AP EAMCET · PHYSICS · Magnetic Effects of Current
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
- 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 \(\mathrm{B}_{\mathrm{A}}=-2 \pi \mathrm{B}_{\mathrm{E}}\)
Answer & Solution
Correct Answer
(A) \(\mathrm{B}_{\mathrm{A}}=2 \mathrm{~B}_{\mathrm{E}}\)
Step-by-step Solution
Detailed explanation
\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…
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