TS EAMCET · Physics · Magnetic Properties of Matter
The magnitude of axial field due to a bar magnet at a distance of \(1 \mathrm{~m}\), is found to be \(5 \times 10^{-8} \mathrm{~T}\). The magnetic moment of the bar magnet is \(\left(\mu_0=4 \pi \times 10^{-7}\right)\)
- A \(0.20 \mathrm{Am}^2\)
- B \(0.25 \mathrm{Am}^2\)
- C \(0.50 \mathrm{Am}^2\)
- D \(0.40 \mathrm{Am}^2\)
Answer & Solution
Correct Answer
(B) \(0.25 \mathrm{Am}^2\)
Step-by-step Solution
Detailed explanation
We have \(\begin{aligned} & \mathrm{B}_{\text {axial }}=\frac{\mu_0}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{d}^3} \\ & \Rightarrow \quad 5 \times 10^{-8}=10^{-7} \times \frac{2 \times \mathrm{M}}{1^3} \\ & \Rightarrow \quad \mathrm{M}=0.25 \mathrm{Am}^2\end{aligned}\)
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