If the current in the toroidal solenoid increases uniformly from zero to 6.0 A in `3.0 mus` Self-inductance of the toroidal solenoid is `40mu` H The magnitude of self-induced emf is

Options

A. 80 V
B. 160 V
C. 24 V
D. 48 V

Answer

To solve the problem step by step, we will use the formula for self-induced electromotive force (emf) in a solenoid:

### Step 1: Identify the given values
- Final current (I) = 6.0 A
- Initial current (I₀) = 0 A
- Time interval (Δt) = 3.0 µs = 3.0 × 10⁻⁶ s
- Self-inductance (L) = 40 µH = 40 × 10⁻⁶ H

### Step 2: Calculate the change in current (ΔI)
The change in current (ΔI) is given by:
\[ \Delta I = I - I_0 = 6.0 \, \text{A} - 0 \, \text{A} = 6.0 \, \text{A} \]

### Step 3: Calculate the rate of change of current (dI/dt)
The rate of change of current (dI/dt) is calculated as:
\[ \frac{dI}{dt} = \frac{\Delta I}{\Delta t} = \frac{6.0 \, \text{A}}{3.0 \times 10^{-6} \, \text{s}} = 2.0 \times 10^{6} \, \text{A/s} \]

### Step 4: Use the formula for self-induced emf
The magnitude of self-induced emf (E) is given by:
\[ |E| = L \left| \frac{dI}{dt} \right| \]
Substituting the values:
\[ |E| = 40 \times 10^{-6} \, \text{H} \times 2.0 \times 10^{6} \, \text{A/s} \]

### Step 5: Calculate the self-induced emf
\[ |E| = 40 \times 2 = 80 \, \text{V} \]

### Final Answer
The magnitude of self-induced emf is **80 V**.

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