A wire of length 15 cm carrying current of 5 Ais balanced in air by a uniform magnetic field, B. The mass of wire is` 0.5` grams.
Calculate the value of B.
Use g = `10 ms^(-2)`.

Answer

To solve the problem, we need to find the value of the magnetic field \( B \) that balances the weight of the wire carrying a current. Here are the steps to derive the solution:

### Step 1: Identify the forces acting on the wire
The wire is subjected to two main forces:
1. The gravitational force acting downwards, which is given by \( F_g = mg \).
2. The magnetic force acting upwards, which is given by \( F_B = ILB \).

### Step 2: Write the equation for balance of forces
For the wire to be balanced in the magnetic field, the magnetic force must equal the gravitational force:
\[ F_B = F_g \]
Thus, we can write:
\[ ILB = mg \]

### Step 3: Substitute known values into the equation
We know:
- Current \( I = 5 \, \text{A} \)
- Length of the wire \( L = 15 \, \text{cm} = 0.15 \, \text{m} \)
- Mass of the wire \( m = 0.5 \, \text{g} = 0.5 \times 10^{-3} \, \text{kg} \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

Substituting these values into the equation gives:
\[ 5 \times 0.15 \times B = 0.5 \times 10^{-3} \times 10 \]

### Step 4: Simplify the equation
Calculating the right-hand side:
\[ 0.5 \times 10^{-3} \times 10 = 5 \times 10^{-3} \, \text{N} \]
Now we have:
\[ 5 \times 0.15 \times B = 5 \times 10^{-3} \]

### Step 5: Solve for \( B \)
Now, simplify the left-hand side:
\[ 0.75B = 5 \times 10^{-3} \]
To find \( B \), divide both sides by \( 0.75 \):
\[ B = \frac{5 \times 10^{-3}}{0.75} \]

Calculating this gives:
\[ B = \frac{5}{0.75} \times 10^{-3} = \frac{20}{3} \times 10^{-3} \approx 6.67 \times 10^{-3} \, \text{T} \]

### Step 6: Convert to milliTesla
Since \( 1 \, \text{T} = 1000 \, \text{mT} \):
\[ B \approx 6.67 \, \text{mT} \]

### Final Answer
The value of the magnetic field \( B \) is approximately \( 6.67 \, \text{mT} \).

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