In an ac dynamo, the peak value of emf is `60V`. The induced emf (in V) in the position when the armature makes an angle of `30^(circ)` with the magnetic ficld perpendicular to the coil, will be

Answer

To find the induced emf in an AC dynamo when the armature makes an angle of \(30^\circ\) with the magnetic field, we can follow these steps:

### Step 1: Understand the relationship between peak emf and induced emf
The peak value of the induced emf (\(E_0\)) is given as \(60V\). The induced emf at any angle can be calculated using the formula:
\[ E = E_0 \sin(\theta) \]
where:
- \(E\) is the induced emf,
- \(E_0\) is the peak value of the emf,
- \(\theta\) is the angle between the magnetic field and the normal to the coil.

### Step 2: Identify the angle
In this case, the angle \(\theta\) is given as \(30^\circ\).

### Step 3: Substitute the values into the formula
Now we can substitute the known values into the formula:
\[ E = 60 \sin(30^\circ) \]

### Step 4: Calculate \(\sin(30^\circ)\)
We know that:
\[ \sin(30^\circ) = \frac{1}{2} \]

### Step 5: Compute the induced emf
Now, substituting \(\sin(30^\circ)\) into the equation:
\[ E = 60 \times \frac{1}{2} = 30V \]

### Conclusion
Thus, the induced emf when the armature makes an angle of \(30^\circ\) with the magnetic field is \(30V\).

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