A long straight wire carries a current of $10 A$ directed along the $n e g a t i v e y - a x i s$ , as shown in figure. A uniform magnetic field $B_{0}$ of magnitude $10^{- 6} T$ is directed parallel to the $x - a x i s$ . What is the resultant magnetic field at $x = 0, z = 2 m$ ?

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$1.2 μ T$

$1.7 μ T$

zero

0.5 $μ T$

answer is D.

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Detailed Solution

The magnetic field at a distance of 2m from the wire

$B = \frac{μ_{0}}{2 π} . \frac{i}{r} = 2 \times 10^{- 7} \times \frac{10}{2} = 10^{- 6} T$

The net magnetic field at $x = 0 a n d z = 2 m$

$\vec{B} = B_{1} \hat{i} - B_{2} \hat{i} = 10^{- 6} \hat{i} - 10^{- 6} \hat{i} = 0$

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