A long straight wire carrying current of 30 A is placed in an external uniform magnetic field of induction $4\times {10}^{-4}\mathrm{T}$. The magnetic field is acting parallel to the direction of current. The magnitude of the resultant magnetic induction in tesla at a point 2.0 cm away form the wire is

Given data:

In the presence of an external uniform magnetic field with induction $4\times {10}^{-4}T$, a long straight wire is carrying a current of $30 A$ in the same direction as the magnetic field.

Formula used: $B=\frac{{\mu }_{0}i}{2\pi r}$

Detailed Solution:

From the question, we get as magnetic field due to the wire is perpendicular to the applied field.

$$\begin{gathered} B_1=\frac{{\mu }_{0}i}{2\pi r} \\ =\frac{\left(2\right)\left({10}^{-7}\right)\left(30\right)}{2\left({10}^{-2}\right)} \\ =3\left({10}^{-4}\right)T \end{gathered}$$

Hence, the magnetic induction B is 

$$\begin{gathered} B=\sqrt{{B_1}^2+{B_2}^2} \\ =\sqrt{3^2+4^2}\times {10}^{-4} \\ =5\times {10}^{-4} \end{gathered}$$

Ans: Therefore, the magnetic induction will be $5\times {10}^{-4}$.