JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Magnetic Effects of Current

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}}

Answer: 3

Solution

Distance from first wire:

X_{1} = \frac{X _{0}}{3}

. Distance from second wire:

X_{2} = X_{0} - X_{1} = \frac{2 X _{0}}{3}

.

Magnetic field from a wire at distance

r

B = \frac{μ _{0} I}{2 π r}

.

Same direction currents: Fields oppose. Net field:

B_{1} = \frac{μ _{0} I}{2 π} (\frac{1}{X _{1}} - \frac{1}{X _{2}}) = \frac{μ _{0} I}{2 π} (\frac{3}{X _{0}} - \frac{3}{2 X _{0}}) = \frac{3 μ _{0} I}{4 π X _{0}}

.

Opposite direction currents: Fields add. Net field:

B_{2} = \frac{μ _{0} I}{2 π} (\frac{1}{X _{1}} + \frac{1}{X _{2}}) = \frac{μ _{0} I}{2 π} (\frac{3}{X _{0}} + \frac{3}{2 X _{0}}) = \frac{9 μ _{0} I}{4 π X _{0}}

.

Radius of curvature:

R = \frac{m u}{q B}

, so

\frac{R _{1}}{R _{2}} = \frac{B _{2}}{B _{1}} = \frac{9}{3} = 3

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Integer answer