Question

A long horizontal straight wire \(P\) carrying a current of \(120\,\mathrm{A}\) is fixed and another horizontal straight wire \(Q\) of linear mass density \(1.2\times10^{-2}\,\mathrm{kg\,m^{-1}}\) is placed \(2.5\,\mathrm{cm}\) below wire \(P\). If wire \(Q\) remains suspended in equilibrium in air, then the current through it is \[ \left(g=10\,\mathrm{m\,s^{-2}}\right) \]

• A. \(225\,\mathrm{A}\)
• B. \(75\,\mathrm{A}\)
• C. \(250\,\mathrm{A}\)
• D. \(125\,\mathrm{A}\)

Hint:

For two long parallel wires, \[ \boxed{ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} } \] For equilibrium, \[ \boxed{ \frac{F}{L} = \lambda g. } \]

Answer 1

The Correct Option is D

Step 1: Balance the magnetic force with the weight of the wire. The magnetic force per unit length between two parallel conductors is \[ \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}. \] For equilibrium, \[ \frac{\mu_0 I_1 I_2}{2\pi d} = \lambda g, \] where \[ I_1=120\,\mathrm{A}, \] \[ \lambda=1.2\times10^{-2}\,\mathrm{kg\,m^{-1}}, \] \[ d=2.5\times10^{-2}\,\mathrm{m}. \]

Step 2: Substitute the values. Using \[ \mu_0=4\pi\times10^{-7}\,\mathrm{H\,m^{-1}}, \] \[ \frac{2\times10^{-7}\times120\times I}{2.5\times10^{-2}} = 1.2\times10^{-2}\times10. \] Simplifying, \[ 9.6\times10^{-4}I = 0.12, \] \[ I = 125\,\mathrm{A}. \] Hence, \[ \boxed{125\,\mathrm{A}} \] Therefore, \[ \boxed{(D)} \] is the correct answer.