Figure 6 below shows a long straight conductor X carrying a current . A point P is at a…
Figure 6 below shows a long straight conductor X carrying a current $I_{1}$. A point P is at a perpendicular distance '$r_{1}$' from it.
(i)[1.5]
How much force conductor X exerts on a short conductor Y of length 'l' which is carrying a current 'I' and is kept at point P parallel to the conductor X?
(ii)[1.5]
Another long wire Z carrying a current $I_{2}$ is now kept parallel to X and Y at a distance '$r_{2}$' from Y such that the conductor Y remains at rest. Obtain the relation between the currents $I_{1}$ and $I_{2}$.
Answer
Answer (i)
AIWritten by AI - it can contain mistakes.
The magnetic field at point P produced by long conductor X carrying current $I_1$ is:
$B_1 = \frac{\mu_0 I_1}{2\pi r_1}$
The magnitude of force exerted on conductor Y of length $l$ carrying current $I$ is:
$F = I l B_1 = \frac{\mu_0 I_1 I l}{2\pi r_1}$
The force is attractive, directed towards conductor X.
Answer (ii)
AIWritten by AI - it can contain mistakes.
For conductor Y to remain at rest, the attractive force exerted by conductor X on Y (directed towards X) must be equal and opposite to the attractive force exerted by conductor Z on Y (directed towards Z):
$F_{XY} = F_{ZY}$
$\frac{\mu_0 I_1 I l}{2\pi r_1} = \frac{\mu_0 I_2 I l}{2\pi r_2}$
Cancelling common terms $\frac{\mu_0 I l}{2\pi}$:
$\frac{I_1}{r_1} = \frac{I_2}{r_2} \implies \frac{I_1}{I_2} = \frac{r_1}{r_2}$
From ISC 2026 Physics Paper 1, question 12.