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Electro Magnetic Induction

154695 Magnetic flux passing through a coil is initially $4 \times 10^{- 4} Wb$ . It reduces to $10 %$ of its original value in ' $t$ ' second. If the e.m.f. induced is 0.72 $m V$ then ' $t$ ' in second is

1 0.3

2 0.4

3 0.5

4 0.6

Explanation:

C $ϕ_{initial} = 4 \times 10^{- 4} Wb$ ,
$emf = 0.72 \times 10^{- 3} V$
$ϕ_{final} = 0.9 \times ϕ_{initial} = 0.9 \times 4 \times 10^{- 4} Wb$ .
$emf = \frac{| Δ ϕ |}{Δ t}$
$0.72 \times 10^{- 3} = \frac{0.36 \times 10^{- 3}}{Δ t}$
$Δ t = \frac{0.36}{0.72} = \frac{1}{2}$
$Δ t = 0.5 \sec$

MHT-CET 2016

Electro Magnetic Induction

154696 A helicopter rises vertically with a speed of 100 ${ms}^{- 1}$ . If helicopter has length $10 m$ and horizontal component of earth's magnetic field is $5 \times 10^{- 3} {Wbm}^{- 2}$ , then the induced emf between the tip of nose and tail of helicopter is

1

50 V

2

0.5 V

3

5 V

4

25 V

Explanation:

C Given:-
Velocity, $v = 100 m . / s$
Length, $l = 10 m$
Magnetic field, $B = 5 \times 10^{- 3} Wb / m^{2}$
$emf = B v l = 5 \times 10^{- 3} \times 100 \times 10$
$emf = 5 V$

MHT-CET 2004

Electro Magnetic Induction

154699 A solenoid has 2000 turns wound over a length of $0.30 m$ . The area of its cross-section is $1.2 \times 10^{- 3} m^{2}$ . Around its central section, a coil of 300 turn is wound. If an initial current of $2 A$ in the solenoid is reversed in $0.25 s$ , then the emf induced in the coil is

1

6 \times 10^{- 4} V

2

4.8 \times 10^{- 3} V

3

6 \times 10^{- 2} V

4

48 mV

Explanation:

D Given:-
$N_{1} = 2000 turns, l = 0.3 m$
$A = 1.2 \times 10^{- 3} m^{2}$
$N_{2} = 300 turns, I_{1} = 2 A, I_{2} = - 2 A$
$t = 0.25 \sec .$
$emf induced = M \cdot \frac{dI}{dt} = \frac{μ_{0} N_{1} N_{2} A}{l} \frac{dI}{dt}$
$emf = \frac{4 π \times 10^{- 7} \times 2000 \times 300 \times 1.2 \times 10^{- 3}}{0.3} \times \frac{[2 - (- 2)]}{0.25}$
$emf = 48 mV$

VITEEE-2011

Electro Magnetic Induction

154700 A rectangular coil of 300 turns has an average area of $25 cm \times 10 cm$ . The coil rotates with a speed of 50 cps in uniform magnetic field of strength $4 \times 10^{- 2} T$ about an axis perpendicular to the field. The peak value of the induced emf is (in volt) :

1

300 π

2

3000 π

3

3 π

4

30 π

Explanation:

D Given- N $= 300$ turns
Area, $A = 25 cm \times 10 cm$
Frequency, $f = 50 cps$
Magnetic field, $B = 4 \times 10^{- 2} T$
Angular velocity, $ω = 2 π f$
$= 2 π \times 50 = 100 π$
Peak value of induced emf $=$ NAB $ω$ $emf = 300 \times 25 \times 10 \times 10^{- 4} \times 4 \times 10^{- 2} \times 100 π$ emf $= 30 π$ volt

Karnataka CET-2004

Electro Magnetic Induction

154695 Magnetic flux passing through a coil is initially $4 \times 10^{- 4} Wb$ . It reduces to $10 %$ of its original value in ' $t$ ' second. If the e.m.f. induced is 0.72 $m V$ then ' $t$ ' in second is

1 0.3

2 0.4

3 0.5

4 0.6

Explanation:

C $ϕ_{initial} = 4 \times 10^{- 4} Wb$ ,
$emf = 0.72 \times 10^{- 3} V$
$ϕ_{final} = 0.9 \times ϕ_{initial} = 0.9 \times 4 \times 10^{- 4} Wb$ .
$emf = \frac{| Δ ϕ |}{Δ t}$
$0.72 \times 10^{- 3} = \frac{0.36 \times 10^{- 3}}{Δ t}$
$Δ t = \frac{0.36}{0.72} = \frac{1}{2}$
$Δ t = 0.5 \sec$

MHT-CET 2016

Electro Magnetic Induction

154696 A helicopter rises vertically with a speed of 100 ${ms}^{- 1}$ . If helicopter has length $10 m$ and horizontal component of earth's magnetic field is $5 \times 10^{- 3} {Wbm}^{- 2}$ , then the induced emf between the tip of nose and tail of helicopter is

1

50 V

2

0.5 V

3

5 V

4

25 V

Explanation:

C Given:-
Velocity, $v = 100 m . / s$
Length, $l = 10 m$
Magnetic field, $B = 5 \times 10^{- 3} Wb / m^{2}$
$emf = B v l = 5 \times 10^{- 3} \times 100 \times 10$
$emf = 5 V$

MHT-CET 2004

Electro Magnetic Induction

154699 A solenoid has 2000 turns wound over a length of $0.30 m$ . The area of its cross-section is $1.2 \times 10^{- 3} m^{2}$ . Around its central section, a coil of 300 turn is wound. If an initial current of $2 A$ in the solenoid is reversed in $0.25 s$ , then the emf induced in the coil is

1

6 \times 10^{- 4} V

2

4.8 \times 10^{- 3} V

3

6 \times 10^{- 2} V

4

48 mV

Explanation:

D Given:-
$N_{1} = 2000 turns, l = 0.3 m$
$A = 1.2 \times 10^{- 3} m^{2}$
$N_{2} = 300 turns, I_{1} = 2 A, I_{2} = - 2 A$
$t = 0.25 \sec .$
$emf induced = M \cdot \frac{dI}{dt} = \frac{μ_{0} N_{1} N_{2} A}{l} \frac{dI}{dt}$
$emf = \frac{4 π \times 10^{- 7} \times 2000 \times 300 \times 1.2 \times 10^{- 3}}{0.3} \times \frac{[2 - (- 2)]}{0.25}$
$emf = 48 mV$

VITEEE-2011

Electro Magnetic Induction

154700 A rectangular coil of 300 turns has an average area of $25 cm \times 10 cm$ . The coil rotates with a speed of 50 cps in uniform magnetic field of strength $4 \times 10^{- 2} T$ about an axis perpendicular to the field. The peak value of the induced emf is (in volt) :

1

300 π

2

3000 π

3

3 π

4

30 π

Explanation:

D Given- N $= 300$ turns
Area, $A = 25 cm \times 10 cm$
Frequency, $f = 50 cps$
Magnetic field, $B = 4 \times 10^{- 2} T$
Angular velocity, $ω = 2 π f$
$= 2 π \times 50 = 100 π$
Peak value of induced emf $=$ NAB $ω$ $emf = 300 \times 25 \times 10 \times 10^{- 4} \times 4 \times 10^{- 2} \times 100 π$ emf $= 30 π$ volt

Karnataka CET-2004

Electro Magnetic Induction

154695 Magnetic flux passing through a coil is initially $4 \times 10^{- 4} Wb$ . It reduces to $10 %$ of its original value in ' $t$ ' second. If the e.m.f. induced is 0.72 $m V$ then ' $t$ ' in second is

1 0.3

2 0.4

3 0.5

4 0.6

Explanation:

C $ϕ_{initial} = 4 \times 10^{- 4} Wb$ ,
$emf = 0.72 \times 10^{- 3} V$
$ϕ_{final} = 0.9 \times ϕ_{initial} = 0.9 \times 4 \times 10^{- 4} Wb$ .
$emf = \frac{| Δ ϕ |}{Δ t}$
$0.72 \times 10^{- 3} = \frac{0.36 \times 10^{- 3}}{Δ t}$
$Δ t = \frac{0.36}{0.72} = \frac{1}{2}$
$Δ t = 0.5 \sec$

MHT-CET 2016

Electro Magnetic Induction

154696 A helicopter rises vertically with a speed of 100 ${ms}^{- 1}$ . If helicopter has length $10 m$ and horizontal component of earth's magnetic field is $5 \times 10^{- 3} {Wbm}^{- 2}$ , then the induced emf between the tip of nose and tail of helicopter is

1

50 V

2

0.5 V

3

5 V

4

25 V

Explanation:

C Given:-
Velocity, $v = 100 m . / s$
Length, $l = 10 m$
Magnetic field, $B = 5 \times 10^{- 3} Wb / m^{2}$
$emf = B v l = 5 \times 10^{- 3} \times 100 \times 10$
$emf = 5 V$

MHT-CET 2004

Electro Magnetic Induction

154699 A solenoid has 2000 turns wound over a length of $0.30 m$ . The area of its cross-section is $1.2 \times 10^{- 3} m^{2}$ . Around its central section, a coil of 300 turn is wound. If an initial current of $2 A$ in the solenoid is reversed in $0.25 s$ , then the emf induced in the coil is

1

6 \times 10^{- 4} V

2

4.8 \times 10^{- 3} V

3

6 \times 10^{- 2} V

4

48 mV

Explanation:

D Given:-
$N_{1} = 2000 turns, l = 0.3 m$
$A = 1.2 \times 10^{- 3} m^{2}$
$N_{2} = 300 turns, I_{1} = 2 A, I_{2} = - 2 A$
$t = 0.25 \sec .$
$emf induced = M \cdot \frac{dI}{dt} = \frac{μ_{0} N_{1} N_{2} A}{l} \frac{dI}{dt}$
$emf = \frac{4 π \times 10^{- 7} \times 2000 \times 300 \times 1.2 \times 10^{- 3}}{0.3} \times \frac{[2 - (- 2)]}{0.25}$
$emf = 48 mV$

VITEEE-2011

Electro Magnetic Induction

154700 A rectangular coil of 300 turns has an average area of $25 cm \times 10 cm$ . The coil rotates with a speed of 50 cps in uniform magnetic field of strength $4 \times 10^{- 2} T$ about an axis perpendicular to the field. The peak value of the induced emf is (in volt) :

1

300 π

2

3000 π

3

3 π

4

30 π

Explanation:

D Given- N $= 300$ turns
Area, $A = 25 cm \times 10 cm$
Frequency, $f = 50 cps$
Magnetic field, $B = 4 \times 10^{- 2} T$
Angular velocity, $ω = 2 π f$
$= 2 π \times 50 = 100 π$
Peak value of induced emf $=$ NAB $ω$ $emf = 300 \times 25 \times 10 \times 10^{- 4} \times 4 \times 10^{- 2} \times 100 π$ emf $= 30 π$ volt

Karnataka CET-2004

Electro Magnetic Induction

154695 Magnetic flux passing through a coil is initially $4 \times 10^{- 4} Wb$ . It reduces to $10 %$ of its original value in ' $t$ ' second. If the e.m.f. induced is 0.72 $m V$ then ' $t$ ' in second is

1 0.3

2 0.4

3 0.5

4 0.6

Explanation:

C $ϕ_{initial} = 4 \times 10^{- 4} Wb$ ,
$emf = 0.72 \times 10^{- 3} V$
$ϕ_{final} = 0.9 \times ϕ_{initial} = 0.9 \times 4 \times 10^{- 4} Wb$ .
$emf = \frac{| Δ ϕ |}{Δ t}$
$0.72 \times 10^{- 3} = \frac{0.36 \times 10^{- 3}}{Δ t}$
$Δ t = \frac{0.36}{0.72} = \frac{1}{2}$
$Δ t = 0.5 \sec$

MHT-CET 2016

Electro Magnetic Induction

154696 A helicopter rises vertically with a speed of 100 ${ms}^{- 1}$ . If helicopter has length $10 m$ and horizontal component of earth's magnetic field is $5 \times 10^{- 3} {Wbm}^{- 2}$ , then the induced emf between the tip of nose and tail of helicopter is

1

50 V

2

0.5 V

3

5 V

4

25 V

Explanation:

C Given:-
Velocity, $v = 100 m . / s$
Length, $l = 10 m$
Magnetic field, $B = 5 \times 10^{- 3} Wb / m^{2}$
$emf = B v l = 5 \times 10^{- 3} \times 100 \times 10$
$emf = 5 V$

MHT-CET 2004

Electro Magnetic Induction

154699 A solenoid has 2000 turns wound over a length of $0.30 m$ . The area of its cross-section is $1.2 \times 10^{- 3} m^{2}$ . Around its central section, a coil of 300 turn is wound. If an initial current of $2 A$ in the solenoid is reversed in $0.25 s$ , then the emf induced in the coil is

1

6 \times 10^{- 4} V

2

4.8 \times 10^{- 3} V

3

6 \times 10^{- 2} V

4

48 mV

Explanation:

D Given:-
$N_{1} = 2000 turns, l = 0.3 m$
$A = 1.2 \times 10^{- 3} m^{2}$
$N_{2} = 300 turns, I_{1} = 2 A, I_{2} = - 2 A$
$t = 0.25 \sec .$
$emf induced = M \cdot \frac{dI}{dt} = \frac{μ_{0} N_{1} N_{2} A}{l} \frac{dI}{dt}$
$emf = \frac{4 π \times 10^{- 7} \times 2000 \times 300 \times 1.2 \times 10^{- 3}}{0.3} \times \frac{[2 - (- 2)]}{0.25}$
$emf = 48 mV$

VITEEE-2011

Electro Magnetic Induction

154700 A rectangular coil of 300 turns has an average area of $25 cm \times 10 cm$ . The coil rotates with a speed of 50 cps in uniform magnetic field of strength $4 \times 10^{- 2} T$ about an axis perpendicular to the field. The peak value of the induced emf is (in volt) :

1

300 π

2

3000 π

3

3 π

4

30 π

Explanation:

D Given- N $= 300$ turns
Area, $A = 25 cm \times 10 cm$
Frequency, $f = 50 cps$
Magnetic field, $B = 4 \times 10^{- 2} T$
Angular velocity, $ω = 2 π f$
$= 2 π \times 50 = 100 π$
Peak value of induced emf $=$ NAB $ω$ $emf = 300 \times 25 \times 10 \times 10^{- 4} \times 4 \times 10^{- 2} \times 100 π$ emf $= 30 π$ volt

Karnataka CET-2004

Question Bank Series

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