Question 247: Electromagnetic Induction

Question

If current of 4 A produces magnetic flux of $3 	imes 10^{-3} \mathrm{~Wb}$ through a coil of 400 turns, the energy stored in the coil will be

Accepted Answer

To find the energy stored in the coil, we can use the formula for the energy stored in a magnetic field of an inductor:
$$ E = \frac{1}{2} L I^2 $$
where $ E $ is the energy, $ L $ is the inductance, and $ I $ is the current.
First, we need to determine the inductance $ L $ of the coil. The relationship between magnetic flux ($\Phi$), inductance ($L$), and current ($I$) is given by:
$$ \Phi = L \cdot I $$
Given:
$\Phi = 3 	imes 10^{-3} \, 	ext{Wb}$,
$I = 4 \, 	ext{A}$,
Number of turns ($N$) = 400.
The total flux linkage is $N \cdot \Phi$:
$$ N \cdot \Phi = L \cdot I $$
Plug in the values:
$$ 400 	imes 3 	imes 10^{-3} = L 	imes 4 $$
Calculate $L$:
$$ L = \frac{400 	imes 3 	imes 10^{-3}}{4} = \frac{1.2}{4} = 0.3 \, 	ext{H} $$
Now, substitute $L = 0.3 \, 	ext{H}$ and $I = 4 \, 	ext{A}$ into the energy formula:
$$ E = \frac{1}{2} 	imes 0.3 	imes 4^2 = \frac{1}{2} 	imes 0.3 	imes 16 = 0.15 	imes 16 = 2.4 \, 	ext{J} $$
Therefore, the energy stored in the coil is
2.4 J
.
The correct answer is
Option B
: 2.4 J.

Answer

1.2 J

Answer

2.4 J

Answer

24 J

Answer

240 J

Electromagnetic Induction

If current of 4 A produces magnetic flux of 3 \times 10^{-3} \mathrm{~Wb} through a coil of 400 turns, the energy stored in the coil will be

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