COMEDK · Physics · 21. Magnetic Effects of Current
A long horizontal wire \(\mathrm{P}\) carries current of \(50 \mathrm{~A}\) from left to right. It is rigidly fixed. Another fine wire \(\mathrm{Q}\) is placed directly above and parallel to \(\mathrm{P}\). The mass of the wire is '\(\mathrm{m}\)' \(\mathrm{kg}\) and carries a current of '\(\mathrm{I}\)' A. The direction of current in \(\mathrm{Q}\) and position of wire \(\mathrm{Q}\) from \(\mathrm{P}\) so that the wire \(\mathrm{Q}\) remains suspended are
- A \(\text { Right to left, } \quad \dfrac{25 I \mu_0}{\pi m g}\)
- B \(\text { Right to left, } \quad \dfrac{I \mu_0}{4 \pi m g}\)
- C \(\text { Right to left, } \quad \dfrac{50 I \mu_0}{\pi m g}\)
- D \(\text { Left to right, } \quad \dfrac{25 I \mu_0}{\pi m g}\)
Answer & Solution
Correct Answer
(A) \(\text { Right to left, } \quad \dfrac{25 I \mu_0}{\pi m g}\)
Step-by-step Solution
Detailed explanation
For the wire Q to remain suspended, the magnetic force acting on it must balance the gravitational force acting downwards. The magnetic force per unit length between two parallel wires carrying currents \(I_1\) and \(I_2\) separated by a distance \(r\) is given by…
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