A wire carries a current of 10 A in south to north direction. The mag…

The correct answer is Option (4) → \( 1.8 \times 10^{-9}\,\mathrm{T} \) downwards

A 1 cm piece is a small current element, so the field is given by the Biot-Savart law:

    \( \displaystyle dB = \frac{\mu_0}{4\pi}\,\frac{I\,dl\,\sin\theta}{r^{2}} \)

The current points north and the point lies to the north-east, so the angle between \( d\vec{l} \) and \( \vec{r} \) is \( 45^\circ \).

    \( I = 10\,\mathrm{A} \)
    \( dl = 1\,\mathrm{cm} = 0.01\,\mathrm{m} \)
    \( r = 200\,\mathrm{cm} = 2\,\mathrm{m} \)
    \( \mu_0/4\pi = 10^{-7}\,\mathrm{T\,m\,A^{-1}} \)

    \( \displaystyle dB = \frac{10^{-7} \times 10 \times 0.01 \times \sin 45^\circ}{2^{2}} = \frac{10^{-8} \times 0.707}{4} \approx 1.8\times10^{-9}\,\mathrm{T} \)

The direction is that of \( d\vec{l} \times \vec{r} \). Take east as \( \hat{\imath} \), north as \( \hat{\jmath} \) and vertically up as \( \hat{k} \):

    \( \displaystyle d\vec{l} \times \vec{r} \propto \hat{\jmath} \times (\hat{\imath} + \hat{\jmath}) = \hat{\jmath} \times \hat{\imath} = -\hat{k} \)

So the field at the point is vertically downwards.