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प्रश्न
The figure shows three infinitely long straight parallel current carrying conductors. Find the
- magnitude and direction of the net magnetic field at point A lying on conductor 1,
- magnetic force on conductor 2.
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उत्तर
(i) Magnetic field due to a straight infinite long straight wire carrying current is given by
Current in wire 2 is 3I
\[B_2 = \frac{\mu_o}{4\pi}\frac{2 \times 6I}{r}\]
\[ B_2 = \frac{\mu_o}{4\pi}(\frac{6 I}{r})\]
\[\text { The direction of magnetic field at point A due to wire 2 is in inward direction}.\]
Magnetic field (B3) at point A due to charge carrying wire 3
Current in wire 3 is 4I
\[B_3 = \frac{\mu_o}{4\pi}\frac{2 \times 4 I}{r + 2r}\]
\[ B_3 = \frac{\mu_o}{4\pi}(\frac{8 I}{3r})\]
\[\text { The direction of magnetic field at point A due to wire 3 is in the outward direction } . \]
\[\text { Net Magnetic field at point A }\]
\[B = B_2 - B_3 \]
\[ = \frac{\mu_o}{4\pi}(\frac{3 I}{r}) - \frac{\mu_o}{4\pi}(\frac{8 I}{3r})\]
\[ = \frac{\mu_o}{4\pi}(\frac{I}{3r}) \]
\[\text { The direction of net magnetic field at point A is in inward direction } . \]
(ii)
Magnitude of force on wire 2 due to wire 1
\[F_{21} = \frac{\mu_o}{4\pi} \frac{2 I_2 I_1}{r}\]
\[ = \frac{\mu_o}{4\pi}\frac{2 \times 3I \times I}{r} = \frac{\mu_o}{4\pi}\frac{6 I^2}{r}\]
\[\text { Magnitude of force on wire 2 due to wire } 3\]
\[ F_{23} = \frac{\mu_o}{4\pi} \frac{2 I_2 I_3}{r}\]
\[ = \frac{\mu_o}{4\pi}\frac{2 \times 3I \times 4I}{2r} = \frac{\mu_o}{4\pi}\frac{12 I^2}{r}\]
\[\text { Net force on wire 2 } = F_{23} - F_{21} \]
\[ = \frac{\mu_o}{4\pi}\frac{12 I^2}{r} - \frac{\mu_o}{4\pi}\frac{6 I^2}{r}\]
\[ = \frac{\mu_o}{4\pi}\frac{6 I^2}{r}\]
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