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JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism

Two very long, straight and insulated wires are kept at \(90^o\) angle from each other In \(xy -\) plane as shown in the figure. These wires carry current of equal magnitude \(I\), whose directions are shown in the figure. The net magnetic field at point \(P\) will be

  1. A \(\frac{{{\mu _0}I}}{{2\pi d}}\left( {\hat x + \hat y} \right)\)
  2. B \(\frac{{ + {\mu _0}I}}{{\pi d}}\left( {\hat z} \right)\)
  3. C Zero
  4. D \(-\frac{{{\mu _0}I}}{{2\pi d}}\left( {\hat x + \hat y} \right)\)
Verified Solution

Answer & Solution

Correct Answer

(C) Zero

Step-by-step Solution

Detailed explanation

Magnetic field at point \(P\) \(\overrightarrow{\mathrm{B}}_{\mathrm{net}}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{d}}(-\hat{\mathrm{k}})+\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{d}}(\hat{\mathrm{k}})=0\)
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