AP EAMCET · PHYSICS · Magnetic Effects of Current
At a place where the magnitude of the earth's magnetic field is \(4 \times 10^{-5} \mathrm{~T}\), a short bar magnet is placed with its axis perpendicular to the earth's magnetic field direction. If the resultant magnetic field at a point at a distance of 40 cm from the centre of the magnet on the normal bisector of the magnet is inclined at \(45^{\circ}\) with the earth's field, then the magnetic moment of the magnet is
- A \(38.4 \mathrm{Am}^2\)
- B \(51.2 \mathrm{Am}^2\)
- C 12.8 Am \({ }^2\)
- D \(25.6 \mathrm{Am}^2\)
Answer & Solution
Correct Answer
(D) \(25.6 \mathrm{Am}^2\)
Step-by-step Solution
Detailed explanation
\(\mathrm{B}_{\mathrm{e}}=4 \times 10^{-5} \mathrm{~T}, \mathrm{r}=40 \mathrm{~cm}\) Magnetic field at a point on the normal bisector of the magnetic is…
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