A straight horizontal conducting rod of length ' $\mathrm{L}$ ' and mass 'M' is suspended by two vertical…

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)
  1. $\frac{\text { IL }}{\mathrm{Mg}}$
  2. $\frac{\mathrm{Mg}}{\mathrm{IL}^{2}}$
  3. $\frac{\mathrm{Mg}}{\mathrm{I}^{2} \mathrm{~L}}$
  4. $\frac{\mathrm{Mg}}{\mathrm{IL}}$

Solution

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

Asked in: MHT CET 2020 (20 Oct Shift 2)

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