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PHXII06:ELECTROMAGNETIC INDUCTION

358474 An aeroplane having a wing space of $35 m$ flies due north with the speed of $90 m s^{- 1}$ , given $B = 4 \times 10^{- 5} T$ . The potential difference between the tips of the wings will be:

1

0.013 V

2

1.26 V

3

12.6 V

4

0.126 V

Explanation:

The induced emf is given by
$= B v l = 4 \times 10^{- 5} \times 90 \times 35 = 0.126 V$

PHXII06:ELECTROMAGNETIC INDUCTION

358475 Statement A :
An emf can be induced by moving a conductor into a magnetic field.
Statement B :
An emf can be induced by changing the magnetic field.

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

358476 Assertion :
When a conducting rod moves in uniform transverse magnetic field with uniform speed which is perpendicular to its length, then potential difference may be developed across its ends.
Reason :
In any conductor, free electrons are available.

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.

PHXII06:ELECTROMAGNETIC INDUCTION

358477 A rectangular loop has a sliding connector $P Q$ of length $l$ and resistance $R$ ohm and it is moving with a speed $v$ as shown. The set up is placed in a uniform magnetic field going into the the plane of the paper.
supporting img

1

I_{1} = - I_{2} = \frac{B l v}{R}, I = \frac{2 B l v}{R}

2

I_{1} = I_{2} = \frac{B l v}{3 R}, I = \frac{2 B l v}{3 R}

3

I_{1} = I_{2} = I = \frac{B l v}{R}

4

I_{1} = I_{2} = \frac{B l v}{6 R}, I = \frac{B l v}{3 R}

Explanation:

A moving conductor is equivalent to a battery of emf $v Bl$ (motional emf)
$I = I_{1} + I_{2}$
supporting img

Applying Kirchhoff's law, we get
$\begin{aligned} (i) I_{1} R + I R - v B l = 0 \\ (ii) I_{2} R + I R - v B l = 0 \end{aligned}$
Adding (i) and (ii), we get
$\begin{aligned} 2 I R + I R = 2 v B l \\ I = \frac{2 v B l}{3 R} \Rightarrow I_{1} = I_{2} = \frac{v B l}{3 R} \end{aligned}$

PHXII06:ELECTROMAGNETIC INDUCTION

358478 A metallic rod of length $1 m$ held along east west direction is allowed to fall down freely. Given horizontal component of earth's magnetic field $B_{H} = 3 \times 10^{- 5} T$ . The $e m f$ induced in the rod at an instant $t = 2 s$ after it is released is (Take $g = 10 m s^{- 2}$ ).

1

3 \times 10^{- 4} V

2

6 \times 10^{- 3} V

3

6 \times 10^{- 4} V

4

3 \times 10^{- 3} V

Explanation:

$B_{H} = 3 \times 10^{- 5} T$
$u = 0$
$\Rightarrow u = g t = 10 \times 2 = 20 m / \sec$
$\Rightarrow V_{induced} = B_{H} l v$
$= 3 \times 10^{- 5} \times 1 \times 20 = 6 \times 10^{- 4} V o l t s .$
Correct option is (3).

KECT - 2023

PHXII06:ELECTROMAGNETIC INDUCTION

358474 An aeroplane having a wing space of $35 m$ flies due north with the speed of $90 m s^{- 1}$ , given $B = 4 \times 10^{- 5} T$ . The potential difference between the tips of the wings will be:

1

0.013 V

2

1.26 V

3

12.6 V

4

0.126 V

Explanation:

The induced emf is given by
$= B v l = 4 \times 10^{- 5} \times 90 \times 35 = 0.126 V$

PHXII06:ELECTROMAGNETIC INDUCTION

358475 Statement A :
An emf can be induced by moving a conductor into a magnetic field.
Statement B :
An emf can be induced by changing the magnetic field.

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

358476 Assertion :
When a conducting rod moves in uniform transverse magnetic field with uniform speed which is perpendicular to its length, then potential difference may be developed across its ends.
Reason :
In any conductor, free electrons are available.

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.

PHXII06:ELECTROMAGNETIC INDUCTION

358477 A rectangular loop has a sliding connector $P Q$ of length $l$ and resistance $R$ ohm and it is moving with a speed $v$ as shown. The set up is placed in a uniform magnetic field going into the the plane of the paper.
supporting img

1

I_{1} = - I_{2} = \frac{B l v}{R}, I = \frac{2 B l v}{R}

2

I_{1} = I_{2} = \frac{B l v}{3 R}, I = \frac{2 B l v}{3 R}

3

I_{1} = I_{2} = I = \frac{B l v}{R}

4

I_{1} = I_{2} = \frac{B l v}{6 R}, I = \frac{B l v}{3 R}

Explanation:

A moving conductor is equivalent to a battery of emf $v Bl$ (motional emf)
$I = I_{1} + I_{2}$
supporting img

Applying Kirchhoff's law, we get
$\begin{aligned} (i) I_{1} R + I R - v B l = 0 \\ (ii) I_{2} R + I R - v B l = 0 \end{aligned}$
Adding (i) and (ii), we get
$\begin{aligned} 2 I R + I R = 2 v B l \\ I = \frac{2 v B l}{3 R} \Rightarrow I_{1} = I_{2} = \frac{v B l}{3 R} \end{aligned}$

PHXII06:ELECTROMAGNETIC INDUCTION

358478 A metallic rod of length $1 m$ held along east west direction is allowed to fall down freely. Given horizontal component of earth's magnetic field $B_{H} = 3 \times 10^{- 5} T$ . The $e m f$ induced in the rod at an instant $t = 2 s$ after it is released is (Take $g = 10 m s^{- 2}$ ).

1

3 \times 10^{- 4} V

2

6 \times 10^{- 3} V

3

6 \times 10^{- 4} V

4

3 \times 10^{- 3} V

Explanation:

$B_{H} = 3 \times 10^{- 5} T$
$u = 0$
$\Rightarrow u = g t = 10 \times 2 = 20 m / \sec$
$\Rightarrow V_{induced} = B_{H} l v$
$= 3 \times 10^{- 5} \times 1 \times 20 = 6 \times 10^{- 4} V o l t s .$
Correct option is (3).

KECT - 2023

PHXII06:ELECTROMAGNETIC INDUCTION

358474 An aeroplane having a wing space of $35 m$ flies due north with the speed of $90 m s^{- 1}$ , given $B = 4 \times 10^{- 5} T$ . The potential difference between the tips of the wings will be:

1

0.013 V

2

1.26 V

3

12.6 V

4

0.126 V

Explanation:

The induced emf is given by
$= B v l = 4 \times 10^{- 5} \times 90 \times 35 = 0.126 V$

PHXII06:ELECTROMAGNETIC INDUCTION

358475 Statement A :
An emf can be induced by moving a conductor into a magnetic field.
Statement B :
An emf can be induced by changing the magnetic field.

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

358476 Assertion :
When a conducting rod moves in uniform transverse magnetic field with uniform speed which is perpendicular to its length, then potential difference may be developed across its ends.
Reason :
In any conductor, free electrons are available.

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.

PHXII06:ELECTROMAGNETIC INDUCTION

358477 A rectangular loop has a sliding connector $P Q$ of length $l$ and resistance $R$ ohm and it is moving with a speed $v$ as shown. The set up is placed in a uniform magnetic field going into the the plane of the paper.
supporting img

1

I_{1} = - I_{2} = \frac{B l v}{R}, I = \frac{2 B l v}{R}

2

I_{1} = I_{2} = \frac{B l v}{3 R}, I = \frac{2 B l v}{3 R}

3

I_{1} = I_{2} = I = \frac{B l v}{R}

4

I_{1} = I_{2} = \frac{B l v}{6 R}, I = \frac{B l v}{3 R}

Explanation:

A moving conductor is equivalent to a battery of emf $v Bl$ (motional emf)
$I = I_{1} + I_{2}$
supporting img

Applying Kirchhoff's law, we get
$\begin{aligned} (i) I_{1} R + I R - v B l = 0 \\ (ii) I_{2} R + I R - v B l = 0 \end{aligned}$
Adding (i) and (ii), we get
$\begin{aligned} 2 I R + I R = 2 v B l \\ I = \frac{2 v B l}{3 R} \Rightarrow I_{1} = I_{2} = \frac{v B l}{3 R} \end{aligned}$

PHXII06:ELECTROMAGNETIC INDUCTION

358478 A metallic rod of length $1 m$ held along east west direction is allowed to fall down freely. Given horizontal component of earth's magnetic field $B_{H} = 3 \times 10^{- 5} T$ . The $e m f$ induced in the rod at an instant $t = 2 s$ after it is released is (Take $g = 10 m s^{- 2}$ ).

1

3 \times 10^{- 4} V

2

6 \times 10^{- 3} V

3

6 \times 10^{- 4} V

4

3 \times 10^{- 3} V

Explanation:

$B_{H} = 3 \times 10^{- 5} T$
$u = 0$
$\Rightarrow u = g t = 10 \times 2 = 20 m / \sec$
$\Rightarrow V_{induced} = B_{H} l v$
$= 3 \times 10^{- 5} \times 1 \times 20 = 6 \times 10^{- 4} V o l t s .$
Correct option is (3).

KECT - 2023

PHXII06:ELECTROMAGNETIC INDUCTION

358474 An aeroplane having a wing space of $35 m$ flies due north with the speed of $90 m s^{- 1}$ , given $B = 4 \times 10^{- 5} T$ . The potential difference between the tips of the wings will be:

1

0.013 V

2

1.26 V

3

12.6 V

4

0.126 V

Explanation:

The induced emf is given by
$= B v l = 4 \times 10^{- 5} \times 90 \times 35 = 0.126 V$

PHXII06:ELECTROMAGNETIC INDUCTION

358475 Statement A :
An emf can be induced by moving a conductor into a magnetic field.
Statement B :
An emf can be induced by changing the magnetic field.

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

358476 Assertion :
When a conducting rod moves in uniform transverse magnetic field with uniform speed which is perpendicular to its length, then potential difference may be developed across its ends.
Reason :
In any conductor, free electrons are available.

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.

PHXII06:ELECTROMAGNETIC INDUCTION

358477 A rectangular loop has a sliding connector $P Q$ of length $l$ and resistance $R$ ohm and it is moving with a speed $v$ as shown. The set up is placed in a uniform magnetic field going into the the plane of the paper.
supporting img

1

I_{1} = - I_{2} = \frac{B l v}{R}, I = \frac{2 B l v}{R}

2

I_{1} = I_{2} = \frac{B l v}{3 R}, I = \frac{2 B l v}{3 R}

3

I_{1} = I_{2} = I = \frac{B l v}{R}

4

I_{1} = I_{2} = \frac{B l v}{6 R}, I = \frac{B l v}{3 R}

Explanation:

A moving conductor is equivalent to a battery of emf $v Bl$ (motional emf)
$I = I_{1} + I_{2}$
supporting img

Applying Kirchhoff's law, we get
$\begin{aligned} (i) I_{1} R + I R - v B l = 0 \\ (ii) I_{2} R + I R - v B l = 0 \end{aligned}$
Adding (i) and (ii), we get
$\begin{aligned} 2 I R + I R = 2 v B l \\ I = \frac{2 v B l}{3 R} \Rightarrow I_{1} = I_{2} = \frac{v B l}{3 R} \end{aligned}$

PHXII06:ELECTROMAGNETIC INDUCTION

358478 A metallic rod of length $1 m$ held along east west direction is allowed to fall down freely. Given horizontal component of earth's magnetic field $B_{H} = 3 \times 10^{- 5} T$ . The $e m f$ induced in the rod at an instant $t = 2 s$ after it is released is (Take $g = 10 m s^{- 2}$ ).

1

3 \times 10^{- 4} V

2

6 \times 10^{- 3} V

3

6 \times 10^{- 4} V

4

3 \times 10^{- 3} V

Explanation:

$B_{H} = 3 \times 10^{- 5} T$
$u = 0$
$\Rightarrow u = g t = 10 \times 2 = 20 m / \sec$
$\Rightarrow V_{induced} = B_{H} l v$
$= 3 \times 10^{- 5} \times 1 \times 20 = 6 \times 10^{- 4} V o l t s .$
Correct option is (3).

KECT - 2023

PHXII06:ELECTROMAGNETIC INDUCTION

358474 An aeroplane having a wing space of $35 m$ flies due north with the speed of $90 m s^{- 1}$ , given $B = 4 \times 10^{- 5} T$ . The potential difference between the tips of the wings will be:

1

0.013 V

2

1.26 V

3

12.6 V

4

0.126 V

Explanation:

The induced emf is given by
$= B v l = 4 \times 10^{- 5} \times 90 \times 35 = 0.126 V$

PHXII06:ELECTROMAGNETIC INDUCTION

358475 Statement A :
An emf can be induced by moving a conductor into a magnetic field.
Statement B :
An emf can be induced by changing the magnetic field.

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

358476 Assertion :
When a conducting rod moves in uniform transverse magnetic field with uniform speed which is perpendicular to its length, then potential difference may be developed across its ends.
Reason :
In any conductor, free electrons are available.

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.

PHXII06:ELECTROMAGNETIC INDUCTION

358477 A rectangular loop has a sliding connector $P Q$ of length $l$ and resistance $R$ ohm and it is moving with a speed $v$ as shown. The set up is placed in a uniform magnetic field going into the the plane of the paper.
supporting img

1

I_{1} = - I_{2} = \frac{B l v}{R}, I = \frac{2 B l v}{R}

2

I_{1} = I_{2} = \frac{B l v}{3 R}, I = \frac{2 B l v}{3 R}

3

I_{1} = I_{2} = I = \frac{B l v}{R}

4

I_{1} = I_{2} = \frac{B l v}{6 R}, I = \frac{B l v}{3 R}

Explanation:

A moving conductor is equivalent to a battery of emf $v Bl$ (motional emf)
$I = I_{1} + I_{2}$
supporting img

Applying Kirchhoff's law, we get
$\begin{aligned} (i) I_{1} R + I R - v B l = 0 \\ (ii) I_{2} R + I R - v B l = 0 \end{aligned}$
Adding (i) and (ii), we get
$\begin{aligned} 2 I R + I R = 2 v B l \\ I = \frac{2 v B l}{3 R} \Rightarrow I_{1} = I_{2} = \frac{v B l}{3 R} \end{aligned}$

PHXII06:ELECTROMAGNETIC INDUCTION

358478 A metallic rod of length $1 m$ held along east west direction is allowed to fall down freely. Given horizontal component of earth's magnetic field $B_{H} = 3 \times 10^{- 5} T$ . The $e m f$ induced in the rod at an instant $t = 2 s$ after it is released is (Take $g = 10 m s^{- 2}$ ).

1

3 \times 10^{- 4} V

2

6 \times 10^{- 3} V

3

6 \times 10^{- 4} V

4

3 \times 10^{- 3} V

Explanation:

$B_{H} = 3 \times 10^{- 5} T$
$u = 0$
$\Rightarrow u = g t = 10 \times 2 = 20 m / \sec$
$\Rightarrow V_{induced} = B_{H} l v$
$= 3 \times 10^{- 5} \times 1 \times 20 = 6 \times 10^{- 4} V o l t s .$
Correct option is (3).

KECT - 2023

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