The magnetic force per unit length on a wire carrying a current of 10 A and making an angle of `45^@` with the direction of a uniform magnetic field of 0.20 T is

Options

A. `2sqrt2Nm^(-1)`
B. `2/sqrt2Nm^(-1)`
C. `sqrt2/2Nm^(-1)`
D. `4sqrt2Nm^(-1)`

Explanation

To find the magnetic force per unit length on a wire carrying a current in a magnetic field, we can use the formula:

\[ F = BIL \sin \theta \]

Where:
- \( F \) is the magnetic force,
- \( B \) is the magnetic field strength,
- \( I \) is the current,
- \( L \) is the length of the wire,
- \( \theta \) is the angle between the wire and the magnetic field.

Since we are looking for the force per unit length (\( \frac{F}{L} \)), we can rearrange the formula:

\[ \frac{F}{L} = BI \sin \theta \]

### Step 1: Identify the given values
- Current (\( I \)) = 10 A
- Magnetic field (\( B \)) = 0.20 T
- Angle (\( \theta \)) = 45°

### Step 2: Calculate \( \sin \theta \)
For \( \theta = 45° \):
\[ \sin 45° = \frac{1}{\sqrt{2}} \]

### Step 3: Substitute the values into the formula
Now substituting the values into the formula for force per unit length:
\[ \frac{F}{L} = B \cdot I \cdot \sin \theta \]
\[ \frac{F}{L} = 0.20 \, \text{T} \cdot 10 \, \text{A} \cdot \frac{1}{\sqrt{2}} \]

### Step 4: Perform the calculation
Calculating the right side:
\[ \frac{F}{L} = 0.20 \cdot 10 \cdot \frac{1}{\sqrt{2}} \]
\[ \frac{F}{L} = 2 \cdot \frac{1}{\sqrt{2}} \]
\[ \frac{F}{L} = \frac{2}{\sqrt{2}} \]

### Step 5: Simplify the expression
To simplify \( \frac{2}{\sqrt{2}} \):
\[ \frac{2}{\sqrt{2}} = \sqrt{2} \]

### Final Answer
Thus, the magnetic force per unit length on the wire is:
\[ \frac{F}{L} = \sqrt{2} \, \text{N/m} \]