Q151 / 160 · AP EAMCET 2023 (17 May Shift 2) · Physics

The axial field $\left(\mathrm{B}_{\mathrm{A}}\right)$ and the equatorial field $\left(\mathrm{B}_{\mathrm{E}}\right)$ due to a short bar magnet at equal distances are related as

  • Correct answer
    A
    $\mathrm{B}_{\mathrm{A}}=2 \mathrm{~B}_{\mathrm{E}}$
  • B
    $\mathrm{B}_{\mathrm{A}}=-2 \mathrm{~B}_{\mathrm{E}}$
  • C
    $\mathrm{B}_{\mathrm{A}}=-\mathrm{B}_{\mathrm{E}}$
  • D
    $\mathrm{B}_{\mathrm{A}}=-2 \pi \mathrm{B}_{\mathrm{E}}$

Solution

$\begin{aligned} & \mathrm{B}_E=\frac{\mu_0 \mathrm{~m}}{4 \pi \mathrm{r}^3} \\ & \mathrm{~B}_{\mathrm{A}}=\frac{2 \mu_0 \mathrm{~m}}{4 \pi \mathrm{r}^3} \\ & \Rightarrow \frac{\mathrm{B}_{\mathrm{E}}}{\mathrm{B}_{\mathrm{A}}}=\frac{1}{2} \Rightarrow \mathrm{B}_{\mathrm{A}}=2 \mathrm{~B}_{\mathrm{E}}\end{aligned}$

Hence, the correct answer is option: 'A'.

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