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PHXII06:ELECTROMAGNETIC INDUCTION

358607 A rectangular coil is held horizontally and a magnet is allowed to fall vertically through it. The acceleration of the magnet while passing through the coil is :

1 equal to

g

2 less than

g

3 more than

g

4 sometimes less and sometimes more than

g

Explanation:

According to Faraday's law there will be change in flux across the rectangular coil which induces a current in the coil as well as an emf will also be induced.
Now, according to Lenz's law, magnet will experience a force in opposite direction of its motion. Hence acceleration of the magnet will be less than $g$ .
So correct option is (2)

PHXII06:ELECTROMAGNETIC INDUCTION

358608 A ring rotates with angular velocity $ω$ about an axis in the plane of the ring and which passes through the center of the ring. A constant magnetic field $B$ exists perpendicular to the plane of the ring. Find the emf induced in the ring as a function of time.

1

B A \sin ω t

2

B A ω \sin ω t

3

B A ω N \sin ω t

4

\frac{B A ω \sin ω t}{\sqrt{2}}

Explanation:

At any time t, $ϕ = N B A \cos θ = N B A \cos ω t$
Now induced emf in the ring
$ε = \frac{- d ϕ}{d t} = N B A ω \sin (ω t)$

PHXII06:ELECTROMAGNETIC INDUCTION

358609 An elasticized conducting band is around a spherical balloon. Its plane passes through the centre of the balloon. A uniform magnetic field of magnitude $0.04 T$ is directed perpendicular to the plane of the band. Air is let out of the balloon at $100 c m^{3} / s$ at an instant when the radius of the balloon is $6 c m$ . What is the induced emf in the band?
supporting img

1

333 μ V

2

236 μ V

3

186 μ V

4

276 μ V

Explanation:

Volume of the balloon at any instant, when radius is $r$ ,
$V = \frac{4}{3} π r^{3}$
Time rate of change of volume,
$\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$
Time rate of change of radius of the balloon,
$\frac{d r}{d t} = \frac{1}{4 π r^{2}} \frac{d V}{d t}$
Flux through rubber band at the given instant,
$ϕ = B (π r^{2})$
Induced emf $= - \frac{d ϕ}{d t} = - \frac{d}{d t} (B π r^{2})$
$\begin{matrix} = - 2 π r B \frac{d r}{d t} = - 2 π r B (\frac{1}{4 π r^{2}} \frac{d V}{d t}) \\ = - \frac{B}{2 r} \frac{d V}{d t} \end{matrix}$
As volume of the balloon is decreasing, $\frac{d V}{d t}$ is negative.
$ε_{induced} = - \frac{(0.4)}{2 \times 6 \times 10^{- 2}} \times (- 100 \times 10^{- 6})$
$= 333 μ V$

PHXII06:ELECTROMAGNETIC INDUCTION

358610 Faraday's laws are consequence of conservation of

1 Charge

2 Energy and magnetic field

3 Energy

4 Magnetic field

PHXII06:ELECTROMAGNETIC INDUCTION

358607 A rectangular coil is held horizontally and a magnet is allowed to fall vertically through it. The acceleration of the magnet while passing through the coil is :

1 equal to

g

2 less than

g

3 more than

g

4 sometimes less and sometimes more than

g

Explanation:

According to Faraday's law there will be change in flux across the rectangular coil which induces a current in the coil as well as an emf will also be induced.
Now, according to Lenz's law, magnet will experience a force in opposite direction of its motion. Hence acceleration of the magnet will be less than $g$ .
So correct option is (2)

PHXII06:ELECTROMAGNETIC INDUCTION

358608 A ring rotates with angular velocity $ω$ about an axis in the plane of the ring and which passes through the center of the ring. A constant magnetic field $B$ exists perpendicular to the plane of the ring. Find the emf induced in the ring as a function of time.

1

B A \sin ω t

2

B A ω \sin ω t

3

B A ω N \sin ω t

4

\frac{B A ω \sin ω t}{\sqrt{2}}

Explanation:

At any time t, $ϕ = N B A \cos θ = N B A \cos ω t$
Now induced emf in the ring
$ε = \frac{- d ϕ}{d t} = N B A ω \sin (ω t)$

PHXII06:ELECTROMAGNETIC INDUCTION

358609 An elasticized conducting band is around a spherical balloon. Its plane passes through the centre of the balloon. A uniform magnetic field of magnitude $0.04 T$ is directed perpendicular to the plane of the band. Air is let out of the balloon at $100 c m^{3} / s$ at an instant when the radius of the balloon is $6 c m$ . What is the induced emf in the band?
supporting img

1

333 μ V

2

236 μ V

3

186 μ V

4

276 μ V

Explanation:

Volume of the balloon at any instant, when radius is $r$ ,
$V = \frac{4}{3} π r^{3}$
Time rate of change of volume,
$\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$
Time rate of change of radius of the balloon,
$\frac{d r}{d t} = \frac{1}{4 π r^{2}} \frac{d V}{d t}$
Flux through rubber band at the given instant,
$ϕ = B (π r^{2})$
Induced emf $= - \frac{d ϕ}{d t} = - \frac{d}{d t} (B π r^{2})$
$\begin{matrix} = - 2 π r B \frac{d r}{d t} = - 2 π r B (\frac{1}{4 π r^{2}} \frac{d V}{d t}) \\ = - \frac{B}{2 r} \frac{d V}{d t} \end{matrix}$
As volume of the balloon is decreasing, $\frac{d V}{d t}$ is negative.
$ε_{induced} = - \frac{(0.4)}{2 \times 6 \times 10^{- 2}} \times (- 100 \times 10^{- 6})$
$= 333 μ V$

PHXII06:ELECTROMAGNETIC INDUCTION

358610 Faraday's laws are consequence of conservation of

1 Charge

2 Energy and magnetic field

3 Energy

4 Magnetic field

PHXII06:ELECTROMAGNETIC INDUCTION

358607 A rectangular coil is held horizontally and a magnet is allowed to fall vertically through it. The acceleration of the magnet while passing through the coil is :

1 equal to

g

2 less than

g

3 more than

g

4 sometimes less and sometimes more than

g

Explanation:

According to Faraday's law there will be change in flux across the rectangular coil which induces a current in the coil as well as an emf will also be induced.
Now, according to Lenz's law, magnet will experience a force in opposite direction of its motion. Hence acceleration of the magnet will be less than $g$ .
So correct option is (2)

PHXII06:ELECTROMAGNETIC INDUCTION

358608 A ring rotates with angular velocity $ω$ about an axis in the plane of the ring and which passes through the center of the ring. A constant magnetic field $B$ exists perpendicular to the plane of the ring. Find the emf induced in the ring as a function of time.

1

B A \sin ω t

2

B A ω \sin ω t

3

B A ω N \sin ω t

4

\frac{B A ω \sin ω t}{\sqrt{2}}

Explanation:

At any time t, $ϕ = N B A \cos θ = N B A \cos ω t$
Now induced emf in the ring
$ε = \frac{- d ϕ}{d t} = N B A ω \sin (ω t)$

PHXII06:ELECTROMAGNETIC INDUCTION

358609 An elasticized conducting band is around a spherical balloon. Its plane passes through the centre of the balloon. A uniform magnetic field of magnitude $0.04 T$ is directed perpendicular to the plane of the band. Air is let out of the balloon at $100 c m^{3} / s$ at an instant when the radius of the balloon is $6 c m$ . What is the induced emf in the band?
supporting img

1

333 μ V

2

236 μ V

3

186 μ V

4

276 μ V

Explanation:

Volume of the balloon at any instant, when radius is $r$ ,
$V = \frac{4}{3} π r^{3}$
Time rate of change of volume,
$\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$
Time rate of change of radius of the balloon,
$\frac{d r}{d t} = \frac{1}{4 π r^{2}} \frac{d V}{d t}$
Flux through rubber band at the given instant,
$ϕ = B (π r^{2})$
Induced emf $= - \frac{d ϕ}{d t} = - \frac{d}{d t} (B π r^{2})$
$\begin{matrix} = - 2 π r B \frac{d r}{d t} = - 2 π r B (\frac{1}{4 π r^{2}} \frac{d V}{d t}) \\ = - \frac{B}{2 r} \frac{d V}{d t} \end{matrix}$
As volume of the balloon is decreasing, $\frac{d V}{d t}$ is negative.
$ε_{induced} = - \frac{(0.4)}{2 \times 6 \times 10^{- 2}} \times (- 100 \times 10^{- 6})$
$= 333 μ V$

PHXII06:ELECTROMAGNETIC INDUCTION

358610 Faraday's laws are consequence of conservation of

1 Charge

2 Energy and magnetic field

3 Energy

4 Magnetic field

PHXII06:ELECTROMAGNETIC INDUCTION

358607 A rectangular coil is held horizontally and a magnet is allowed to fall vertically through it. The acceleration of the magnet while passing through the coil is :

1 equal to

g

2 less than

g

3 more than

g

4 sometimes less and sometimes more than

g

Explanation:

According to Faraday's law there will be change in flux across the rectangular coil which induces a current in the coil as well as an emf will also be induced.
Now, according to Lenz's law, magnet will experience a force in opposite direction of its motion. Hence acceleration of the magnet will be less than $g$ .
So correct option is (2)

PHXII06:ELECTROMAGNETIC INDUCTION

358608 A ring rotates with angular velocity $ω$ about an axis in the plane of the ring and which passes through the center of the ring. A constant magnetic field $B$ exists perpendicular to the plane of the ring. Find the emf induced in the ring as a function of time.

1

B A \sin ω t

2

B A ω \sin ω t

3

B A ω N \sin ω t

4

\frac{B A ω \sin ω t}{\sqrt{2}}

Explanation:

At any time t, $ϕ = N B A \cos θ = N B A \cos ω t$
Now induced emf in the ring
$ε = \frac{- d ϕ}{d t} = N B A ω \sin (ω t)$

PHXII06:ELECTROMAGNETIC INDUCTION

358609 An elasticized conducting band is around a spherical balloon. Its plane passes through the centre of the balloon. A uniform magnetic field of magnitude $0.04 T$ is directed perpendicular to the plane of the band. Air is let out of the balloon at $100 c m^{3} / s$ at an instant when the radius of the balloon is $6 c m$ . What is the induced emf in the band?
supporting img

1

333 μ V

2

236 μ V

3

186 μ V

4

276 μ V

Explanation:

Volume of the balloon at any instant, when radius is $r$ ,
$V = \frac{4}{3} π r^{3}$
Time rate of change of volume,
$\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$
Time rate of change of radius of the balloon,
$\frac{d r}{d t} = \frac{1}{4 π r^{2}} \frac{d V}{d t}$
Flux through rubber band at the given instant,
$ϕ = B (π r^{2})$
Induced emf $= - \frac{d ϕ}{d t} = - \frac{d}{d t} (B π r^{2})$
$\begin{matrix} = - 2 π r B \frac{d r}{d t} = - 2 π r B (\frac{1}{4 π r^{2}} \frac{d V}{d t}) \\ = - \frac{B}{2 r} \frac{d V}{d t} \end{matrix}$
As volume of the balloon is decreasing, $\frac{d V}{d t}$ is negative.
$ε_{induced} = - \frac{(0.4)}{2 \times 6 \times 10^{- 2}} \times (- 100 \times 10^{- 6})$
$= 333 μ V$

PHXII06:ELECTROMAGNETIC INDUCTION

358610 Faraday's laws are consequence of conservation of

1 Charge

2 Energy and magnetic field

3 Energy

4 Magnetic field