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JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism

Two long current carrying thin wires, both with current \(I\), are held by insulating threads oflength \(L\) and are in equilibrium as shown in the figure, with threads making an angle '\(\theta\)' with the vertical. If wires have mass \(\lambda\) per unit length then the value of \(l\) is 
(\(g =\) gravitational acceleration)

  1. A \(2\)\(sin\)\(\theta \sqrt {\frac{{\pi \lambda gL}}{{{\mu _0}cos\theta }}} \;\;\;\;\;\;\;\;\)
  2. B \(2\)\(\sqrt {\frac{{\pi gL}}{{{\mu _0}}}tan\theta } \)
  3. C \(\;\sqrt {\frac{{\pi \lambda gL}}{{{\mu _0}}}tan\theta } \)
  4. D \(sin\)\(\theta \sqrt {\frac{{\pi \lambda gL}}{{{\mu _0}cos\theta }}} \)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2\)\(sin\)\(\theta \sqrt {\frac{{\pi \lambda gL}}{{{\mu _0}cos\theta }}} \;\;\;\;\;\;\;\;\)

Step-by-step Solution

Detailed explanation

Let us consider \('l'\) length of current carrying wire. At equilibrium \(\mathrm{T} \cos \theta=\lambda \mathrm{g} \ell\) and…
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