A wire of length 50 cm moves with a velocity of 300 m/min, perpendicular to a magnetic field. If the emf induced in the wire is 2 V, then the magnitude of the field in tesla is

Options

A. 2
B. 5
C. `0.8`
D. `2.5`

Explanation

To solve the problem, we will use the formula for the induced electromotive force (emf) in a wire moving through a magnetic field. The formula is given by:

\[ \text{emf} = B \cdot L \cdot v \]

where:
- \( \text{emf} \) is the induced electromotive force (in volts),
- \( B \) is the magnetic field strength (in tesla),
- \( L \) is the length of the wire (in meters),
- \( v \) is the velocity of the wire (in meters per second).

### Step 1: Convert the length of the wire from centimeters to meters
The length of the wire is given as 50 cm. To convert this to meters:

\[ L = 50 \, \text{cm} = \frac{50}{100} \, \text{m} = 0.5 \, \text{m} \]

### Step 2: Convert the velocity from meters per minute to meters per second
The velocity is given as 300 m/min. To convert this to meters per second:

\[ v = 300 \, \text{m/min} = \frac{300}{60} \, \text{m/s} = 5 \, \text{m/s} \]

### Step 3: Substitute the known values into the emf formula
We know that the induced emf is 2 V. We can now substitute the values of \( \text{emf} \), \( L \), and \( v \) into the formula:

\[ 2 = B \cdot 0.5 \cdot 5 \]

### Step 4: Solve for the magnetic field \( B \)
Rearranging the equation to solve for \( B \):

\[ B = \frac{\text{emf}}{L \cdot v} = \frac{2}{0.5 \cdot 5} \]

Calculating the denominator:

\[ 0.5 \cdot 5 = 2.5 \]

Now substituting back:

\[ B = \frac{2}{2.5} = 0.8 \, \text{T} \]

### Final Answer
The magnitude of the magnetic field is \( B = 0.8 \, \text{T} \).

---