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^(-1)`, 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, we need to find the rate of increase of induced emf in a rod moving perpendicular to a magnetic field. Let's break it down step by step.

### Given Data:
- Length of the rod, \( l = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Magnetic field intensity, \( B = 5 \times 10^{-4} \, \text{Wb/m}^2 \)
- Acceleration of the rod, \( a = 5 \, \text{m/s}^2 \)

### Step 1: Find the velocity of the rod
Since the rod is accelerating, we can find the velocity \( v \) of the rod at any time \( t \) using the equation:
\[ v = a \cdot t \]
However, we need to find the rate of change of induced emf, which involves the change in velocity over time.

### Step 2: Induced emf formula
The induced emf \( \mathcal{E} \) in the rod moving in a magnetic field is given by:
\[ \mathcal{E} = B \cdot v \cdot l \]
where:
- \( B \) is the magnetic field,
- \( v \) is the velocity of the rod,
- \( l \) is the length of the rod.

### Step 3: Differentiate the induced emf with respect to time
To find the rate of increase of induced emf, we differentiate \( \mathcal{E} \) with respect to time \( t \):
\[ \frac{d\mathcal{E}}{dt} = B \cdot l \cdot \frac{dv}{dt} \]
Since \( \frac{dv}{dt} \) is the acceleration \( a \):
\[ \frac{d\mathcal{E}}{dt} = B \cdot l \cdot a \]

### Step 4: Substitute the values
Now we can substitute the known values into the equation:
\[ \frac{d\mathcal{E}}{dt} = (5 \times 10^{-4} \, \text{Wb/m}^2) \cdot (0.1 \, \text{m}) \cdot (5 \, \text{m/s}^2) \]

### Step 5: Calculate the induced emf rate
Calculating this gives:
\[ \frac{d\mathcal{E}}{dt} = 5 \times 10^{-4} \cdot 0.1 \cdot 5 = 2.5 \times 10^{-5} \, \text{V/s} \]

### Final Result:
The rate of increase of induced emf is:
\[ \frac{d\mathcal{E}}{dt} = 2.5 \times 10^{-5} \, \text{V/s} \]