Question

The current of 2A passing through a coil of 100 turns gives rise to a magnetic flux of 5 x 10^-5 Wb/turn, the magnetic energy associated with the coil is _________ J

• A. 5x10^-4
• B. 5x10^-3
• C. 5
• D. 5x10^-2

Answer 1

The Correct Option is B

To calculate the magnetic energy associated with the coil, we'll follow these steps:

1. Given Data:
- Current (\(I\)) = 2 A
- Number of turns (\(N\)) = 100
- Magnetic flux per turn (\(\phi\)) = 5 × 10^-5 Wb/turn

2. Calculate Total Flux Linkage:
The total flux linkage (\(\Lambda\)) is given by: \[ \Lambda = N \times \phi = 100 \times 5 \times 10^{-5} = 5 \times 10^{-3} \text{ Wb} \]

3. Determine Self-Inductance (\(L\)):
Using the relation \(\Lambda = LI\): \[ L = \frac{\Lambda}{I} = \frac{5 \times 10^{-3}}{2} = 2.5 \times 10^{-3} \text{ H (Henry)} \]

4. Calculate Magnetic Energy (\(U\)):
The energy stored in the magnetic field is: \[ U = \frac{1}{2}LI^2 = \frac{1}{2} \times 2.5 \times 10^{-3} \times (2)^2 = \frac{1}{2} \times 2.5 \times 10^{-3} \times 4 = 5 \times 10^{-3} \text{ J} \]

5. Final Answer:
The magnetic energy associated with the coil is \(\boxed{5 \times 10^{-3} \text{ J}}\) (5 millijoules).