MHT CET · Physics · Magnetic Effects of Current
Two parallel conducting wires of equal length are placed distance \(d\) apart carry current \(I_1\) and \(I_2\) respectively in opposite directions. The resultant magnetic field at the midpoint of the distance between both the wires is
- A \(\frac{\mu_0\left(I_1-I_2\right)}{2 \pi d}\)
- B \(\frac{\mu_0\left(I_1+I_2\right)}{2 \pi d}\)
- C \(\frac{\mu_0\left(I_1+I_2\right)}{\pi d}\)
- D \(\frac{\mu_0\left(I_1-I_2\right)}{\pi d}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mu_0\left(I_1+I_2\right)}{\pi d}\)
Step-by-step Solution
Detailed explanation
Field due to a current-carrying long wire is given by
\(B=\frac{\mu_0 I}{2 \pi r}\)
The field at the midpoint between the two wires are in the same direction and hence get added.
The distance of the midpoint \(r=\frac{d}{2}\)
\(\begin{aligned} & \therefore B=B_1+B_2=\frac{\mu_0 I_1}{2 \pi\left(\frac{d}{2}\right)}+\frac{\mu_0 I_2}{2 \pi\left(\frac{d}{2}\right)} \\ & =\frac{\mu_0\left(I_1+I_2\right)}{2 \pi\left(\frac{d}{2}\right)}\end{aligned}\)
\(B=\frac{\mu_0 I}{2 \pi r}\)
The field at the midpoint between the two wires are in the same direction and hence get added.
The distance of the midpoint \(r=\frac{d}{2}\)
\(\begin{aligned} & \therefore B=B_1+B_2=\frac{\mu_0 I_1}{2 \pi\left(\frac{d}{2}\right)}+\frac{\mu_0 I_2}{2 \pi\left(\frac{d}{2}\right)} \\ & =\frac{\mu_0\left(I_1+I_2\right)}{2 \pi\left(\frac{d}{2}\right)}\end{aligned}\)
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