A rod of 10 cm length is moving perpendicular to uniform magnetic field of intensity `5xx10^(-4)"Wbm"^(-2)`. If the acceleration of the rod is `5ms^(-2)`, then the rate of increase of induced emf is

Options

A. `25xx10^(-4)Vs^(-1)`
B. `2.5xx10^(-4)Vs^(-1)`
C. `20xx10^(-4)Vs^(-1)`
D. `20xx10^(-4)Vs^(-1)`

Explanation

To solve the problem step by step, we will follow the given parameters and apply the relevant formulas.

### Step 1: Identify the given parameters
- Length of the rod (L) = 10 cm = 0.1 m (conversion from cm to m)
- Magnetic field intensity (B) = \(5 \times 10^{-4}\) Wb/m²
- Acceleration (a) = \(5 \, \text{m/s}^2\)

### Step 2: Write the formula for induced EMF
The formula for the motional EMF (E) induced in a rod moving through a magnetic field is given by:
\[ E = B \cdot V \cdot L \]
where:
- \(E\) = induced EMF
- \(B\) = magnetic field intensity
- \(V\) = velocity of the rod
- \(L\) = length of the rod

### Step 3: Relate velocity to acceleration
Since the rod is accelerating, we can express the velocity (V) in terms of acceleration (a) and time (t):
\[ V = a \cdot t \]

### Step 4: Substitute the expression for velocity into the EMF formula
Substituting \(V = a \cdot t\) into the EMF formula gives:
\[ E = B \cdot (a \cdot t) \cdot L \]

### Step 5: Differentiate the EMF with respect to time to find the rate of increase of EMF
To find the rate of increase of induced EMF, we differentiate \(E\) with respect to time (t):
\[ \frac{dE}{dt} = B \cdot a \cdot L \]

### Step 6: Substitute the known values into the differentiated formula
Now, we can substitute the values of \(B\), \(a\), and \(L\):
- \(B = 5 \times 10^{-4} \, \text{Wb/m}^2\)
- \(a = 5 \, \text{m/s}^2\)
- \(L = 0.1 \, \text{m}\)

Thus,
\[ \frac{dE}{dt} = (5 \times 10^{-4}) \cdot (5) \cdot (0.1) \]

### Step 7: Calculate the result
Calculating the above expression:
\[ \frac{dE}{dt} = 5 \times 10^{-4} \cdot 5 \cdot 0.1 = 2.5 \times 10^{-4} \, \text{V/s} \]

### Final Answer
The rate of increase of induced EMF is:
\[ \frac{dE}{dt} = 2.5 \times 10^{-4} \, \text{V/s} \]

---