- Home
- Standard 12
- Physics
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 direction 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.
$3$
$4$
$5$
$6$
Solution
$B _2=\frac{\mu_0 I }{2 \pi x _1}+\frac{\mu_0 I }{2 \pi\left( x – x _1\right)}$ (opposite )
$B_1=\frac{\mu_0 I }{2 \pi x _1}-\frac{\mu_0 I }{2 \pi\left( x – x _1\right)} \text { (same) }$
Case -$1$ When current is in the same direction
$B=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} $
$R _1=\frac{ mv }{ qB }$
Case-$2$ When current is in oposite direction
$B = B _2=\frac{9 \mu_0 I }{4 \pi x _0} $
$R _2=\frac{ mv }{ qB } $
$\frac{ R _1}{ R _2}=\frac{ B _2}{ B _1}=\frac{9}{3}=3$
