Question

A current of 2 A passes through a coil of 100 turns, producing a magnetic flux of 5 × 10⁻⁵ Wb per turn. What is the magnetic energy associated with the coil?

• A. $5 \times 10^{-3} \, \text{J}$
• B. $6 \times 10^{-3} \, \text{J}$
• C. $5 \times 10^{-6} \, \text{J}$
• D. $7 \times 10^{-3} \, \text{J}$

Hint:

Magnetic energy depends on both the current and the coil's inductance.

Answer 1

The Correct Option is A

Step 1: Understand the Magnetic Energy Formula The magnetic energy stored in an inductor (or coil) is given by: \[ U = \frac{1}{2} L I^2 \] where \(L\) is the inductance of the coil (in henries, H), and \(I\) is the current (in amperes, A). To find \(U\), we need to calculate \(L\). 

Step 2: Calculate the Inductance \(L\) The inductance of a coil is related to the magnetic flux linkage by: \[ \Phi_{\text{total}} = N \Phi = L I \] where: - \(N\) is the number of turns, - \(\Phi\) is the magnetic flux per turn, - \(I\) is the current. Rearranging for \(L\): \[ L = \frac{N \Phi}{I} \] Given: - \(N = 100\) turns, - \(\Phi = 5 \times 10^{-5}\) Wb per turn, - \(I = 2\) A. Substitute the values: \[ L = \frac{100 \times 5 \times 10^{-5}}{2} \] \[ L = \frac{100 \times 5 \times 10^{-5}}{2} = \frac{5 \times 10^{-3}}{2} = 2.5 \times 10^{-3} \, \text{H} \] \[ L = 2.5 \, \text{mH} \] 

Step 3: Calculate the Magnetic Energy \(U\) Now, use the magnetic energy formula: \[ U = \frac{1}{2} L I^2 \] Substitute \(L = 2.5 \times 10^{-3} \, \text{H}\) and \(I = 2 \, \text{A}\): \[ U = \frac{1}{2} \times (2.5 \times 10^{-3}) \times (2)^2 \] \[ U = \frac{1}{2} \times 2.5 \times 10^{-3} \times 4 \] \[ U = \frac{1}{2} \times 10 \times 10^{-3} = 5 \times 10^{-3} \, \text{J} \] \[ U = 5 \, \text{mJ} \] 

Step 4: Final Answer The magnetic energy associated with the coil is: \[ \boxed{5 \times 10^{-3} \, \text{J}} \]