What magnitude of electric current can store 2.5 J of energy in a 3 H induction coil?

Analysis

The question asks for the magnitude of electric current ($I$) that results in a specific amount of energy ($E$) being stored in an inductor with a given inductance ($L$).

Concept

The energy stored in an inductor is given by the formula:

Where:

• $E$ = Energy stored in Joules (J)
• $L$ = Inductance in Henrys (H)
• $I$ = Current in Amperes (A)

Step-by-Step Solution

1. Identify the given values: • $E = 2.5\text{ J}$ • $L = 3\text{ H}$

2. Rearrange the formula to solve for $I$: $$2E = LI^2$$ $$I^2 = \frac{2E}{L}$$ $$I = \sqrt{\frac{2E}{L}}$$

3. Substitute the values into the equation: $$I = \sqrt{\frac{2 \times 2.5}{3}}$$ $$I = \sqrt{\frac{5}{3}}$$ $$I = \sqrt{1.666...}$$

4. Calculate the final value: $$I \approx 1.29099\text{ A}$$

Rounding to two decimal places, we get 1.29 A.

Conclusion

Comparing our result to the provided options, the correct value is 1.29 A.

Exam Tip

When dealing with energy formulas in physics (like $\frac{1}{2}LI^2$ for inductors or $\frac{1}{2}CV^2$ for capacitors), always remember the "1/2" factor. A common mistake students make is forgetting to multiply the energy by 2 before taking the square root. Also, ensure your units are in SI (Joules and Henrys) before calculating to avoid conversion errors.