JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2015
- Hard
- MCQ
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)
Pick an option to check your answer.
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 \(T \sin \theta=\frac{\mu_{0}}{2 \pi} \frac{I \times I l}{2 L \sin \theta}\left[\because \frac{F_{B}}{\ell}=\frac{\mu_{0}}{4 \pi} \frac{2 I \times I}{2 \ell \sin \theta}\right]\)
Therefore, \(I=2 \sin \theta \sqrt{\frac{\pi \lambda g L}{u_{0} \cos \theta}}\)
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