Question

In a coil the current varies from \(-3\,A\) to \(+3\,A\) in \(4\,s\), and induces an emf of \(0.2\,V\). The self inductance of the coil is

• A. \(0.133\,H\)
• B. \(0.266\,H\)
• C. \(0.65\,H\)
• D. \(0.532\,H\)

Hint:

Always calculate total current change carefully: \[ \Delta I = I_f-I_i \] Negative current values are extremely important in inductance problems.

Answer 1

The Correct Option is A

Step 1: Recall the inductor emf rule.
A changing current in a coil makes an induced emf. \[ e = L\frac{dI}{dt} \] where $L$ is the self inductance and $\dfrac{dI}{dt}$ is how fast the current changes.

Step 2: Find the change in current.
The current goes from $-3$ A to $+3$ A. \[ \Delta I = 3 - (-3) = 6\,\text{A} \]
Step 3: Find the rate of change.
This change happens over $4$ s. \[ \frac{dI}{dt} = \frac{6}{4} = 1.5\,\text{A/s} \]
Step 4: Write down the emf.
The induced emf is given as $0.2$ V.
Step 5: Plug into the formula.
\[ 0.2 = L (1.5) \]
Step 6: Solve for the inductance.
\[ L = \frac{0.2}{1.5} = 0.133\,\text{H} \] \[ \boxed{0.133\,\text{H}} \]