Question 35: Electromagnetic Induction

Question

$ 	ext { Match the LIST-I with LIST-II } $
List-I
List-II
A.
Magnetic induction
I.
M
L

T

−

2

A

−

2

M
L

T

−

2

A

−

2

MLT^(-2)A^(-2)
B.
Magnetic flux
II.
M

L

2

T

−

2

A

−

2

M

L

2

T

−

2

A

−

2

ML^(2)T^(-2)A^(-2)
C.
Magnetic permeability
III.
M

L

0

T

−

2

A

−

1

M

L

0

T

−

2

A

−

1

ML^(0)T^(-2)A^(-1)
D.
Self inductance
IV.
M

L

2

T

−

2

A

−

1

M

L

2

T

−

2

A

−

1

ML^(2)T^(-2)A^(-1)
Choose the correct answer from the options given below:

Accepted Answer

The force on a current-carrying wire is given by $\mathrm{F}=\mathrm{BI} \mathrm{I} \sin 	heta$
$ \mathrm{B}=\frac{\mathrm{F}}{\mathrm{I} \cdot \mathrm{l} \sin 	heta} $
Dimensions of force $[\mathrm{F}]=\left[\mathrm{MLT}^{-2}ight]$
Dimensions of current $[\mathrm{I}]=[\mathrm{A}]$
Dimension of length $[\mathrm{l}]=[\mathrm{L}]$
Dimension of trigonometric function is $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0ight]$, so $[\sin 	heta]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0ight]$
So, the dimension of magnetic field is,
$ [\mathrm{B}]=\left[\mathrm{MLT}^{-2}ight][\mathrm{A}][\mathrm{L}]=\left[\mathrm{M} \mathrm{~L}^{1-1} \mathrm{~T}^{-2} \mathrm{~A}^{-1}ight] $
$\Rightarrow $ $ [\mathrm{B}]=\left[\mathrm{M} \mathrm{~L}^0 \mathrm{~T}^{-2} \mathrm{~A}^{-1}ight] ightarrow(\mathrm{III}) $
Magnetic flux is the product of magnetic field and area: $\Phi=\mathrm{B} \cdot \mathrm{A}$
Dimensions of magnetic induction $[\mathrm{B}]=\left[\mathrm{M} \mathrm{T}^{-2} \mathrm{~A}^{-1}ight]$
Dimensions of area $[\mathrm{A}]=\left[\mathrm{L}^2ight]$
So, the dimension of magnetic flux is,
$ [\Phi]=\left[\mathrm{M} \mathrm{~T}^{-2} \mathrm{~A}^{-1}ight] 	imes\left[\mathrm{L}^2ight] $
$\Rightarrow $ $[\Phi]=\left[\mathrm{M} \mathrm{L}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-1}ight] ightarrow(\mathrm{IV})$
From Ampere's Law or the force between two parallel wires :
$\mathrm{F}=\frac{\mu \mathrm{I}_1 \mathrm{I}_2 \mathrm{l}}{2 \pi \mathrm{r}}$.
$\Rightarrow $$ \mu=\frac{\mathrm{F} \cdot 2 \pi \mathrm{r}}{\mathrm{I}_1 \mathrm{I}_2 \cdot \mathrm{l}} $
Dimensions of Force $[\mathrm{F}]=\left[\mathrm{MLT}^{-2}ight]$
Dimensions of distance $[\mathrm{r}]=[\mathrm{L}]$
Dimensions of 2 and $\pi$ is $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0ight]$
Dimensions of current $[\mathrm{I}]=[\mathrm{A}]$
Dimensions of length $[\mathrm{l}]=[\mathrm{L}]$
So, the dimensions of magnetic permeability is,
$ [\mu]=\frac{\left[\mathrm{MLT}^{-2}ight][\mathrm{L}]}{[\mathrm{A}][\mathrm{A}][\mathrm{L}]} $
$\Rightarrow $ $[\mu]=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}ight]}{\left[\mathrm{A}^2ight][\mathrm{L}]}=\left[\mathrm{ML}^{2-1} \mathrm{~T}^{-2} \mathrm{~A}^{-2}ight]$
$\Rightarrow $ $ [\mu]=\left[\mathrm{M} \mathrm{~L} \mathrm{~T}^{-2} \mathrm{~A}^{-2}ight] ightarrow(\mathrm{I}) $
The energy stored in an inductor is $\mathrm{U}=\frac{1}{2} \mathrm{LI}^2$.
$ \mathrm{L}=\frac{2 \mathrm{U}}{\mathrm{I}^2} $
Dimensions of energy $[\mathrm{U}]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}ight]$
Dimensions of current $[\mathrm{I}]=[\mathrm{A}]$
So, the dimensions of self-inductance is,
$[\mathrm{L}]=\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}ight]}{\left[\mathrm{A}^2ight]}$
$\Rightarrow $ $ [\mathrm{L}]=\left[\mathrm{M} \mathrm{~L}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-2}ight] ightarrow(\mathrm{II}) $
Hence, the correct match is A-III, B-IV, C-I, D-II. Therefore, the correct option is (D).

Answer

A-III, B-IV, C-II, D-I

Answer

A-I, B-III, C-IV, D-II

Answer

A-IV, B-III, C-I, D-II

Answer

A-III, B-IV, C-I, D-II

Electromagnetic Induction

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