Ncert Exercises & Solutions

A long straight wire carrying a current of 25A rests on a table as shown in the figure. Another wire PQ of length 1m, mass 2.5 g carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise?

Hint: The weight of the wire PQ is balanced by the magnetic force.
Step 1: The magnetic field produced by a long straight wire carrying a current of 25A rests on a table on a small wire:

B = \frac{μ_{0} I}{2 πh}

The magnetic force on small conductor is;

F = BILsinθ = BIL

Step 2: Force applied on PQ balances the weight of the small current-carrying wire.

\begin{array}{l} F = mg = \frac{μ_{0} I^{2} L}{2 πh} \\ h = \frac{μ_{0} I^{2} L}{2 πmg} = \frac{4 π \times 10^{- 7} \times 25 \times 25 \times 1}{2 π \times 2 .5 \times 10^{- 3} \times 9 .8} = 51 \times 10^{- 4} \\ h = 0 .51 cm \end{array}

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