MHT CET · Physics · Magnetic Effects of Current

A straight horizontal conducting rod of length ' \(\mathrm{L}\) ' and mass 'M' is suspended by two vertical wires at its ends. If 'I' is the current passing through the rod, then in order that tension in the wire is zero, the magnetic field set up normal to the conductor is
(Neglect the mass of wire, \(\mathrm{g}=\) acceleration due to gravity)

A \(\frac{\text { IL }}{\mathrm{Mg}}\)
B \(\frac{\mathrm{Mg}}{\mathrm{IL}^{2}}\)
C \(\frac{\mathrm{Mg}}{\mathrm{I}^{2} \mathrm{~L}}\)
D \(\frac{\mathrm{Mg}}{\mathrm{IL}}\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{\mathrm{Mg}}{\mathrm{IL}}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{F}=\mathrm{B} \mathrm{qv}=\mathrm{Bit} \times \frac{\ell}{\mathrm{t}}=\mathrm{Bi} \ell\)
\(\mathrm{F}=\mathrm{BIL}=\mathrm{Mg}\)
\(\therefore \mathrm{B}=\frac{\mathrm{Mg}}{\mathrm{IL}}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From MHT CET

Explore more questions on app