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 of finding the magnitude of self-induced EMF in a toroidal solenoid, we can follow these steps:

### Step 1: Understand the relationship between self-induced EMF, inductance, and current change.
The self-induced EMF (E) in a solenoid can be expressed using the formula:
\[ E = -L \frac{di}{dt} \]
where:
- \(E\) is the self-induced EMF,
- \(L\) is the self-inductance of the solenoid,
- \(\frac{di}{dt}\) is the rate of change of current.

### Step 2: Identify the given values.
From the problem, we have:
- Self-inductance \(L = 40 \, \mu H = 40 \times 10^{-6} \, H\)
- The current increases from \(0 \, A\) to \(6 \, A\) in \(3.0 \, \mu s = 3.0 \times 10^{-6} \, s\).

### Step 3: Calculate the rate of change of current (\(\frac{di}{dt}\)).
The change in current (\(di\)) is:
\[ di = 6 \, A - 0 \, A = 6 \, A \]
The time interval (\(dt\)) is:
\[ dt = 3.0 \, \mu s = 3.0 \times 10^{-6} \, s \]
Now, we can calculate \(\frac{di}{dt}\):
\[ \frac{di}{dt} = \frac{di}{dt} = \frac{6 \, A}{3.0 \times 10^{-6} \, s} = 2 \times 10^6 \, A/s \]

### Step 4: Substitute the values into the EMF formula.
Now we can substitute \(L\) and \(\frac{di}{dt}\) into the formula for self-induced EMF:
\[ E = -L \frac{di}{dt} = - (40 \times 10^{-6} \, H) \times (2 \times 10^6 \, A/s) \]

### Step 5: Calculate the magnitude of the self-induced EMF.
Calculating the above expression:
\[ E = - (40 \times 10^{-6}) \times (2 \times 10^6) = -80 \, V \]
The negative sign indicates the direction of the induced EMF (according to Lenz's law), but we are interested in the magnitude:
\[ |E| = 80 \, V \]

### Final Answer
The magnitude of the self-induced EMF is \(80 \, V\).

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