Question

A horizontal metallic rod of mass $m$ and length $l$ is supported by two vertical identical springs of spring constant, $K$ each and natural length $l_{0}$ . A current $i$ is flowing in the rod in the direction shown. If the rod is in equilibrium then the length of each spring in this state is:

• A. $l_{0}+\frac{i l B - m g}{K}$
• B. $l_{0}+\frac{\textit{ilB} - \textit{mg}}{2 K}$
• C. $l_{0}+\frac{m g - i l B}{2 K}$
• D. $l_{0}+\frac{m g - i l B}{K}$

Answer 1

Answer: (B)

The force on the rod due to the magnetic field and gravity is
$ilB-mg\left(\right.upwards\left.\right)$
Hence the extension in the springs is $\frac{i l B - m g}{2 k}$
$\left(\right.Note$ that effective spring constant is 2k)
Therefore the length of the spring is $l_{0}+\frac{i l B - m g}{2 k}$