Magnetics 2 Question 20
20. Two parallel wires in the plane of the paper are distance $x _0$ apart. A point charge is moving with speed $u$ between the wires in the same plane at a distance $x _1$ from one of the wires. When the wires carry current of magnitude I in the same direction, the radius of curvature of the path of the point charge is $R _1$. In contrast, if the currents I in the two wires have directions opposite to each other, the radius of curvature of the path is $R _2$. If $\frac{x _0}{x _1}=3$, and value of $\frac{R _1}{R _2}$ is
(2014 Adv.)
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Answer:
Correct Answer: 20. 3
Solution:
- $B _2=\frac{\mu _0 I}{2 \pi x _1}+\frac{\mu _0 I}{2 \pi\left(x _0-x _1\right)}$ (when currents are in opposite directions)
$$ B _1=\frac{\mu _0 I}{2 \pi x _1}-\frac{\mu _0 I}{2 \pi\left(x _0-x _1\right)} $$
(when currents are in same direction)
$$ \begin{aligned} \text { Substituting } x _1 & =\frac{x _0}{3} \quad\left(\text { as } \frac{x _0}{x _1}=3\right) \\ B _1 & =\frac{3 \mu _0 I}{2 \pi x _0}-\frac{3 \mu _0 I}{4 \pi x _0}=\frac{3 \mu _0 I}{4 \pi x _0} \\ R _1 & =\frac{m v}{q B _1} \text { and } B _2=\frac{9 \mu _0 I}{4 \pi x _0} \\ R _2 & =\frac{m v}{q B _2} \\ \Rightarrow \quad \frac{R _1}{R _2} & =\frac{B _2}{B _1}=\frac{9}{3}=3 \end{aligned} $$
