Chapter 22, Problem 71PE

To determine

The required current in the top wire in the given figure to produce a field of zero at the point equidistant from the wires, provided currents in the bottom two wires are both 10.0 A into the page.

Answer to Problem 71PE

  $(1+\sqrt{3})5A$ into the page.

Explanation of Solution

Given:

Current carrying wire is placed at the vertices of an equilateral triangle of side 10.0 cm= ${10}^{-1}m$. Magnitude and direction of current in each wire is mentioned in the figure. Dark circle represents current out of page while circle enclose cross represents current into the page.

Formula used:

Magnetic field due to a long current carrying wire at perpendicular distance $d$ from the wire is $B=\frac{{\mu }_{o}I}{2\pi d}$ ----(1)

Where $I$ is the current in the wire and ${\mu }_o$ is permeability in free space.

Calculation:

Label the wires at the vertices of triangle as P, Q and R.The equidistant point from all the three wires is centroid of the triangle. Let call the centroid point as O.

The perpendicular distance from the three wires to centroid of the triangle is

  $d=\frac{2}{3}\sqrt{75}=\frac{10}{\sqrt{3}}cm=\frac{{10}^{-1}}{\sqrt{3}}m=0.06m$

Let the current in P wire be $I$ A coming out of page. The magnetic field at centroid O due to all the three wires are calculated using equation (1) as,

  $B_P=\frac{{\mu }_{o}I}{2\pi d}=2\times {10}^{-7}\times \frac{I}{\frac{{10}^{-1}}{\sqrt{3}}}=2\sqrt{3}I\times {10}^{-6}T=2\sqrt{3}I\mu T$

(Note the value of $\frac{{\mu }_o}{4\pi }={10}^{-7}$ )

  $B_Q=\frac{{\mu }_{o}I}{2\pi d}=2\times {10}^{-7}\times \frac{10}{\frac{{10}^{-1}}{\sqrt{3}}}=2\sqrt{3}\times {10}^{-5}T=20\sqrt{3}\mu T$ and

  $B_R=\frac{{\mu }_{o}I}{2\pi d}=2\times {10}^{-7}\times \frac{10}{\frac{{10}^{-1}}{\sqrt{3}}}=2\sqrt{3}\times {10}^{-5}T=20\sqrt{3}\mu T$

Direction of fields at the centroid is shown by blue arrow in the figure below.

The resultant of vector $B_Q$ and $B_R$ along vector $B_P$ are

  $B=B_Q\cos 60°+B_R\cos 30°=10\sqrt{3}+(10\sqrt{3})\sqrt{3}=(1+\sqrt{3})10\sqrt{3}T$

Now the net magnetic field at the centroid is

  $B_{net}=B+B_P=(1+\sqrt{3})10\sqrt{3}+2\sqrt{3}I=(10+10\sqrt{3}+2I)\sqrt{3}\mu T$

We want zero field at centroid. So, $B_{net}$ =0 at centroid. This gives

  $(10+10\sqrt{3}+2I)\sqrt{3}=0$

Or $I=-(1+\sqrt{3})5A$.

We are getting negative sign because what we have assumed the direction of current in wire P is not right to get zero field at centroid. So the new direction of current in top wire is to be into the page and its magnitude is $(1+\sqrt{3})5A$.

Conclusion:

Current in the top wire to produce a field of zero at the point equidistant from the wires is $(1+\sqrt{3})5A$ into the page.