JEE Main 2015PhysicsElectrostaticsMediumMCQ

JEE Main 2015 — Electrostatics Question with Solution

JEE Main 2015 (04 Apr)

Question



Two long currents 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 a mass λ per unit length then the value of I is:
(g= gravitational acceleration)

Choose an option

Show full solutionCorrect option: C
Correct answer
C2sinθπλgLμ0cosθ

Step-by-step explanation



Two wires will repel each other due to magnetic force, then the magnetic force per unit length is, dfdl=μ0I22π2Lsinθ=μ0I24πLsinθ. And mass per unit length of each wire=dmdl=λ. So, the magnetic force on the total length L of the wire is fm=μ0I2L4πLsinθ, and weight =λLg. By equilibrium of wire,  Tsinθ=fm  &  Tcosθ = W = λl'g⇒TsinθTcosθ=fmmg⇒fm=λl'gtanθ
⇒μ0I24πLsinθL=λLgsinθcosθ

⇒I2=λgπLμ0cosθ4sin2θ⇒I=2sinθλπgLμ0cosθ

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About this question

This is a previous-year question from JEE Main 2015, covering the Electrostatics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.