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Electro Magnetic Induction

154532 The magnetic flux linked with a closed coil is increased to a maximum value in $2 s$ and its relation with time is $ϕ = a^{2} + b t + c$ then the relation between $a, b$ and $c$ is

1

a = - b

2

a = \frac{- b}{4}

3

a + b = c

4

ac = \frac{b}{2}

Explanation:

B Given, $ϕ = a t^{2} + b t + c$
$\frac{d ϕ}{dt} = 2 at + b$
At, $t = 2$
$\frac{d ϕ}{dt} = 2 \times a \times 2 + b$
(for maximum, $\frac{d ϕ}{dt} = 0$ )
$0 = 4 a + b$
$a = - b / 4$

AP EAMCET-25.04.2017

Electro Magnetic Induction

154533 If $B_{H} = 4 \times 10^{- 5} T$ and $B_{v} = 2 \times 10^{- 5} T$ , then the
Earth's total field at the place is (in T)

1

6 \times 10^{- 5} T

2

2 \sqrt{5} \times 10^{- 5} T

3

4 \times 10^{- 5} T

4

3 \times 10^{- 5} T

Explanation:

B Given that,
$B_{H} = 4 \times 10^{- 5} T$
$B_{V} = 2 \times 10^{- 5} T$
We know that,
The total earth's field at the place-
$B = \sqrt{B_{V}^{2} + B_{H}^{2}}$
$= \sqrt{{(4 \times 10^{- 5})}^{2} + {(2 \times 10^{- 5})}^{2}}$
$= \sqrt{16 \times 10^{- 10} + 4 \times 10^{- 10}}$
$= \sqrt{20} \times 10^{- 5} = 2 \sqrt{5} \times 10^{- 5} T$

J and K CET-2017

Electro Magnetic Induction

154534 A coil of circular cross-section having 1000 turns and $4 {cm}^{2}$ face area is placed with its axis parallel to a magnetic field which decreases by $10^{- 2} Wb m^{- 2}$ in $0.01 s$ . The e.m.f. induced in the coil is :

1

400 mV

2

200 mV

3

4 mV

4

0.4 mV

Explanation:

A Given that,
No. of turns $(N) = 1000$
Face area $(A) = 4 {cm}^{2} = 4 \times 10^{- 4} m^{2}$
Change in magnetic field -
$Δ B = 10^{- 2} {Wbm}^{- 2}$
Time taken, $t = 0.01 s = 10^{- 2} s$
We know that,
Induced emf $(ε) = \frac{- d ϕ}{dt}$
$ε = N (\frac{Δ B}{Δ t}) A$
$ε = \frac{1000 \times 10^{- 2} \times 4 \times 10^{- 4}}{10^{- 2}}$
$ε = 400 mV .$

AIIMS-2017

Electro Magnetic Induction

154536 The magnetic flux through a coil varies with time as $ϕ = 5 t^{2} + 6 t + 9$ . The ratio of emf at $t =$ $3 s$ to $t = 0$ s will be

1

1 : 9

2

1 : 6

3

6 : 1

4

9 : 1

Explanation:

C We know that,
$| ε | = \frac{d ϕ}{dt}$
$ε = \frac{d}{dt} (5 t^{2} + 6 t + 9)$
$ε = 10 t + 6$
$At t = 3 \sec$
$ε_{1} = 10 \times 3 + 6 = 36 V$
$At t = 0 \sec$
$ε_{2} = 10 \times 0 + 6 = 6 V$
$Therefore, \frac{ε_{1}}{ε_{2}} = \frac{36}{6} = \frac{6}{1} = 6 : 1 V$

JCECE-2017

Electro Magnetic Induction

154532 The magnetic flux linked with a closed coil is increased to a maximum value in $2 s$ and its relation with time is $ϕ = a^{2} + b t + c$ then the relation between $a, b$ and $c$ is

1

a = - b

2

a = \frac{- b}{4}

3

a + b = c

4

ac = \frac{b}{2}

Explanation:

B Given, $ϕ = a t^{2} + b t + c$
$\frac{d ϕ}{dt} = 2 at + b$
At, $t = 2$
$\frac{d ϕ}{dt} = 2 \times a \times 2 + b$
(for maximum, $\frac{d ϕ}{dt} = 0$ )
$0 = 4 a + b$
$a = - b / 4$

AP EAMCET-25.04.2017

Electro Magnetic Induction

154533 If $B_{H} = 4 \times 10^{- 5} T$ and $B_{v} = 2 \times 10^{- 5} T$ , then the
Earth's total field at the place is (in T)

1

6 \times 10^{- 5} T

2

2 \sqrt{5} \times 10^{- 5} T

3

4 \times 10^{- 5} T

4

3 \times 10^{- 5} T

Explanation:

B Given that,
$B_{H} = 4 \times 10^{- 5} T$
$B_{V} = 2 \times 10^{- 5} T$
We know that,
The total earth's field at the place-
$B = \sqrt{B_{V}^{2} + B_{H}^{2}}$
$= \sqrt{{(4 \times 10^{- 5})}^{2} + {(2 \times 10^{- 5})}^{2}}$
$= \sqrt{16 \times 10^{- 10} + 4 \times 10^{- 10}}$
$= \sqrt{20} \times 10^{- 5} = 2 \sqrt{5} \times 10^{- 5} T$

J and K CET-2017

Electro Magnetic Induction

154534 A coil of circular cross-section having 1000 turns and $4 {cm}^{2}$ face area is placed with its axis parallel to a magnetic field which decreases by $10^{- 2} Wb m^{- 2}$ in $0.01 s$ . The e.m.f. induced in the coil is :

1

400 mV

2

200 mV

3

4 mV

4

0.4 mV

Explanation:

A Given that,
No. of turns $(N) = 1000$
Face area $(A) = 4 {cm}^{2} = 4 \times 10^{- 4} m^{2}$
Change in magnetic field -
$Δ B = 10^{- 2} {Wbm}^{- 2}$
Time taken, $t = 0.01 s = 10^{- 2} s$
We know that,
Induced emf $(ε) = \frac{- d ϕ}{dt}$
$ε = N (\frac{Δ B}{Δ t}) A$
$ε = \frac{1000 \times 10^{- 2} \times 4 \times 10^{- 4}}{10^{- 2}}$
$ε = 400 mV .$

AIIMS-2017

Electro Magnetic Induction

154536 The magnetic flux through a coil varies with time as $ϕ = 5 t^{2} + 6 t + 9$ . The ratio of emf at $t =$ $3 s$ to $t = 0$ s will be

1

1 : 9

2

1 : 6

3

6 : 1

4

9 : 1

Explanation:

C We know that,
$| ε | = \frac{d ϕ}{dt}$
$ε = \frac{d}{dt} (5 t^{2} + 6 t + 9)$
$ε = 10 t + 6$
$At t = 3 \sec$
$ε_{1} = 10 \times 3 + 6 = 36 V$
$At t = 0 \sec$
$ε_{2} = 10 \times 0 + 6 = 6 V$
$Therefore, \frac{ε_{1}}{ε_{2}} = \frac{36}{6} = \frac{6}{1} = 6 : 1 V$

JCECE-2017

Electro Magnetic Induction

154532 The magnetic flux linked with a closed coil is increased to a maximum value in $2 s$ and its relation with time is $ϕ = a^{2} + b t + c$ then the relation between $a, b$ and $c$ is

1

a = - b

2

a = \frac{- b}{4}

3

a + b = c

4

ac = \frac{b}{2}

Explanation:

B Given, $ϕ = a t^{2} + b t + c$
$\frac{d ϕ}{dt} = 2 at + b$
At, $t = 2$
$\frac{d ϕ}{dt} = 2 \times a \times 2 + b$
(for maximum, $\frac{d ϕ}{dt} = 0$ )
$0 = 4 a + b$
$a = - b / 4$

AP EAMCET-25.04.2017

Electro Magnetic Induction

154533 If $B_{H} = 4 \times 10^{- 5} T$ and $B_{v} = 2 \times 10^{- 5} T$ , then the
Earth's total field at the place is (in T)

1

6 \times 10^{- 5} T

2

2 \sqrt{5} \times 10^{- 5} T

3

4 \times 10^{- 5} T

4

3 \times 10^{- 5} T

Explanation:

B Given that,
$B_{H} = 4 \times 10^{- 5} T$
$B_{V} = 2 \times 10^{- 5} T$
We know that,
The total earth's field at the place-
$B = \sqrt{B_{V}^{2} + B_{H}^{2}}$
$= \sqrt{{(4 \times 10^{- 5})}^{2} + {(2 \times 10^{- 5})}^{2}}$
$= \sqrt{16 \times 10^{- 10} + 4 \times 10^{- 10}}$
$= \sqrt{20} \times 10^{- 5} = 2 \sqrt{5} \times 10^{- 5} T$

J and K CET-2017

Electro Magnetic Induction

154534 A coil of circular cross-section having 1000 turns and $4 {cm}^{2}$ face area is placed with its axis parallel to a magnetic field which decreases by $10^{- 2} Wb m^{- 2}$ in $0.01 s$ . The e.m.f. induced in the coil is :

1

400 mV

2

200 mV

3

4 mV

4

0.4 mV

Explanation:

A Given that,
No. of turns $(N) = 1000$
Face area $(A) = 4 {cm}^{2} = 4 \times 10^{- 4} m^{2}$
Change in magnetic field -
$Δ B = 10^{- 2} {Wbm}^{- 2}$
Time taken, $t = 0.01 s = 10^{- 2} s$
We know that,
Induced emf $(ε) = \frac{- d ϕ}{dt}$
$ε = N (\frac{Δ B}{Δ t}) A$
$ε = \frac{1000 \times 10^{- 2} \times 4 \times 10^{- 4}}{10^{- 2}}$
$ε = 400 mV .$

AIIMS-2017

Electro Magnetic Induction

154536 The magnetic flux through a coil varies with time as $ϕ = 5 t^{2} + 6 t + 9$ . The ratio of emf at $t =$ $3 s$ to $t = 0$ s will be

1

1 : 9

2

1 : 6

3

6 : 1

4

9 : 1

Explanation:

C We know that,
$| ε | = \frac{d ϕ}{dt}$
$ε = \frac{d}{dt} (5 t^{2} + 6 t + 9)$
$ε = 10 t + 6$
$At t = 3 \sec$
$ε_{1} = 10 \times 3 + 6 = 36 V$
$At t = 0 \sec$
$ε_{2} = 10 \times 0 + 6 = 6 V$
$Therefore, \frac{ε_{1}}{ε_{2}} = \frac{36}{6} = \frac{6}{1} = 6 : 1 V$

JCECE-2017

Electro Magnetic Induction

154532 The magnetic flux linked with a closed coil is increased to a maximum value in $2 s$ and its relation with time is $ϕ = a^{2} + b t + c$ then the relation between $a, b$ and $c$ is

1

a = - b

2

a = \frac{- b}{4}

3

a + b = c

4

ac = \frac{b}{2}

Explanation:

B Given, $ϕ = a t^{2} + b t + c$
$\frac{d ϕ}{dt} = 2 at + b$
At, $t = 2$
$\frac{d ϕ}{dt} = 2 \times a \times 2 + b$
(for maximum, $\frac{d ϕ}{dt} = 0$ )
$0 = 4 a + b$
$a = - b / 4$

AP EAMCET-25.04.2017

Electro Magnetic Induction

154533 If $B_{H} = 4 \times 10^{- 5} T$ and $B_{v} = 2 \times 10^{- 5} T$ , then the
Earth's total field at the place is (in T)

1

6 \times 10^{- 5} T

2

2 \sqrt{5} \times 10^{- 5} T

3

4 \times 10^{- 5} T

4

3 \times 10^{- 5} T

Explanation:

B Given that,
$B_{H} = 4 \times 10^{- 5} T$
$B_{V} = 2 \times 10^{- 5} T$
We know that,
The total earth's field at the place-
$B = \sqrt{B_{V}^{2} + B_{H}^{2}}$
$= \sqrt{{(4 \times 10^{- 5})}^{2} + {(2 \times 10^{- 5})}^{2}}$
$= \sqrt{16 \times 10^{- 10} + 4 \times 10^{- 10}}$
$= \sqrt{20} \times 10^{- 5} = 2 \sqrt{5} \times 10^{- 5} T$

J and K CET-2017

Electro Magnetic Induction

154534 A coil of circular cross-section having 1000 turns and $4 {cm}^{2}$ face area is placed with its axis parallel to a magnetic field which decreases by $10^{- 2} Wb m^{- 2}$ in $0.01 s$ . The e.m.f. induced in the coil is :

1

400 mV

2

200 mV

3

4 mV

4

0.4 mV

Explanation:

A Given that,
No. of turns $(N) = 1000$
Face area $(A) = 4 {cm}^{2} = 4 \times 10^{- 4} m^{2}$
Change in magnetic field -
$Δ B = 10^{- 2} {Wbm}^{- 2}$
Time taken, $t = 0.01 s = 10^{- 2} s$
We know that,
Induced emf $(ε) = \frac{- d ϕ}{dt}$
$ε = N (\frac{Δ B}{Δ t}) A$
$ε = \frac{1000 \times 10^{- 2} \times 4 \times 10^{- 4}}{10^{- 2}}$
$ε = 400 mV .$

AIIMS-2017

Electro Magnetic Induction

154536 The magnetic flux through a coil varies with time as $ϕ = 5 t^{2} + 6 t + 9$ . The ratio of emf at $t =$ $3 s$ to $t = 0$ s will be

1

1 : 9

2

1 : 6

3

6 : 1

4

9 : 1

Explanation:

C We know that,
$| ε | = \frac{d ϕ}{dt}$
$ε = \frac{d}{dt} (5 t^{2} + 6 t + 9)$
$ε = 10 t + 6$
$At t = 3 \sec$
$ε_{1} = 10 \times 3 + 6 = 36 V$
$At t = 0 \sec$
$ε_{2} = 10 \times 0 + 6 = 6 V$
$Therefore, \frac{ε_{1}}{ε_{2}} = \frac{36}{6} = \frac{6}{1} = 6 : 1 V$

JCECE-2017

Question Bank Series

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