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PHXII06:ELECTROMAGNETIC INDUCTION

358602 Shown in the figure is a circular loop of radius $r$ and resistance $R$ . A variable magnetic field of induction $B = B_{0} e^{- t}$ is established inside the coil. If the key $(K)$ is closed, the electrical power developed right after closing the switch is equal to
supporting img

1

\frac{B_{0}^{2} π r^{2}}{R}

2

\frac{B_{0}^{2} π r^{3}}{R}

3

\frac{B_{0}^{2} π^{2} r^{4} R}{5}

4

\frac{B_{0}^{2} π^{2} r^{4}}{R}

Explanation:

$P = \frac{e^{2}}{R}; e = - \frac{d}{d t} (B A)$
$= A \frac{d}{d t} (B_{0} e^{- t}) = A B_{0} e^{- t}$
$\Rightarrow P = \frac{1}{R} {(A B_{0} e^{- t})}^{2} = \frac{A^{2} B_{0}^{2} e^{- 2 t}}{R}$
At the time of starting $t = 0$ , and $P = \frac{A^{2} B_{0}^{2}}{R}$
$\Rightarrow P = \frac{{(π r^{2})}^{2} B_{0}^{2}}{R} = \frac{B_{0}^{2} π^{2} r^{4}}{R}$

PHXII06:ELECTROMAGNETIC INDUCTION

358603 Assertion :
An induced emf is generated when magnet is withdrawn from the solenoid.
Reason :
The relative motion between magnet and solenoid induces $e m f .$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

These are correct by Faraday's law of electro-magnetic induction. So correct option is $(1)$ .

PHXII06:ELECTROMAGNETIC INDUCTION

358604 A conducting ring of radius $r$ is placed perpendicularly inside a time varying magnetic field given by $B = B_{0} + α t$ , as shown in the figure. $B_{o}$ and $α$ are positive constants. Find the emf produced in the ring.
supporting img

1

- π α r^{2}

2

- π α r

3

- π α^{2} r^{2}

4

- π α^{2} r

Explanation:

Given that, the magnetic field vary with time as
$B = B_{0} + α t (1)$
Here, $B_{0}$ and $α$ are positive constants,
We know the induced emf is
$e = - \frac{d ϕ}{d t} . (2)$
$e = - \frac{d}{d t} (B A)$
On putting the value of B from Eq. (1)
$So, \begin{aligned} e & = - \frac{d}{d t} [(B_{0} + α t) A] \\ = - A \frac{d}{d t} [B_{0} + α t] = - A [0 + α] = - A α \end{aligned}$
But $A = π r^{2}$ is the area of ring, here $r$ is the radius of ring, so, $e = - π r^{2} α \Rightarrow e = - π α r^{2}$

PHXII06:ELECTROMAGNETIC INDUCTION

358605 A 800 turn coil of effective area $0.05 m^{2}$ is kept perpendicular to a magnetic field $5 \times 10^{- 5} T$ . When the plane of the coil is rotated by $90^{\circ}$ around any of its coplanar axis in $0.1 s$ , the emf induced in the coil will be:

1

2 V

2

0.2 V

3

2 \times 10^{- 3} V

4

0.02 V

Explanation:

Given that $B = 5 \times 10^{- 5} T$
Number of turns $N = 800$
$A = 0.05 m^{2}$
$Δ t = 0.1 s$
Change in magnetic flux $Δ ϕ$
$\begin{aligned} = N B A \cos 90^{\circ} - N B A \cos 0^{\circ} \\ = - N B A \\ = - 2 \times 10^{- 3} weber \end{aligned}$
$ε = - \frac{Δ ϕ}{Δ t} = \frac{- (-) 2 \times 10^{- 3} W b}{0.1 s} = 0.02 V$

NEET - 2019

PHXII06:ELECTROMAGNETIC INDUCTION

358606 The four separate figures below involve a cylindrical magnet and a tiny light bulb connected to the ends of a loop of copper wire. These figures are to be used in the following question. The plane of the wire loop is perpendicular to the reference axis. The states of motion of the magnet and of the loop of wire indicated in the diagram, Speed will be represented by $v$ and CCW represents counter clockwise.
In which of the below figure will the light bulb be glowing?
supporting img

1 I, II, IV

2 I, III, IV

3 I, II, III

4 II, III, IV

PHXII06:ELECTROMAGNETIC INDUCTION

358602 Shown in the figure is a circular loop of radius $r$ and resistance $R$ . A variable magnetic field of induction $B = B_{0} e^{- t}$ is established inside the coil. If the key $(K)$ is closed, the electrical power developed right after closing the switch is equal to
supporting img

1

\frac{B_{0}^{2} π r^{2}}{R}

2

\frac{B_{0}^{2} π r^{3}}{R}

3

\frac{B_{0}^{2} π^{2} r^{4} R}{5}

4

\frac{B_{0}^{2} π^{2} r^{4}}{R}

Explanation:

$P = \frac{e^{2}}{R}; e = - \frac{d}{d t} (B A)$
$= A \frac{d}{d t} (B_{0} e^{- t}) = A B_{0} e^{- t}$
$\Rightarrow P = \frac{1}{R} {(A B_{0} e^{- t})}^{2} = \frac{A^{2} B_{0}^{2} e^{- 2 t}}{R}$
At the time of starting $t = 0$ , and $P = \frac{A^{2} B_{0}^{2}}{R}$
$\Rightarrow P = \frac{{(π r^{2})}^{2} B_{0}^{2}}{R} = \frac{B_{0}^{2} π^{2} r^{4}}{R}$

PHXII06:ELECTROMAGNETIC INDUCTION

358603 Assertion :
An induced emf is generated when magnet is withdrawn from the solenoid.
Reason :
The relative motion between magnet and solenoid induces $e m f .$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

These are correct by Faraday's law of electro-magnetic induction. So correct option is $(1)$ .

PHXII06:ELECTROMAGNETIC INDUCTION

358604 A conducting ring of radius $r$ is placed perpendicularly inside a time varying magnetic field given by $B = B_{0} + α t$ , as shown in the figure. $B_{o}$ and $α$ are positive constants. Find the emf produced in the ring.
supporting img

1

- π α r^{2}

2

- π α r

3

- π α^{2} r^{2}

4

- π α^{2} r

Explanation:

Given that, the magnetic field vary with time as
$B = B_{0} + α t (1)$
Here, $B_{0}$ and $α$ are positive constants,
We know the induced emf is
$e = - \frac{d ϕ}{d t} . (2)$
$e = - \frac{d}{d t} (B A)$
On putting the value of B from Eq. (1)
$So, \begin{aligned} e & = - \frac{d}{d t} [(B_{0} + α t) A] \\ = - A \frac{d}{d t} [B_{0} + α t] = - A [0 + α] = - A α \end{aligned}$
But $A = π r^{2}$ is the area of ring, here $r$ is the radius of ring, so, $e = - π r^{2} α \Rightarrow e = - π α r^{2}$

PHXII06:ELECTROMAGNETIC INDUCTION

358605 A 800 turn coil of effective area $0.05 m^{2}$ is kept perpendicular to a magnetic field $5 \times 10^{- 5} T$ . When the plane of the coil is rotated by $90^{\circ}$ around any of its coplanar axis in $0.1 s$ , the emf induced in the coil will be:

1

2 V

2

0.2 V

3

2 \times 10^{- 3} V

4

0.02 V

Explanation:

Given that $B = 5 \times 10^{- 5} T$
Number of turns $N = 800$
$A = 0.05 m^{2}$
$Δ t = 0.1 s$
Change in magnetic flux $Δ ϕ$
$\begin{aligned} = N B A \cos 90^{\circ} - N B A \cos 0^{\circ} \\ = - N B A \\ = - 2 \times 10^{- 3} weber \end{aligned}$
$ε = - \frac{Δ ϕ}{Δ t} = \frac{- (-) 2 \times 10^{- 3} W b}{0.1 s} = 0.02 V$

NEET - 2019

PHXII06:ELECTROMAGNETIC INDUCTION

358606 The four separate figures below involve a cylindrical magnet and a tiny light bulb connected to the ends of a loop of copper wire. These figures are to be used in the following question. The plane of the wire loop is perpendicular to the reference axis. The states of motion of the magnet and of the loop of wire indicated in the diagram, Speed will be represented by $v$ and CCW represents counter clockwise.
In which of the below figure will the light bulb be glowing?
supporting img

1 I, II, IV

2 I, III, IV

3 I, II, III

4 II, III, IV

PHXII06:ELECTROMAGNETIC INDUCTION

358602 Shown in the figure is a circular loop of radius $r$ and resistance $R$ . A variable magnetic field of induction $B = B_{0} e^{- t}$ is established inside the coil. If the key $(K)$ is closed, the electrical power developed right after closing the switch is equal to
supporting img

1

\frac{B_{0}^{2} π r^{2}}{R}

2

\frac{B_{0}^{2} π r^{3}}{R}

3

\frac{B_{0}^{2} π^{2} r^{4} R}{5}

4

\frac{B_{0}^{2} π^{2} r^{4}}{R}

Explanation:

$P = \frac{e^{2}}{R}; e = - \frac{d}{d t} (B A)$
$= A \frac{d}{d t} (B_{0} e^{- t}) = A B_{0} e^{- t}$
$\Rightarrow P = \frac{1}{R} {(A B_{0} e^{- t})}^{2} = \frac{A^{2} B_{0}^{2} e^{- 2 t}}{R}$
At the time of starting $t = 0$ , and $P = \frac{A^{2} B_{0}^{2}}{R}$
$\Rightarrow P = \frac{{(π r^{2})}^{2} B_{0}^{2}}{R} = \frac{B_{0}^{2} π^{2} r^{4}}{R}$

PHXII06:ELECTROMAGNETIC INDUCTION

358603 Assertion :
An induced emf is generated when magnet is withdrawn from the solenoid.
Reason :
The relative motion between magnet and solenoid induces $e m f .$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

These are correct by Faraday's law of electro-magnetic induction. So correct option is $(1)$ .

PHXII06:ELECTROMAGNETIC INDUCTION

358604 A conducting ring of radius $r$ is placed perpendicularly inside a time varying magnetic field given by $B = B_{0} + α t$ , as shown in the figure. $B_{o}$ and $α$ are positive constants. Find the emf produced in the ring.
supporting img

1

- π α r^{2}

2

- π α r

3

- π α^{2} r^{2}

4

- π α^{2} r

Explanation:

Given that, the magnetic field vary with time as
$B = B_{0} + α t (1)$
Here, $B_{0}$ and $α$ are positive constants,
We know the induced emf is
$e = - \frac{d ϕ}{d t} . (2)$
$e = - \frac{d}{d t} (B A)$
On putting the value of B from Eq. (1)
$So, \begin{aligned} e & = - \frac{d}{d t} [(B_{0} + α t) A] \\ = - A \frac{d}{d t} [B_{0} + α t] = - A [0 + α] = - A α \end{aligned}$
But $A = π r^{2}$ is the area of ring, here $r$ is the radius of ring, so, $e = - π r^{2} α \Rightarrow e = - π α r^{2}$

PHXII06:ELECTROMAGNETIC INDUCTION

358605 A 800 turn coil of effective area $0.05 m^{2}$ is kept perpendicular to a magnetic field $5 \times 10^{- 5} T$ . When the plane of the coil is rotated by $90^{\circ}$ around any of its coplanar axis in $0.1 s$ , the emf induced in the coil will be:

1

2 V

2

0.2 V

3

2 \times 10^{- 3} V

4

0.02 V

Explanation:

Given that $B = 5 \times 10^{- 5} T$
Number of turns $N = 800$
$A = 0.05 m^{2}$
$Δ t = 0.1 s$
Change in magnetic flux $Δ ϕ$
$\begin{aligned} = N B A \cos 90^{\circ} - N B A \cos 0^{\circ} \\ = - N B A \\ = - 2 \times 10^{- 3} weber \end{aligned}$
$ε = - \frac{Δ ϕ}{Δ t} = \frac{- (-) 2 \times 10^{- 3} W b}{0.1 s} = 0.02 V$

NEET - 2019

PHXII06:ELECTROMAGNETIC INDUCTION

358606 The four separate figures below involve a cylindrical magnet and a tiny light bulb connected to the ends of a loop of copper wire. These figures are to be used in the following question. The plane of the wire loop is perpendicular to the reference axis. The states of motion of the magnet and of the loop of wire indicated in the diagram, Speed will be represented by $v$ and CCW represents counter clockwise.
In which of the below figure will the light bulb be glowing?
supporting img

1 I, II, IV

2 I, III, IV

3 I, II, III

4 II, III, IV

PHXII06:ELECTROMAGNETIC INDUCTION

358602 Shown in the figure is a circular loop of radius $r$ and resistance $R$ . A variable magnetic field of induction $B = B_{0} e^{- t}$ is established inside the coil. If the key $(K)$ is closed, the electrical power developed right after closing the switch is equal to
supporting img

1

\frac{B_{0}^{2} π r^{2}}{R}

2

\frac{B_{0}^{2} π r^{3}}{R}

3

\frac{B_{0}^{2} π^{2} r^{4} R}{5}

4

\frac{B_{0}^{2} π^{2} r^{4}}{R}

Explanation:

$P = \frac{e^{2}}{R}; e = - \frac{d}{d t} (B A)$
$= A \frac{d}{d t} (B_{0} e^{- t}) = A B_{0} e^{- t}$
$\Rightarrow P = \frac{1}{R} {(A B_{0} e^{- t})}^{2} = \frac{A^{2} B_{0}^{2} e^{- 2 t}}{R}$
At the time of starting $t = 0$ , and $P = \frac{A^{2} B_{0}^{2}}{R}$
$\Rightarrow P = \frac{{(π r^{2})}^{2} B_{0}^{2}}{R} = \frac{B_{0}^{2} π^{2} r^{4}}{R}$

PHXII06:ELECTROMAGNETIC INDUCTION

358603 Assertion :
An induced emf is generated when magnet is withdrawn from the solenoid.
Reason :
The relative motion between magnet and solenoid induces $e m f .$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

These are correct by Faraday's law of electro-magnetic induction. So correct option is $(1)$ .

PHXII06:ELECTROMAGNETIC INDUCTION

358604 A conducting ring of radius $r$ is placed perpendicularly inside a time varying magnetic field given by $B = B_{0} + α t$ , as shown in the figure. $B_{o}$ and $α$ are positive constants. Find the emf produced in the ring.
supporting img

1

- π α r^{2}

2

- π α r

3

- π α^{2} r^{2}

4

- π α^{2} r

Explanation:

Given that, the magnetic field vary with time as
$B = B_{0} + α t (1)$
Here, $B_{0}$ and $α$ are positive constants,
We know the induced emf is
$e = - \frac{d ϕ}{d t} . (2)$
$e = - \frac{d}{d t} (B A)$
On putting the value of B from Eq. (1)
$So, \begin{aligned} e & = - \frac{d}{d t} [(B_{0} + α t) A] \\ = - A \frac{d}{d t} [B_{0} + α t] = - A [0 + α] = - A α \end{aligned}$
But $A = π r^{2}$ is the area of ring, here $r$ is the radius of ring, so, $e = - π r^{2} α \Rightarrow e = - π α r^{2}$

PHXII06:ELECTROMAGNETIC INDUCTION

358605 A 800 turn coil of effective area $0.05 m^{2}$ is kept perpendicular to a magnetic field $5 \times 10^{- 5} T$ . When the plane of the coil is rotated by $90^{\circ}$ around any of its coplanar axis in $0.1 s$ , the emf induced in the coil will be:

1

2 V

2

0.2 V

3

2 \times 10^{- 3} V

4

0.02 V

Explanation:

Given that $B = 5 \times 10^{- 5} T$
Number of turns $N = 800$
$A = 0.05 m^{2}$
$Δ t = 0.1 s$
Change in magnetic flux $Δ ϕ$
$\begin{aligned} = N B A \cos 90^{\circ} - N B A \cos 0^{\circ} \\ = - N B A \\ = - 2 \times 10^{- 3} weber \end{aligned}$
$ε = - \frac{Δ ϕ}{Δ t} = \frac{- (-) 2 \times 10^{- 3} W b}{0.1 s} = 0.02 V$

NEET - 2019

PHXII06:ELECTROMAGNETIC INDUCTION

358606 The four separate figures below involve a cylindrical magnet and a tiny light bulb connected to the ends of a loop of copper wire. These figures are to be used in the following question. The plane of the wire loop is perpendicular to the reference axis. The states of motion of the magnet and of the loop of wire indicated in the diagram, Speed will be represented by $v$ and CCW represents counter clockwise.
In which of the below figure will the light bulb be glowing?
supporting img

1 I, II, IV

2 I, III, IV

3 I, II, III

4 II, III, IV

PHXII06:ELECTROMAGNETIC INDUCTION

358602 Shown in the figure is a circular loop of radius $r$ and resistance $R$ . A variable magnetic field of induction $B = B_{0} e^{- t}$ is established inside the coil. If the key $(K)$ is closed, the electrical power developed right after closing the switch is equal to
supporting img

1

\frac{B_{0}^{2} π r^{2}}{R}

2

\frac{B_{0}^{2} π r^{3}}{R}

3

\frac{B_{0}^{2} π^{2} r^{4} R}{5}

4

\frac{B_{0}^{2} π^{2} r^{4}}{R}

Explanation:

$P = \frac{e^{2}}{R}; e = - \frac{d}{d t} (B A)$
$= A \frac{d}{d t} (B_{0} e^{- t}) = A B_{0} e^{- t}$
$\Rightarrow P = \frac{1}{R} {(A B_{0} e^{- t})}^{2} = \frac{A^{2} B_{0}^{2} e^{- 2 t}}{R}$
At the time of starting $t = 0$ , and $P = \frac{A^{2} B_{0}^{2}}{R}$
$\Rightarrow P = \frac{{(π r^{2})}^{2} B_{0}^{2}}{R} = \frac{B_{0}^{2} π^{2} r^{4}}{R}$

PHXII06:ELECTROMAGNETIC INDUCTION

358603 Assertion :
An induced emf is generated when magnet is withdrawn from the solenoid.
Reason :
The relative motion between magnet and solenoid induces $e m f .$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

These are correct by Faraday's law of electro-magnetic induction. So correct option is $(1)$ .

PHXII06:ELECTROMAGNETIC INDUCTION

358604 A conducting ring of radius $r$ is placed perpendicularly inside a time varying magnetic field given by $B = B_{0} + α t$ , as shown in the figure. $B_{o}$ and $α$ are positive constants. Find the emf produced in the ring.
supporting img

1

- π α r^{2}

2

- π α r

3

- π α^{2} r^{2}

4

- π α^{2} r

Explanation:

Given that, the magnetic field vary with time as
$B = B_{0} + α t (1)$
Here, $B_{0}$ and $α$ are positive constants,
We know the induced emf is
$e = - \frac{d ϕ}{d t} . (2)$
$e = - \frac{d}{d t} (B A)$
On putting the value of B from Eq. (1)
$So, \begin{aligned} e & = - \frac{d}{d t} [(B_{0} + α t) A] \\ = - A \frac{d}{d t} [B_{0} + α t] = - A [0 + α] = - A α \end{aligned}$
But $A = π r^{2}$ is the area of ring, here $r$ is the radius of ring, so, $e = - π r^{2} α \Rightarrow e = - π α r^{2}$

PHXII06:ELECTROMAGNETIC INDUCTION

358605 A 800 turn coil of effective area $0.05 m^{2}$ is kept perpendicular to a magnetic field $5 \times 10^{- 5} T$ . When the plane of the coil is rotated by $90^{\circ}$ around any of its coplanar axis in $0.1 s$ , the emf induced in the coil will be:

1

2 V

2

0.2 V

3

2 \times 10^{- 3} V

4

0.02 V

Explanation:

Given that $B = 5 \times 10^{- 5} T$
Number of turns $N = 800$
$A = 0.05 m^{2}$
$Δ t = 0.1 s$
Change in magnetic flux $Δ ϕ$
$\begin{aligned} = N B A \cos 90^{\circ} - N B A \cos 0^{\circ} \\ = - N B A \\ = - 2 \times 10^{- 3} weber \end{aligned}$
$ε = - \frac{Δ ϕ}{Δ t} = \frac{- (-) 2 \times 10^{- 3} W b}{0.1 s} = 0.02 V$

NEET - 2019

PHXII06:ELECTROMAGNETIC INDUCTION

358606 The four separate figures below involve a cylindrical magnet and a tiny light bulb connected to the ends of a loop of copper wire. These figures are to be used in the following question. The plane of the wire loop is perpendicular to the reference axis. The states of motion of the magnet and of the loop of wire indicated in the diagram, Speed will be represented by $v$ and CCW represents counter clockwise.
In which of the below figure will the light bulb be glowing?
supporting img

1 I, II, IV

2 I, III, IV

3 I, II, III

4 II, III, IV

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Question Bank Series

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