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PHXII06:ELECTROMAGNETIC INDUCTION

358491 A square loop of side $15 c m$ being moved towards right at a constant speed of $2 c m / s$ as shown in figure. The front edge enters the $50 c m$ wide magnetic field at $t = 0$ . The value of induced emf in the loop at $t = 10 s$ will be
supporting img

1

0.3 m V

2

4.5 m V

3

3 m V

4 zero

Explanation:

Speed $= 2 c m / s$
Đistance $= 50 c m$
Time taken to cover magnetic field,
$t = \frac{50}{2} = 25 s$
So, the loop is completely in magnetic field, no flux change and hence no induced emf.
As per Faraday's law, $ε = \frac{- d ϕ}{d t}$
$∴ ε = 0$ . Correct option is (4).

JEE - 2024

PHXII06:ELECTROMAGNETIC INDUCTION

358492 A conducting rod $P Q$ of length $L = 1.0 m$ is moving with a uniform speed $v = 2.0 m s^{- 1}$ in a uniform magnetic field $B = 4.0 T$ directed into the paper. A capacitor of capacity $C = 10 μ F$ is connected as shown in figure. Then
supporting img

1

q_{A} = 0 = q_{B}

2

q_{A} = + 80 μ C

and

q_{B} = - 80 μ C

3

q_{A} = - 80 μ C

and

q_{B} = + 80 μ C

4 Charge stored in the capacitor increase exponentially with time

Explanation:

Motional emf across $P Q$
$V = B l v = 4 (1) (2) = 8 v o l t$
This is the potential to which the capacitor is charged.
As
$q = C V$
$∴ q = (10 \times 10^{- 6}) 8 = 8 \times 10^{- 5} C = 80 μ C$
As magnetic force on electron in the conducting $rod$
$P Q$ is towards $Q$ , therefore, $A$ is positively charged and $B$ is negatively charged
$i . e ., q_{A} = + 80 μ C & q_{B} = - 80 μ C$

PHXII06:ELECTROMAGNETIC INDUCTION

358493 In figure, there are two sliders and they can slide on two frictionless parallel wires in uniform magnetic field $B$ , which is present everywhere. The mass of each slider is $m$ , resistance $R$ and initially these are at rest. Now, if one slider is given a velocity $v_{0} = 16 m s^{- 1}$ , what will be the velocity of other slider after long time? (neglect the self-induction)
supporting img

1

4 m s^{- 1}

2

10 m s^{- 1}

3

8 m s^{- 1}

4

5 m s^{- 1}

Explanation:

After long time, from conservation of momentum.
$m v_{0} = 2 m v; v = \frac{v_{0}}{2} = 8 m s^{- 1}$

PHXII06:ELECTROMAGNETIC INDUCTION

358494 In the circuit shown in figure, a conducting wire $H E$ is moved with a constant speed $v$ towards left. The complete circuit is placed in a uniform magnetic field $B$ perpendicular to the plane of circuit. The current in HKDE is
supporting img

1 Alternating

2 Anti-clockwise

3 Clockwise

4 Zero

Explanation:

Potential difference across the capacitor $=$ emf induced across $H E = B l v$ which is constant. Therefore, charge stored in the capacitor is constant. Hence current in the circuit $H K D E$ is zero.

PHXII06:ELECTROMAGNETIC INDUCTION

358495 A wire of length $1 m$ moving with velocity $8 m / s$ at right angles to a magnetic field of $2 T$ . The magnitude of induced emf, between the ends of wire will be

1

12 V

2

8 V

3

16 V

4

20 V

Explanation:

Here, $l = 1 m, v = 8 m / s, θ = 90^{\circ}, B = 2 T$ Induced $e m f,$
$e = (\vec{V} \times \vec{B}) \cdot \vec{l} = B v l \sin θ$
$= 2 \times 8 \times 1 \times 1 = 16 V$

JEE - 2023

PHXII06:ELECTROMAGNETIC INDUCTION

358491 A square loop of side $15 c m$ being moved towards right at a constant speed of $2 c m / s$ as shown in figure. The front edge enters the $50 c m$ wide magnetic field at $t = 0$ . The value of induced emf in the loop at $t = 10 s$ will be
supporting img

1

0.3 m V

2

4.5 m V

3

3 m V

4 zero

Explanation:

Speed $= 2 c m / s$
Đistance $= 50 c m$
Time taken to cover magnetic field,
$t = \frac{50}{2} = 25 s$
So, the loop is completely in magnetic field, no flux change and hence no induced emf.
As per Faraday's law, $ε = \frac{- d ϕ}{d t}$
$∴ ε = 0$ . Correct option is (4).

JEE - 2024

PHXII06:ELECTROMAGNETIC INDUCTION

358492 A conducting rod $P Q$ of length $L = 1.0 m$ is moving with a uniform speed $v = 2.0 m s^{- 1}$ in a uniform magnetic field $B = 4.0 T$ directed into the paper. A capacitor of capacity $C = 10 μ F$ is connected as shown in figure. Then
supporting img

1

q_{A} = 0 = q_{B}

2

q_{A} = + 80 μ C

and

q_{B} = - 80 μ C

3

q_{A} = - 80 μ C

and

q_{B} = + 80 μ C

4 Charge stored in the capacitor increase exponentially with time

Explanation:

Motional emf across $P Q$
$V = B l v = 4 (1) (2) = 8 v o l t$
This is the potential to which the capacitor is charged.
As
$q = C V$
$∴ q = (10 \times 10^{- 6}) 8 = 8 \times 10^{- 5} C = 80 μ C$
As magnetic force on electron in the conducting $rod$
$P Q$ is towards $Q$ , therefore, $A$ is positively charged and $B$ is negatively charged
$i . e ., q_{A} = + 80 μ C & q_{B} = - 80 μ C$

PHXII06:ELECTROMAGNETIC INDUCTION

358493 In figure, there are two sliders and they can slide on two frictionless parallel wires in uniform magnetic field $B$ , which is present everywhere. The mass of each slider is $m$ , resistance $R$ and initially these are at rest. Now, if one slider is given a velocity $v_{0} = 16 m s^{- 1}$ , what will be the velocity of other slider after long time? (neglect the self-induction)
supporting img

1

4 m s^{- 1}

2

10 m s^{- 1}

3

8 m s^{- 1}

4

5 m s^{- 1}

Explanation:

After long time, from conservation of momentum.
$m v_{0} = 2 m v; v = \frac{v_{0}}{2} = 8 m s^{- 1}$

PHXII06:ELECTROMAGNETIC INDUCTION

358494 In the circuit shown in figure, a conducting wire $H E$ is moved with a constant speed $v$ towards left. The complete circuit is placed in a uniform magnetic field $B$ perpendicular to the plane of circuit. The current in HKDE is
supporting img

1 Alternating

2 Anti-clockwise

3 Clockwise

4 Zero

Explanation:

Potential difference across the capacitor $=$ emf induced across $H E = B l v$ which is constant. Therefore, charge stored in the capacitor is constant. Hence current in the circuit $H K D E$ is zero.

PHXII06:ELECTROMAGNETIC INDUCTION

358495 A wire of length $1 m$ moving with velocity $8 m / s$ at right angles to a magnetic field of $2 T$ . The magnitude of induced emf, between the ends of wire will be

1

12 V

2

8 V

3

16 V

4

20 V

Explanation:

Here, $l = 1 m, v = 8 m / s, θ = 90^{\circ}, B = 2 T$ Induced $e m f,$
$e = (\vec{V} \times \vec{B}) \cdot \vec{l} = B v l \sin θ$
$= 2 \times 8 \times 1 \times 1 = 16 V$

JEE - 2023

PHXII06:ELECTROMAGNETIC INDUCTION

358491 A square loop of side $15 c m$ being moved towards right at a constant speed of $2 c m / s$ as shown in figure. The front edge enters the $50 c m$ wide magnetic field at $t = 0$ . The value of induced emf in the loop at $t = 10 s$ will be
supporting img

1

0.3 m V

2

4.5 m V

3

3 m V

4 zero

Explanation:

Speed $= 2 c m / s$
Đistance $= 50 c m$
Time taken to cover magnetic field,
$t = \frac{50}{2} = 25 s$
So, the loop is completely in magnetic field, no flux change and hence no induced emf.
As per Faraday's law, $ε = \frac{- d ϕ}{d t}$
$∴ ε = 0$ . Correct option is (4).

JEE - 2024

PHXII06:ELECTROMAGNETIC INDUCTION

358492 A conducting rod $P Q$ of length $L = 1.0 m$ is moving with a uniform speed $v = 2.0 m s^{- 1}$ in a uniform magnetic field $B = 4.0 T$ directed into the paper. A capacitor of capacity $C = 10 μ F$ is connected as shown in figure. Then
supporting img

1

q_{A} = 0 = q_{B}

2

q_{A} = + 80 μ C

and

q_{B} = - 80 μ C

3

q_{A} = - 80 μ C

and

q_{B} = + 80 μ C

4 Charge stored in the capacitor increase exponentially with time

Explanation:

Motional emf across $P Q$
$V = B l v = 4 (1) (2) = 8 v o l t$
This is the potential to which the capacitor is charged.
As
$q = C V$
$∴ q = (10 \times 10^{- 6}) 8 = 8 \times 10^{- 5} C = 80 μ C$
As magnetic force on electron in the conducting $rod$
$P Q$ is towards $Q$ , therefore, $A$ is positively charged and $B$ is negatively charged
$i . e ., q_{A} = + 80 μ C & q_{B} = - 80 μ C$

PHXII06:ELECTROMAGNETIC INDUCTION

358493 In figure, there are two sliders and they can slide on two frictionless parallel wires in uniform magnetic field $B$ , which is present everywhere. The mass of each slider is $m$ , resistance $R$ and initially these are at rest. Now, if one slider is given a velocity $v_{0} = 16 m s^{- 1}$ , what will be the velocity of other slider after long time? (neglect the self-induction)
supporting img

1

4 m s^{- 1}

2

10 m s^{- 1}

3

8 m s^{- 1}

4

5 m s^{- 1}

Explanation:

After long time, from conservation of momentum.
$m v_{0} = 2 m v; v = \frac{v_{0}}{2} = 8 m s^{- 1}$

PHXII06:ELECTROMAGNETIC INDUCTION

358494 In the circuit shown in figure, a conducting wire $H E$ is moved with a constant speed $v$ towards left. The complete circuit is placed in a uniform magnetic field $B$ perpendicular to the plane of circuit. The current in HKDE is
supporting img

1 Alternating

2 Anti-clockwise

3 Clockwise

4 Zero

Explanation:

Potential difference across the capacitor $=$ emf induced across $H E = B l v$ which is constant. Therefore, charge stored in the capacitor is constant. Hence current in the circuit $H K D E$ is zero.

PHXII06:ELECTROMAGNETIC INDUCTION

358495 A wire of length $1 m$ moving with velocity $8 m / s$ at right angles to a magnetic field of $2 T$ . The magnitude of induced emf, between the ends of wire will be

1

12 V

2

8 V

3

16 V

4

20 V

Explanation:

Here, $l = 1 m, v = 8 m / s, θ = 90^{\circ}, B = 2 T$ Induced $e m f,$
$e = (\vec{V} \times \vec{B}) \cdot \vec{l} = B v l \sin θ$
$= 2 \times 8 \times 1 \times 1 = 16 V$

JEE - 2023

PHXII06:ELECTROMAGNETIC INDUCTION

358491 A square loop of side $15 c m$ being moved towards right at a constant speed of $2 c m / s$ as shown in figure. The front edge enters the $50 c m$ wide magnetic field at $t = 0$ . The value of induced emf in the loop at $t = 10 s$ will be
supporting img

1

0.3 m V

2

4.5 m V

3

3 m V

4 zero

Explanation:

Speed $= 2 c m / s$
Đistance $= 50 c m$
Time taken to cover magnetic field,
$t = \frac{50}{2} = 25 s$
So, the loop is completely in magnetic field, no flux change and hence no induced emf.
As per Faraday's law, $ε = \frac{- d ϕ}{d t}$
$∴ ε = 0$ . Correct option is (4).

JEE - 2024

PHXII06:ELECTROMAGNETIC INDUCTION

358492 A conducting rod $P Q$ of length $L = 1.0 m$ is moving with a uniform speed $v = 2.0 m s^{- 1}$ in a uniform magnetic field $B = 4.0 T$ directed into the paper. A capacitor of capacity $C = 10 μ F$ is connected as shown in figure. Then
supporting img

1

q_{A} = 0 = q_{B}

2

q_{A} = + 80 μ C

and

q_{B} = - 80 μ C

3

q_{A} = - 80 μ C

and

q_{B} = + 80 μ C

4 Charge stored in the capacitor increase exponentially with time

Explanation:

Motional emf across $P Q$
$V = B l v = 4 (1) (2) = 8 v o l t$
This is the potential to which the capacitor is charged.
As
$q = C V$
$∴ q = (10 \times 10^{- 6}) 8 = 8 \times 10^{- 5} C = 80 μ C$
As magnetic force on electron in the conducting $rod$
$P Q$ is towards $Q$ , therefore, $A$ is positively charged and $B$ is negatively charged
$i . e ., q_{A} = + 80 μ C & q_{B} = - 80 μ C$

PHXII06:ELECTROMAGNETIC INDUCTION

358493 In figure, there are two sliders and they can slide on two frictionless parallel wires in uniform magnetic field $B$ , which is present everywhere. The mass of each slider is $m$ , resistance $R$ and initially these are at rest. Now, if one slider is given a velocity $v_{0} = 16 m s^{- 1}$ , what will be the velocity of other slider after long time? (neglect the self-induction)
supporting img

1

4 m s^{- 1}

2

10 m s^{- 1}

3

8 m s^{- 1}

4

5 m s^{- 1}

Explanation:

After long time, from conservation of momentum.
$m v_{0} = 2 m v; v = \frac{v_{0}}{2} = 8 m s^{- 1}$

PHXII06:ELECTROMAGNETIC INDUCTION

358494 In the circuit shown in figure, a conducting wire $H E$ is moved with a constant speed $v$ towards left. The complete circuit is placed in a uniform magnetic field $B$ perpendicular to the plane of circuit. The current in HKDE is
supporting img

1 Alternating

2 Anti-clockwise

3 Clockwise

4 Zero

Explanation:

Potential difference across the capacitor $=$ emf induced across $H E = B l v$ which is constant. Therefore, charge stored in the capacitor is constant. Hence current in the circuit $H K D E$ is zero.

PHXII06:ELECTROMAGNETIC INDUCTION

358495 A wire of length $1 m$ moving with velocity $8 m / s$ at right angles to a magnetic field of $2 T$ . The magnitude of induced emf, between the ends of wire will be

1

12 V

2

8 V

3

16 V

4

20 V

Explanation:

Here, $l = 1 m, v = 8 m / s, θ = 90^{\circ}, B = 2 T$ Induced $e m f,$
$e = (\vec{V} \times \vec{B}) \cdot \vec{l} = B v l \sin θ$
$= 2 \times 8 \times 1 \times 1 = 16 V$

JEE - 2023

PHXII06:ELECTROMAGNETIC INDUCTION

358491 A square loop of side $15 c m$ being moved towards right at a constant speed of $2 c m / s$ as shown in figure. The front edge enters the $50 c m$ wide magnetic field at $t = 0$ . The value of induced emf in the loop at $t = 10 s$ will be
supporting img

1

0.3 m V

2

4.5 m V

3

3 m V

4 zero

Explanation:

Speed $= 2 c m / s$
Đistance $= 50 c m$
Time taken to cover magnetic field,
$t = \frac{50}{2} = 25 s$
So, the loop is completely in magnetic field, no flux change and hence no induced emf.
As per Faraday's law, $ε = \frac{- d ϕ}{d t}$
$∴ ε = 0$ . Correct option is (4).

JEE - 2024

PHXII06:ELECTROMAGNETIC INDUCTION

358492 A conducting rod $P Q$ of length $L = 1.0 m$ is moving with a uniform speed $v = 2.0 m s^{- 1}$ in a uniform magnetic field $B = 4.0 T$ directed into the paper. A capacitor of capacity $C = 10 μ F$ is connected as shown in figure. Then
supporting img

1

q_{A} = 0 = q_{B}

2

q_{A} = + 80 μ C

and

q_{B} = - 80 μ C

3

q_{A} = - 80 μ C

and

q_{B} = + 80 μ C

4 Charge stored in the capacitor increase exponentially with time

Explanation:

Motional emf across $P Q$
$V = B l v = 4 (1) (2) = 8 v o l t$
This is the potential to which the capacitor is charged.
As
$q = C V$
$∴ q = (10 \times 10^{- 6}) 8 = 8 \times 10^{- 5} C = 80 μ C$
As magnetic force on electron in the conducting $rod$
$P Q$ is towards $Q$ , therefore, $A$ is positively charged and $B$ is negatively charged
$i . e ., q_{A} = + 80 μ C & q_{B} = - 80 μ C$

PHXII06:ELECTROMAGNETIC INDUCTION

358493 In figure, there are two sliders and they can slide on two frictionless parallel wires in uniform magnetic field $B$ , which is present everywhere. The mass of each slider is $m$ , resistance $R$ and initially these are at rest. Now, if one slider is given a velocity $v_{0} = 16 m s^{- 1}$ , what will be the velocity of other slider after long time? (neglect the self-induction)
supporting img

1

4 m s^{- 1}

2

10 m s^{- 1}

3

8 m s^{- 1}

4

5 m s^{- 1}

Explanation:

After long time, from conservation of momentum.
$m v_{0} = 2 m v; v = \frac{v_{0}}{2} = 8 m s^{- 1}$

PHXII06:ELECTROMAGNETIC INDUCTION

358494 In the circuit shown in figure, a conducting wire $H E$ is moved with a constant speed $v$ towards left. The complete circuit is placed in a uniform magnetic field $B$ perpendicular to the plane of circuit. The current in HKDE is
supporting img

1 Alternating

2 Anti-clockwise

3 Clockwise

4 Zero

Explanation:

Potential difference across the capacitor $=$ emf induced across $H E = B l v$ which is constant. Therefore, charge stored in the capacitor is constant. Hence current in the circuit $H K D E$ is zero.

PHXII06:ELECTROMAGNETIC INDUCTION

358495 A wire of length $1 m$ moving with velocity $8 m / s$ at right angles to a magnetic field of $2 T$ . The magnitude of induced emf, between the ends of wire will be

1

12 V

2

8 V

3

16 V

4

20 V

Explanation:

Here, $l = 1 m, v = 8 m / s, θ = 90^{\circ}, B = 2 T$ Induced $e m f,$
$e = (\vec{V} \times \vec{B}) \cdot \vec{l} = B v l \sin θ$
$= 2 \times 8 \times 1 \times 1 = 16 V$

JEE - 2023

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