91Ó°ÊÓ

$B_{ext}=5.0×{10}^{−3}\text{â€Í¿}$, the field lines are perpendicular to the wire i=100A.

We can use the equation for the field produced by the long current-carrying wire at a point away from the wire. The distance should be such that the field produced by the wire is exactly the same as the given field.

Formula:

Magnetic field due to a long straight wire at distance r from the wire carrying current i

$B_r=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}$

Since the long wire is kept in an external magnetic field, the field due to wire (B_w) and external magnetic field (B_ext) will cancel out when their magnitudes are the same, and the direction is opposite. So, the set points which will satisfy this condition lie on the line parallel to the wire at distance r.

$\begin{array}{c}{B}_r=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{Ï€}r}=B_{ext}\\ r=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{Ï€}{B}_{ext}}\\ r=\frac{1.26×{10}^{−6}×100}{2×\mathrm{Ï€}×5.0×{10}^{−3}}\\ r=4.0×{10}^{−3}\text{″¾}\end{array}$

Hence, the Point on the line parallel to the wire at a distance$r=4.0×{10}^{-3}\text{m}$