A long straight wire carrying a current of `25A` is placed in an external uniform magnetic field `3*0xx10^-4T` parallel to the current. Find the magnitude of the resultant magnetic field at a point `1*5cm` away from the wire.

Explanation

To find the magnitude of the resultant magnetic field at a point 1.5 cm away from a long straight wire carrying a current of 25 A in an external magnetic field of \(3 \times 10^{-4} \, \text{T}\), we can follow these steps:

### Step 1: Identify the magnetic field due to the wire
The magnetic field \(B_2\) created by a long straight wire at a distance \(d\) from the wire can be calculated using the formula:

\[ B_2 = \frac{\mu_0 I}{2 \pi d} \]

Where:
- \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\) (permeability of free space)
- \(I = 25 \, \text{A}\) (current)
- \(d = 1.5 \, \text{cm} = 0.015 \, \text{m}\)

### Step 2: Substitute the values into the formula
Substituting the values into the formula gives:

\[ B_2 = \frac{4\pi \times 10^{-7} \times 25}{2 \pi \times 0.015} \]

### Step 3: Simplify the expression
The \(\pi\) terms cancel out:

\[ B_2 = \frac{4 \times 10^{-7} \times 25}{2 \times 0.015} \]

Calculating the denominator:

\[ 2 \times 0.015 = 0.03 \]

Now substituting back:

\[ B_2 = \frac{100 \times 10^{-7}}{0.03} \]

### Step 4: Calculate \(B_2\)
Calculating this gives:

\[ B_2 = \frac{100 \times 10^{-7}}{0.03} = \frac{100}{3} \times 10^{-5} \, \text{T} \approx 3.33 \times 10^{-5} \, \text{T} \]

### Step 5: Identify the external magnetic field
The external magnetic field \(B_1\) is given as:

\[ B_1 = 3 \times 10^{-4} \, \text{T} \]

### Step 6: Determine the direction of the magnetic fields
- \(B_1\) is in the positive \(j\) direction.
- \(B_2\) due to the wire will be in the negative \(k\) direction (using the right-hand rule).

### Step 7: Calculate the resultant magnetic field
Since \(B_1\) and \(B_2\) are perpendicular to each other, the magnitude of the resultant magnetic field \(B_R\) can be calculated using the Pythagorean theorem:

\[ B_R = \sqrt{B_1^2 + B_2^2} \]

Substituting the values:

\[ B_R = \sqrt{(3 \times 10^{-4})^2 + (3.33 \times 10^{-5})^2} \]

Calculating each term:

\[ (3 \times 10^{-4})^2 = 9 \times 10^{-8} \]
\[ (3.33 \times 10^{-5})^2 \approx 1.11 \times 10^{-9} \]

Now substituting back:

\[ B_R = \sqrt{9 \times 10^{-8} + 1.11 \times 10^{-9}} \approx \sqrt{9.11 \times 10^{-8}} \approx 3.02 \times 10^{-4} \, \text{T} \]

### Final Result
The magnitude of the resultant magnetic field at a point 1.5 cm away from the wire is approximately:

\[ B_R \approx 3.02 \times 10^{-4} \, \text{T} \]