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78P

Expert-verifiedFound in: Page 863

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A long wire carrying 100A is perpendicular to the magnetic field lines of a uniform magnetic field of magnitude 5.0 mT. At what distance from the wire is the net magnetic field equal to zero?**

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

${B}_{ext}=5.0\times {10}^{-3}\text{\hspace{0.17em}T}$, 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

${B}_{r}=\frac{{\mu}_{0}i}{2\pi r}$

** **

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.

$\begin{array}{c}{B}_{r}=\frac{{\mu}_{0}i}{2\pi r}={B}_{ext}\\ r=\frac{{\mu}_{0}i}{2\pi {B}_{ext}}\\ r=\frac{1.26\times {10}^{-6}\times 100}{2\times \pi \times 5.0\times {10}^{-3}}\\ r=4.0\times {10}^{-3}\text{\hspace{0.17em}m}\end{array}$

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

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