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In Figure 20.115 two long straight wires carrying a large conventional current I are connected by one-and-a-quarter turns of wire of radius $\mathit{R}$. An electron is moving to the right with speed v at the instant that it passes through the center of the arc. You apply an electric field $\stackrel{\mathbf{→}}{\mathbf{E}}$at the center of the arc in such a way that the net force on the electron at this instant is zero. (You can neglect the gravitational force on the electron, which is easily shown to be negligible, and the magnetic field of the coil is much larger than the magnetic field of the Earth.)

Determine the direction and magnitude of the electric field . Be sure to explain your work fully; draw and label any vectors you use.

Step by step solution

01

Given data

Circular loop having radius R

Speed of electron v

02

Concept/ Formula used

The magnetic field runs parallel to the wire in a perpendicular direction. The direction in which the fingers would curl if you wrapped your right hand's fingers around the wire with your thumb pointing in the direction of the current would indicate the direction of the magnetic field.

03

Calculation for Electric field ,Magnetic field at center

$\begin{array}{c}\stackrel{→}{B}=\frac{{\mathrm{μ}}_{0}I}{2R}\left(1+\frac{1}{4}\right)\\ =\frac{5{\mathrm{μ}}_{0}I}{8R}\end{array}$

Magnetic field due ti straight part of current carrying wire is zero.

$\stackrel{→}{B}=\frac{5{\mathrm{μ}}_{0}I}{8R}\left(−\widehat{k}\right)$

For net force on electron to be zero

$\begin{array}{l}{\stackrel{→}{F}}_E+{\stackrel{→}{F}}_M=0\\ q\stackrel{→}{E}+q\left(\stackrel{→}{V}×\stackrel{→}{B}\right)=0\\ \stackrel{→}{E}=−\left(\stackrel{→}{V}×\stackrel{→}{B}\right)\\ =−\left(V\text{ }\widehat{i}\right)×\left(\frac{5{\mathrm{μ}}_{0}I}{8R}\left(−\widehat{k}\right)\right)\end{array}$

$\begin{array}{c}\stackrel{→}{E}=\frac{5{\mathrm{μ}}_{0}I}{8R}\left(\widehat{i}×\widehat{k}\right)\\ =\frac{5{\mathrm{μ}}_{0}I}{8R}\left(−\widehat{j}\right)\end{array}$

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