A current of 0.5 A is passed through the coil of a galvanometer having 500 turns and each turns has an average area of `3xx10^(-4) m^(2)` if a torque of 1.5 N-m is required for this coil carrying same current to set it parallel to a magnetic field calculate the strength of the magnetic field

Options

A. 20T
B. 25 T
C. 23 T
D. 21 T

Answer

To solve the problem, we will use the formula for the torque (\( \tau \)) acting on a coil in a magnetic field, which is given by:

\[ \tau = B \cdot I \cdot N \cdot A \]

Where:
- \( \tau \) = Torque (in N·m)
- \( B \) = Magnetic field strength (in T)
- \( I \) = Current (in A)
- \( N \) = Number of turns in the coil
- \( A \) = Area of each turn (in m²)

Given values:
- \( \tau = 1.5 \, \text{N·m} \)
- \( I = 0.5 \, \text{A} \)
- \( N = 500 \)
- \( A = 3 \times 10^{-4} \, \text{m}^2 \)

### Step 1: Rearranging the formula to solve for \( B \)

We need to isolate \( B \) in the torque equation:

\[ B = \frac{\tau}{I \cdot N \cdot A} \]

### Step 2: Substituting the given values into the formula

Now we substitute the known values into the rearranged formula:

\[ B = \frac{1.5}{0.5 \cdot 500 \cdot (3 \times 10^{-4})} \]

### Step 3: Calculating the denominator

First, we calculate the denominator:

\[ 0.5 \cdot 500 = 250 \]
\[ 250 \cdot (3 \times 10^{-4}) = 250 \cdot 0.0003 = 0.075 \]

### Step 4: Calculating \( B \)

Now we can calculate \( B \):

\[ B = \frac{1.5}{0.075} \]

Calculating this gives:

\[ B = 20 \, \text{T} \]

### Final Answer

The strength of the magnetic field \( B \) is \( 20 \, \text{T} \).

---