Two parallel wires in the plane of the paper are distance apart. A point charge is moving with speed between the wires in the same plane at a distance from one of the wires. When the wires carry current of magnitude in the same direction, the radius o

Question

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 $\frac{X_{0}}{X_{1}} = 3,$ the value of $\frac{R_{1}}{R_{2}}$ is [2014]

Smart Achievers · Accepted Answer

(3)

$\frac{m v^{2}}{R} = q v B \Rightarrow R = \frac{m v}{q B}$

$or R \propto \frac{1}{B}$ $[∵ m, q, v are the same]$

$∴$ $\frac{R_{1}}{R_{2}} = \frac{B_{2}}{B_{1}}$

$or, \frac{R_{1}}{R_{2}} = \frac{\frac{μ_{0}}{4 π} \times 2 I [\frac{1}{X_{1}} + \frac{1}{X_{0} - X_{1}}]}{\frac{μ_{0}}{4 π} \times 2 I [\frac{1}{X_{1}} - \frac{1}{X_{0} - X_{1}}]}$

$= \frac{(X_{0} - X_{1}) + X_{1}}{(X_{0} - X_{1}) - X_{1}} = \frac{X_{0}}{X_{0} - 2 X_{1}}$ $[Given \frac{X_{0}}{X_{1}} = 3]$

$∴$ $\frac{R_{1}}{R_{2}} = \frac{\frac{X_{0}}{X_{1}}}{\frac{X_{0}}{X_{1}} - 2} = \frac{3}{3 - 2} = 3$

Solution

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Solution

Ans. (3) mv2R=qvB⇒R=mvqB or R∝1B [∵ m,q,v are the same] ∴ R1R2=B2B1 or, R1R2=μ04π×2I[1X1+1X0-X1]μ04π×2I[1X1-1X0-X1] =(X0-X1)+X1(X0-X1)-X1=X0X0-2X1 [Given X0X1=3] ∴ R1R2=X0X1X0X1-2=33-2=3

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