A long straight wire carrying current of `30 A` is placed in an external unifrom magnetic field of induction `4xx10^(4) T`. The magnetic field is acting parallel to the direction of current. The maggnetic of the resultant magnetic inuduction in tesla at a point `2.0 cm` away form the wire is

Options

A. `10^(4)`
B. `3 xx 10^(-4)`
C. `5 xx 10^(-4)`
D. `6 xx 10^(-4)`

Answer

To find the resultant magnetic induction at a point 2.0 cm away from a long straight wire carrying a current of 30 A, placed in an external uniform magnetic field of induction \(4 \times 10^{-4} \, \text{T}\), we can follow these steps:

### Step 1: Calculate the magnetic field due to the current-carrying wire
The magnetic field \(B_1\) at a distance \(D\) from a long straight wire carrying current \(I\) is given by the formula:
\[ B_1 = \frac{\mu_0 I}{2 \pi D} \]
where:
- \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\) (permeability of free space),
- \(I = 30 \, \text{A}\) (current),
- \(D = 2.0 \, \text{cm} = 0.02 \, \text{m}\) (distance from the wire).

Substituting the values:
\[ B_1 = \frac{(4\pi \times 10^{-7}) \times 30}{2 \pi \times 0.02} \]
\[ B_1 = \frac{4 \times 30 \times 10^{-7}}{2 \times 0.02} \]
\[ B_1 = \frac{120 \times 10^{-7}}{0.04} \]
\[ B_1 = 3 \times 10^{-4} \, \text{T} \]

### Step 2: Identify the direction of the magnetic fields
- The magnetic field \(B_1\) due to the wire will be directed out of the plane (using the right-hand rule).
- The external magnetic field \(B_2 = 4 \times 10^{-4} \, \text{T}\) is acting parallel to the direction of the current.

### Step 3: Calculate the resultant magnetic field
Since \(B_1\) and \(B_2\) are in the same direction, the resultant magnetic field \(B_R\) is given by:
\[ B_R = B_1 + B_2 \]
\[ B_R = 3 \times 10^{-4} + 4 \times 10^{-4} \]
\[ B_R = 7 \times 10^{-4} \, \text{T} \]

### Step 4: Final Result
Thus, the magnitude of the resultant magnetic induction at a point 2.0 cm away from the wire is:
\[ B_R = 7 \times 10^{-4} \, \text{T} \]

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