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Electro Magnetic Induction

154680 A wheel with radial metal spokes $1 m$ in length is rotated in a magnetic field of $0.5 \times 10^{- 4} T$ normal to the plane of the wheel. If the emf induced between the rim and axle is $π / 3000 V$ , then the rotational speed of the wheel in revolutions per minute is

1 400

2 500

3 600

4 700

Explanation:

A Given:-
Length, $R = 1 m$
Magnetic field, $B = 0.5 \times 10^{- 4} T$
$Emf induced, ε = \frac{ϕ}{3000} V$
We know that induced emf between rim and axle of wheel.
$ε = \frac{1}{2} B ω R^{2}$
$\frac{π}{3000} = \frac{1}{2} \times 0.5 \times 10^{- 4} \times ω \times (1)^{2}$
$ω = \frac{2 \times π}{3000 \times 0.5 \times 10^{- 4}}$
$ω = \frac{2 \times π \times 10000 \times 2}{3000}$
Rotation frequency of the wheel revolution per minute
$\frac{2 π N}{60} = \frac{2 π \times 10000 \times 2}{3000}$
$= \frac{10000 \times 2}{3000} \times 60$
$N = 400 rpm$

TS- EAMCET-05.05.2018

Electro Magnetic Induction

154681 A conducting bar of length $50 cm$ rotates with a constant angular speed of $100 rad / \sec$ about a pivot at one end. If a uniform magnetic field of $0.1 T$ is directed perpendicular to the plane of rotation the emf induced between the ends of the bar is

1

5 V

2

1.25 V

3

4.5 V

4

2.75 V

Explanation:

B Given,
Length of bar $(l) = 50 cm = 0.5 m$
Angular velocity $(ω) = 100 rad / \sec$
Magnetic field $(B) = 0.1 T$ .
Induced emfn $(ε) = \frac{1}{2} B ω l^{2}$
$ε = \frac{1}{2} \times 0.1 \times 100 \times (0.5)^{2}$
$ε = 1.25 V$

TS EAMCET 02.05.2018

Electro Magnetic Induction

154682 A straight conductor of length $2 m$ moves with a velocity of $2 m / s$ in a magnetic field of induction $0.5 Wb / m^{2}$ and perpendicular to it. The emf induced in the conductor is

1

2 V

2

1 V

3

0.5 V

4

3 V

Explanation:

A Given,
Length, $l = 2 m$
Velocity, $v = 2 m / s$ .
Magnetic field, $B = 0.5 Wb / m^{2}$
$emf = B \lor l = 0.5 \times 2 \times 2$
$emf = 2 V$

COMEDK 2018

Electro Magnetic Induction

154684 A generator with a circular coil of 100 turns of area $2 \times 10^{- 2} m^{2}$ is immersed in a $0.01 T$ magnetic field and rotated at a frequency of 50 Hz. The maximum emf which is produced during a cycle is

1

6.28 V

2

3.44 V

3

10 V

4

1.32 V

Explanation:

A Given number of turn (N) $= 100$ turn.
Area $(A) = 2 \times 10^{- 2} m^{2}$ .
Magnetic field $(B) = 0.01 T$ .
Frequency $(f) = 50 Hz$
Angular velocity $(ω) = 2 π f$ .
$ω = 2 π \times 50 = 100 π$
We know that,
Maximum emf $(ε) = NBA ω$
$ε = 100 \times 0.01 \times 2 \times 10^{- 2} \times 100 \times \frac{22}{7}$
$ε = 6.28 V$

TS EAMCET (Engg.)-2017

Electro Magnetic Induction

154680 A wheel with radial metal spokes $1 m$ in length is rotated in a magnetic field of $0.5 \times 10^{- 4} T$ normal to the plane of the wheel. If the emf induced between the rim and axle is $π / 3000 V$ , then the rotational speed of the wheel in revolutions per minute is

1 400

2 500

3 600

4 700

Explanation:

A Given:-
Length, $R = 1 m$
Magnetic field, $B = 0.5 \times 10^{- 4} T$
$Emf induced, ε = \frac{ϕ}{3000} V$
We know that induced emf between rim and axle of wheel.
$ε = \frac{1}{2} B ω R^{2}$
$\frac{π}{3000} = \frac{1}{2} \times 0.5 \times 10^{- 4} \times ω \times (1)^{2}$
$ω = \frac{2 \times π}{3000 \times 0.5 \times 10^{- 4}}$
$ω = \frac{2 \times π \times 10000 \times 2}{3000}$
Rotation frequency of the wheel revolution per minute
$\frac{2 π N}{60} = \frac{2 π \times 10000 \times 2}{3000}$
$= \frac{10000 \times 2}{3000} \times 60$
$N = 400 rpm$

TS- EAMCET-05.05.2018

Electro Magnetic Induction

154681 A conducting bar of length $50 cm$ rotates with a constant angular speed of $100 rad / \sec$ about a pivot at one end. If a uniform magnetic field of $0.1 T$ is directed perpendicular to the plane of rotation the emf induced between the ends of the bar is

1

5 V

2

1.25 V

3

4.5 V

4

2.75 V

Explanation:

B Given,
Length of bar $(l) = 50 cm = 0.5 m$
Angular velocity $(ω) = 100 rad / \sec$
Magnetic field $(B) = 0.1 T$ .
Induced emfn $(ε) = \frac{1}{2} B ω l^{2}$
$ε = \frac{1}{2} \times 0.1 \times 100 \times (0.5)^{2}$
$ε = 1.25 V$

TS EAMCET 02.05.2018

Electro Magnetic Induction

154682 A straight conductor of length $2 m$ moves with a velocity of $2 m / s$ in a magnetic field of induction $0.5 Wb / m^{2}$ and perpendicular to it. The emf induced in the conductor is

1

2 V

2

1 V

3

0.5 V

4

3 V

Explanation:

A Given,
Length, $l = 2 m$
Velocity, $v = 2 m / s$ .
Magnetic field, $B = 0.5 Wb / m^{2}$
$emf = B \lor l = 0.5 \times 2 \times 2$
$emf = 2 V$

COMEDK 2018

Electro Magnetic Induction

154684 A generator with a circular coil of 100 turns of area $2 \times 10^{- 2} m^{2}$ is immersed in a $0.01 T$ magnetic field and rotated at a frequency of 50 Hz. The maximum emf which is produced during a cycle is

1

6.28 V

2

3.44 V

3

10 V

4

1.32 V

Explanation:

A Given number of turn (N) $= 100$ turn.
Area $(A) = 2 \times 10^{- 2} m^{2}$ .
Magnetic field $(B) = 0.01 T$ .
Frequency $(f) = 50 Hz$
Angular velocity $(ω) = 2 π f$ .
$ω = 2 π \times 50 = 100 π$
We know that,
Maximum emf $(ε) = NBA ω$
$ε = 100 \times 0.01 \times 2 \times 10^{- 2} \times 100 \times \frac{22}{7}$
$ε = 6.28 V$

TS EAMCET (Engg.)-2017

Electro Magnetic Induction

154680 A wheel with radial metal spokes $1 m$ in length is rotated in a magnetic field of $0.5 \times 10^{- 4} T$ normal to the plane of the wheel. If the emf induced between the rim and axle is $π / 3000 V$ , then the rotational speed of the wheel in revolutions per minute is

1 400

2 500

3 600

4 700

Explanation:

A Given:-
Length, $R = 1 m$
Magnetic field, $B = 0.5 \times 10^{- 4} T$
$Emf induced, ε = \frac{ϕ}{3000} V$
We know that induced emf between rim and axle of wheel.
$ε = \frac{1}{2} B ω R^{2}$
$\frac{π}{3000} = \frac{1}{2} \times 0.5 \times 10^{- 4} \times ω \times (1)^{2}$
$ω = \frac{2 \times π}{3000 \times 0.5 \times 10^{- 4}}$
$ω = \frac{2 \times π \times 10000 \times 2}{3000}$
Rotation frequency of the wheel revolution per minute
$\frac{2 π N}{60} = \frac{2 π \times 10000 \times 2}{3000}$
$= \frac{10000 \times 2}{3000} \times 60$
$N = 400 rpm$

TS- EAMCET-05.05.2018

Electro Magnetic Induction

154681 A conducting bar of length $50 cm$ rotates with a constant angular speed of $100 rad / \sec$ about a pivot at one end. If a uniform magnetic field of $0.1 T$ is directed perpendicular to the plane of rotation the emf induced between the ends of the bar is

1

5 V

2

1.25 V

3

4.5 V

4

2.75 V

Explanation:

B Given,
Length of bar $(l) = 50 cm = 0.5 m$
Angular velocity $(ω) = 100 rad / \sec$
Magnetic field $(B) = 0.1 T$ .
Induced emfn $(ε) = \frac{1}{2} B ω l^{2}$
$ε = \frac{1}{2} \times 0.1 \times 100 \times (0.5)^{2}$
$ε = 1.25 V$

TS EAMCET 02.05.2018

Electro Magnetic Induction

154682 A straight conductor of length $2 m$ moves with a velocity of $2 m / s$ in a magnetic field of induction $0.5 Wb / m^{2}$ and perpendicular to it. The emf induced in the conductor is

1

2 V

2

1 V

3

0.5 V

4

3 V

Explanation:

A Given,
Length, $l = 2 m$
Velocity, $v = 2 m / s$ .
Magnetic field, $B = 0.5 Wb / m^{2}$
$emf = B \lor l = 0.5 \times 2 \times 2$
$emf = 2 V$

COMEDK 2018

Electro Magnetic Induction

154684 A generator with a circular coil of 100 turns of area $2 \times 10^{- 2} m^{2}$ is immersed in a $0.01 T$ magnetic field and rotated at a frequency of 50 Hz. The maximum emf which is produced during a cycle is

1

6.28 V

2

3.44 V

3

10 V

4

1.32 V

Explanation:

A Given number of turn (N) $= 100$ turn.
Area $(A) = 2 \times 10^{- 2} m^{2}$ .
Magnetic field $(B) = 0.01 T$ .
Frequency $(f) = 50 Hz$
Angular velocity $(ω) = 2 π f$ .
$ω = 2 π \times 50 = 100 π$
We know that,
Maximum emf $(ε) = NBA ω$
$ε = 100 \times 0.01 \times 2 \times 10^{- 2} \times 100 \times \frac{22}{7}$
$ε = 6.28 V$

TS EAMCET (Engg.)-2017

Electro Magnetic Induction

154680 A wheel with radial metal spokes $1 m$ in length is rotated in a magnetic field of $0.5 \times 10^{- 4} T$ normal to the plane of the wheel. If the emf induced between the rim and axle is $π / 3000 V$ , then the rotational speed of the wheel in revolutions per minute is

1 400

2 500

3 600

4 700

Explanation:

A Given:-
Length, $R = 1 m$
Magnetic field, $B = 0.5 \times 10^{- 4} T$
$Emf induced, ε = \frac{ϕ}{3000} V$
We know that induced emf between rim and axle of wheel.
$ε = \frac{1}{2} B ω R^{2}$
$\frac{π}{3000} = \frac{1}{2} \times 0.5 \times 10^{- 4} \times ω \times (1)^{2}$
$ω = \frac{2 \times π}{3000 \times 0.5 \times 10^{- 4}}$
$ω = \frac{2 \times π \times 10000 \times 2}{3000}$
Rotation frequency of the wheel revolution per minute
$\frac{2 π N}{60} = \frac{2 π \times 10000 \times 2}{3000}$
$= \frac{10000 \times 2}{3000} \times 60$
$N = 400 rpm$

TS- EAMCET-05.05.2018

Electro Magnetic Induction

154681 A conducting bar of length $50 cm$ rotates with a constant angular speed of $100 rad / \sec$ about a pivot at one end. If a uniform magnetic field of $0.1 T$ is directed perpendicular to the plane of rotation the emf induced between the ends of the bar is

1

5 V

2

1.25 V

3

4.5 V

4

2.75 V

Explanation:

B Given,
Length of bar $(l) = 50 cm = 0.5 m$
Angular velocity $(ω) = 100 rad / \sec$
Magnetic field $(B) = 0.1 T$ .
Induced emfn $(ε) = \frac{1}{2} B ω l^{2}$
$ε = \frac{1}{2} \times 0.1 \times 100 \times (0.5)^{2}$
$ε = 1.25 V$

TS EAMCET 02.05.2018

Electro Magnetic Induction

154682 A straight conductor of length $2 m$ moves with a velocity of $2 m / s$ in a magnetic field of induction $0.5 Wb / m^{2}$ and perpendicular to it. The emf induced in the conductor is

1

2 V

2

1 V

3

0.5 V

4

3 V

Explanation:

A Given,
Length, $l = 2 m$
Velocity, $v = 2 m / s$ .
Magnetic field, $B = 0.5 Wb / m^{2}$
$emf = B \lor l = 0.5 \times 2 \times 2$
$emf = 2 V$

COMEDK 2018

Electro Magnetic Induction

154684 A generator with a circular coil of 100 turns of area $2 \times 10^{- 2} m^{2}$ is immersed in a $0.01 T$ magnetic field and rotated at a frequency of 50 Hz. The maximum emf which is produced during a cycle is

1

6.28 V

2

3.44 V

3

10 V

4

1.32 V

Explanation:

A Given number of turn (N) $= 100$ turn.
Area $(A) = 2 \times 10^{- 2} m^{2}$ .
Magnetic field $(B) = 0.01 T$ .
Frequency $(f) = 50 Hz$
Angular velocity $(ω) = 2 π f$ .
$ω = 2 π \times 50 = 100 π$
We know that,
Maximum emf $(ε) = NBA ω$
$ε = 100 \times 0.01 \times 2 \times 10^{- 2} \times 100 \times \frac{22}{7}$
$ε = 6.28 V$

TS EAMCET (Engg.)-2017