Two long parallel wires X and Y, separated by a distance of 6 cm, carry currents of 5 A and 4 A, respectively, in opposite directions as shown in the figure. Magnitude of the resultant magnetic field at point P at a distance of 4 cm from wire Y is \( 3 \times 10^{-5} \) T. The value of \( x \), which represents the distance of point P from wire X, is ______ cm. (Take permeability of free space as \( \mu_0 = 4\pi \times 10^{-7} \) SI units.)
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When calculating the magnetic field due to two parallel currents, use the formula \( B = \frac{\mu_0 I}{2\pi r} \) for each wire. Add the magnetic fields vectorially if they are in opposite directions.
The magnetic field produced by a current-carrying wire at a distance \( r \) is determined by the formula:\[B = \frac{\mu_0 I}{2\pi r}.\]At point P, the magnetic fields originating from wires X and Y must be combined vectorially due to their opposing directions. The magnetic field from wire Y at point P is calculated as:\[B_Y = \frac{\mu_0 I_Y}{2\pi r_Y},\]with \( I_Y = 4 \, \text{A} \) and \( r_Y = 4 \, \text{cm} = 0.04 \, \text{m} \). This results in:\[B_Y = \frac{4\pi \times 10^{-7} \times 4}{2\pi \times 0.04} = 2 \times 10^{-5} \, \text{T}.\]Similarly, the magnetic field from wire X at point P is given by:\[B_X = \frac{\mu_0 I_X}{2\pi r_X},\]where \( I_X = 5 \, \text{A} \) and \( r_X = 6 \, \text{cm} = 0.06 \, \text{m} \). Therefore:\[B_X = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.06} = \frac{10^{-5}}{0.06} = 1.67 \times 10^{-5} \, \text{T}.\]The aggregate magnetic field at point P is the sum of these fields:\[B_{\text{total}} = B_X + B_Y = 1 \times 10^{-5} \, \text{T} + 2 \times 10^{-5} \, \text{T} = 3 \times 10^{-5} \, \text{T}.\]Consequently, \( x = 1 \, \text{m} \).The definitive answer is \( \boxed{1} \).
