Question

A long straight wire carrying a current of $30A$ is placed in an external uniform magnetic field of induction $4\times 10^{-4}T$ . The direction of external magnetic field is along the direction of the current. The magnitude of the resultant magnetic induction at a point $2.0cm$ away from the wire is $n\times 10^{-4}T$ . What is the value of $n$ ?

Answer 1

Answer: 5

Magnetic field due to wire
$B=\frac{{\mu }_{0}I}{2\pi r}=\frac{4\pi \times 10^{-7}}{2\pi }\times \frac{30}{2\times 10^{-2}}$
$=3\times 10^{-4}T$
This magnetic field will be perpendicular to external magnetic field.
$\therefore$ Net magnetic field
$B=\sqrt{B^2+B_0^2}$
$=\sqrt{{\left(3\times 10^{-4}\right)}^2+{\left(4\times 10^{-4}\right)}^2}$
$=5\times 10^{-4}T$