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A wire PQ is carrying a steady current of 15A and is lying on the table. Another wire RS carrying current 6A is held vertically above PQ at a height of 2mm.  Find the mass per unit length of the wire RS so that it remains suspended at the position when left Free. Give the direcion of the current flowing RS with respect to that in PQ, g=10ms^-^2

Option: 1

9 \times 10^{-4} \mathrm{kgm}^{-1}


Option: 2

7.5 \times 10^{-2} \mathrm{kgm}^{-1}


Option: 3

8 \times 10^{-2} \mathrm{kgm}^{-1}


Option: 4

9.6 \times 10^{-2} \mathrm{kgm}^{-1}


Answers (1)

Weight per unit length of upper wine= Magnetic force per unit length

\Rightarrow \frac{m g}{l}=\frac{\mu_0}{4 \pi} \frac{2 I_1 I_2}{r}

Mass per unit length

=\frac{m}{l}=\frac{\mu_0}{4 \pi} \frac{2 I_1 I_2}{r g}

=\begin{aligned} \frac{10^{-7} \times 2 \times 15 \times 6}{\left(2 \times 10^{-3}\right) \times 10}=\frac{180}{2} \times 10^{-5} \\ =90 \times 10^{-5}=9 \times 10^{-4} \end{aligned}

Posted by

Ramraj Saini

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