JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
Two long parallel wires \(X\) and \(Y\), separated by a distance of 6 cm , carry currents of 5A and 4A, 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 \(x \times 10^{-5} \mathrm{~T}\). The value of \(x\) is__________ . Take permeability of free space as \(\mu_0=4 \pi \times 10^{-7} \mathrm{SI}\) units.
- A 5
- B 4
- C 3
- D 1
Answer & Solution
Correct Answer
(D) 1
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & B=\frac{\mu_0(5)}{2 \pi \times .01}-\frac{\mu_0 4}{2 \pi \times 0.04} \\ & =-\frac{100 \mu_0}{4 \pi} \\ & =-100 \times 10^{-7} \\ & =-1 \times 10^{-5} \mathrm{~T}\end{aligned}\)
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