There are two infinite parallel current carrying wires in vertical plane. Lower wire is fixed and upper wire is having a linear mass density `lambda`. Two wires are carrying currents `I_(1)` and `I_(2)`. Now upper wire is placed in a magnetic field produced by lower wire. Magnetic field due to lower wire at the location of upper wire is `(mu_(0)I)/(2pid)`, where `I_(1)` current lower wire, `d` separation between wires.
Force on any small portion of upper wire having length `dl` is `dF=(mu_(0)I_(1)I_(2)dl)/(2pid),` where `I_(2)` current in the upper wire. If directions of current in the wires is appropriate then upper wire can be in equilibrium if its weight is balanced by magnetic force. Now answer the following questions.
If upper wire is slightly displaced from its mean position and released it will perform simple harmonic motion. As wire then total mechanical energy of wire

Options

A. Remains constant
B. Decreases
C. We can't say anything about mechanical energy in magnetic field
D. Increases

Answer

A

Explanation

Wire will perform SHM i.e., total energy will constant.