JEE Main · 2015 · 04 Apr 2015 · Offline · Magnetic Effects of Current

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Two long current carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle '

θ

' with the vertical. If wires have mass

λ

per unit length then the value of I is: (

g =

gravitational acceleration)

(A) $2 sin θ \frac{π λ gL}{μ _{0} c o s θ}$ ✓ Correct
(B) $2 \frac{π gL}{μ _{0}} tan θ$
(C) $\frac{π λ g L}{μ _{0}} tan θ$
(D) $sin θ \frac{π λ gL}{μ _{0} c o s θ}$

Answer: (A)

Solution

The two wires carry parallel currents, so they attract, and they settle with their threads at angle

θ

to the vertical, a distance

d = 2 L sin θ

apart. The attractive force per unit length is

\frac{μ _{0} I ^{2}}{2 π d}

. Each wire balances under gravity

λ g

per length, the thread tension, and this magnetic pull, so

tan θ = \frac{magnetic force}{weight} = \frac{μ _{0} I ^{2}}{2 π ( 2 L s i n θ ) λ g}

. Solving,

I^{2} = \frac{4 π λ g L s i n ^{2} θ}{μ _{0} c o s θ}

, so

I = 2 sin θ \frac{π λ g L}{μ _{0} c o s θ}

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option

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JEE Main · 2015 · 04 Apr 2015 · Offline · Magnetic Effects of Current

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