Question

The inductor shown in the figure has inductance 0.50 H and carries a current in the direction shown that is decreasing at a uniform rate $\frac{di}{dt} = -0.03A/s$.

Answer 1

15

Answer 2

The self-induced electromotive force (emf) in an inductor is given by the formula:

$$\epsilon = -L \frac{di}{dt}$$

where:

• $\epsilon$ is the self-induced emf
• $L$ is the inductance of the inductor
• $\frac{di}{dt}$ is the rate of change of current

Given values:

• Inductance, $L = 0.50 \text{ H}$
• Rate of change of current, $\frac{di}{dt} = -0.03 \text{ A/s}$ (The negative sign indicates that the current is decreasing)

Substitute the given values into the formula:

$$\epsilon = -(0.50 \text{ H}) \times (-0.03 \text{ A/s}) = 0.015 \text{ V}$$

The problem states that the self-induced emf is $n \times 10^{-3} \text{ V}$. We need to find the value of $n$. We have $\epsilon = 0.015 \text{ V}$. To express this in the form $n \times 10^{-3} \text{ V}$, we can rewrite $0.015$ as:

$$0.015 = 15 \times 0.001 = 15 \times 10^{-3}$$

So, $0.015 \text{ V} = 15 \times 10^{-3} \text{ V}$

Comparing this with $n \times 10^{-3} \text{ V}$, we find $n = 15$.