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JEE Main 202621 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Evening Shift
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Main 2026 · 21 January · Evening Shift · Physics · Magnetic Effects of Current

An infinitely long straight wire carrying current I is bent in a...

An infinitely long straight wire carrying current II is bent in a planer shape as shown in the diagram. The radius of the circular part is rr. The magnetic field at the centre OO of the circular loop is :
An infinitely long straight wire carrying current I is bent in a planer shape as shown in the diagram. The radius of

Options

  1. A−μ02πIr(π−1)i^-\frac{\mu_{0}}{2 \pi} \frac{I}{r}(\pi-1) \hat{i}
  2. Bμ02πIr(π−1)i^\frac{\mu_{0}}{2 \pi} \frac{I}{r}(\pi-1) \hat{i}
  3. C−μ02πIr(π+1)i^-\frac{\mu_{0}}{2 \pi} \frac{I}{r}(\pi+1) \hat{i}
  4. Dμ02πIr(π+1)i^\frac{\mu_{0}}{2 \pi} \frac{I}{r}(\pi+1) \hat{i}

Correct answer

A. −μ02πIr(π−1)i^-\frac{\mu_{0}}{2 \pi} \frac{I}{r}(\pi-1) \hat{i}

Step-by-step solution

An infinitely long straight wire carrying current I is bent in a planer shape as shown in the diagram. The radius of
B→0=B→AB+B→DE+B→BCD\overrightarrow{\mathrm{B}}_0=\overrightarrow{\mathrm{B}}_{\mathrm{AB}}+\overrightarrow{\mathrm{B}}_{\mathrm{DE}}+\overrightarrow{\mathrm{B}}_{\mathrm{BCD}}
=μ0i4πri^+μ0i4πri^−μ0i2ri^=\frac{\mu_0 \mathrm{i}}{4 \pi \mathrm{r}} \hat{\mathrm{i}}+\frac{\mu_0 \mathrm{i}}{4 \pi \mathrm{r}} \hat{\mathrm{i}}-\frac{\mu_0 \mathrm{i}}{2 \mathrm{r}} \hat{\mathrm{i}}
=μ0i2πri^−μ0i2ri^=\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}} \hat{\mathrm{i}}-\frac{\mu_0 \mathrm{i}}{2 \mathrm{r}} \hat{\mathrm{i}}
=μ0i2πr(1−π)i^=\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}(1-\pi) \hat{\mathrm{i}}
=−μ0i2πr(π−1)i^=-\frac{\mu_0 \mathrm{i}}{2 \pi \mathrm{r}}(\pi-1) \hat{\mathrm{i}}
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