AP EAMCET · PHYSICS · Magnetic Effects of Current
A long wire lies along \(X\)-axis and carries a current of \(40 \mathrm{~A}\) in positive \(x\)-direction. A second long wire is perpendicular to the \(x y\)-plane, passes through point \((3.0 \mathrm{~m}) \hat{j}\) and carries a current along positive \(z\)-direction. If the magnitude of resultant magnetic field at the point \((2.0 \mathrm{~m}) \hat{j}\) \(R=5 \times 10^{-6} \mathrm{~T}\) then the current in the second wire is
(Permeability of free space, \(\mu_0=4 \pi \times 10^{-7}\) SI unit)
- A \(30 \mathrm{~A}\)
- B \(15 \mathrm{~A}\)
- C \(25 \mathrm{~A}\)
- D \(7.5 \mathrm{~A}\)
Answer & Solution
Correct Answer
(B) \(15 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
Given arrangement is There are two magnetic fields at point \(2 \hat{j}\) as shown \(B_1=\) magnetic field due to wire \(A\). \(=\frac{\mu_0 I_1}{2 \pi d}=\frac{\mu_0 \times 40}{2 \pi \times 2} \mathrm{~T}\) and \(B_2=\) magnetic field due to wire \(B\)…
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