Q145 / 160 · TS EAMCET 2003 · Physics
A long straight wire carrying a current of $30 \mathrm{~A}$ 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 $2.0 \mathrm{~cm}$ away from the wire is
$\left(\mu_0=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}\right)$
- A(a) $10^{-4}$
- B$3 \times 10^{-4}$
- Correct answerC$5 \times 10^{-4}$
- D$6 \times 10^{-4}$
Solution
$i=30 \mathrm{~A}, B=4 \times 10^{-4} \mathrm{~T}$
$r=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}$
Initial magnetic field (parallel to the wire),
$B_1=4 \times 10^{-4} \mathrm{~T}$
Magnetic field produced by the straight wire,
$B_2=\frac{\mu_0 i}{2 \pi r}=\frac{2 \times 10^{-7} \times 30}{2 \times 10^{-2}}$
$=3 \times 10^{-4} \mathrm{~T}$
$B_2$ will be in the plane perpendicular to the plane of wire, so $B_1$ and $B_2$ are perpendicular to each other.
$\therefore$ Resultant magnetic field,
$B=\sqrt{B_1^2+B_2^2}$
$=\sqrt{\left(4 \times 10^{-4}\right)^2+\left(3 \times 10^{-4}\right)^2}$
$=5 \times 10^{-4}$
Hence, the correct answer is option: 'C'.