the flux linked with the loop will increase, so in accordance with Len's law the direction of current induced in the loop will be inverse of that in wire, i.e., anticlockwise.

PHXII06:ELECTROMAGNETIC INDUCTION

358704 Statement A :
The presence of large magnetic flux through a coil produces current in the coil.
Statement B :
Magnetic flux is essential to maintain an induced current in the coil.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

PHXII06:ELECTROMAGNETIC INDUCTION

358701 The magnetic flux through a circuit of resistance $R$ changes by an amount $Δ ϕ$ in time $Δ t$ . The total quantity of electric charge $Q$ which passes during this time through any point of circuit is given by

1

Q = \frac{Δ ϕ}{Δ t} \times R

2

Q = - \frac{Δ ϕ}{Δ t} + R

3

Q = \frac{Δ ϕ}{Δ t}

4

Q = \frac{Δ ϕ}{R}

Explanation:

The emf induced $ε = \frac{Δ ϕ}{Δ t}$
$i = \frac{Q}{Δ t} = \frac{ε}{R} = \frac{Δ ϕ}{R Δ t} \Rightarrow Q = \frac{Δ ϕ}{R}$

PHXII06:ELECTROMAGNETIC INDUCTION

358702 As shown in the figure, a magnet is moved with a fast speed towards a coil at rest. Due to this induced electromotive force, induced current and induced charge in the coil is $E$ , $I$ and $Q$ respectively. If the speed of the magnet is doubled, the incorrect statements is
supporting img

1

I

increases

2

E

increases

3

Q

increases

4

Q

remains same

Explanation:

Induced charge is given by
$\begin{aligned} i_{ind} = \frac{d q}{d t} = \frac{ε_{ind}}{R} = \frac{1}{R} \frac{d ϕ}{d t} \\ d q = \frac{d ϕ}{R} \\ Δ q = \frac{Δ ϕ}{R} \end{aligned}$
So the induced charge does not depend on speed of the incoming magnet.

NCERT Exempler

PHXII06:ELECTROMAGNETIC INDUCTION

358703 A current $i = 2.5 (1 + 2 t) \times 10^{3} A$ increases at a steady rate in a long straight wire. A small circular loop of radius $10^{- 3} m$ has its plane parallel to the wire and its centre is placed at a distance of 1 $m$ from the wire. If the resistance of the loop is $5 π \times 10^{- 4} Ω$ , find the magnitude of the induced current in the loop.
supporting img

1

1.0 μ A

2

3.0 μ A

3

5.0 μ A

4

2.0 μ A

Explanation:

The field due to the wire at the centre of loop is
$B = \frac{μ_{0} I}{2 π d} = \frac{4 π \times 10^{- 7} \times I}{2 π \times 1} = 2 I \times 10^{- 7} W / m^{2}$
So the flux linked with the loop wire
$ϕ = B A = B \times π r^{2} = 2 I \times 10^{- 7} \times π \times {(10^{- 3})}^{2}$
$= 2 π I \times 10^{- 13} W$
The magnitude of emf induced in the loop due to change of current
$| e | = \frac{d ϕ}{d t} = 2 π \times 10^{- 13} \frac{d I}{d t} V (1)$
$∵$ $I = 2.5 (1 + 2 t) \times 10^{3}$
$∴ \frac{d I}{d t} = 5.0 \times 10^{3} A / s (2)$
From (1) and (2), we get
$e = 2 π \times 10^{- 13} \times 5.0 \times 10^{3} = 10 π \times 10^{- 10} V$
And the induced current in the loop
$i = \frac{e}{R} = \frac{10 π \times 10^{- 10}}{5 π \times 10^{- 4}} = 2.0 \times 10^{- 6} A = 2.0 μ A$
Due to increase in current in the wire the flux linked with the loop will increase, so in accordance with Len's law the direction of current induced in the loop will be inverse of that in wire, i.e., anticlockwise.

PHXII06:ELECTROMAGNETIC INDUCTION

358704 Statement A :
The presence of large magnetic flux through a coil produces current in the coil.
Statement B :
Magnetic flux is essential to maintain an induced current in the coil.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

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