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← JEE Advanced 2014 · Paper 1
PhysicsInteger+3Question 14 of 60

JEE Advanced 2014 · Paper 1, Physics question 14

Two parallel wires in the plane of the paper are distance X0X_0 apart. A point charge is moving with speed uu between the wires in the same plane at a distance X1X_1 from one of the wires. When the wires carry current of magnitude II in the same direction, the radius of curvature of the path of the point charge is R1R_1. In contrast, if the currents II in the two wires have directions opposite to each other, the radius of curvature of the path is R2R_2. If X0X1=3\dfrac{X_0}{X_1}=3, the value of R1R2\dfrac{R_1}{R_2} is

Answer and worked solution

Official answer: 3

Distances. The charge is X1X_1 from one wire and X0−X1=2X1X_0-X_1=2X_1 from the other.

Fields. B=μ0I2πxB=\dfrac{\mu_0I}{2\pi x} from each wire.

  • Same direction: between the wires the fields oppose: B1=μ0I2π(1X1−12X1)=μ0I2π⋅12X1B_1=\dfrac{\mu_0I}{2\pi}\left(\dfrac1{X_1}-\dfrac1{2X_1}\right)=\dfrac{\mu_0I}{2\pi}\cdot\dfrac1{2X_1}.
  • Opposite directions: they add: B2=μ0I2π⋅32X1B_2=\dfrac{\mu_0I}{2\pi}\cdot\dfrac3{2X_1}.

Radii. R=muqBR=\dfrac{mu}{qB}, so
R1R2=B2B1=3.\frac{R_1}{R_2}=\frac{B_2}{B_1}=3.

Answer: 3