Question 34: Electromagnetic Induction

Question

When a coil is placed in a time dependent magnetic field the power dissipated in it is $P$. The number of turns, area of the coil and radius of the coil wire are $N$, $A$ and $r$ respectively. For a second coil number of turns, area of the coil and radius of the coil wire are $2N$, $2A$ and $3r$ respectively. When the first coil is replaced with second coil the power dissipated in it is $\sqrt{2} \,\alpha P$. The value of $\alpha$ is ______.

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

36

Answer

128 $ \sqrt{2} $

Answer

16

Answer

64

Electromagnetic Induction

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