AP EAMCET · PHYSICS · Magnetic Properties of Matter
A short bar magnet produces a magnetic field of \(6.4 \times 10^{-5} \mathrm{~T}\) at a distance of \(20 \mathrm{~cm}\) from the centre of the magnet on the normal bisector of the magnet. The magnetic field produced by this magnet at a distance of \(40 \mathrm{~cm}\) from the centre of the magnet on the axis, is
- A \(4.8 \times 10^{-5} \mathrm{~T}\)
- B \(3.2 \times 10^{-5} \mathrm{~T}\)
- C \(1.6 \times 10^{-5} \mathrm{~T}\)
- D \(6.4 \times 10^{-5} \mathrm{~T}\)
Answer & Solution
Correct Answer
(C) \(1.6 \times 10^{-5} \mathrm{~T}\)
Step-by-step Solution
Detailed explanation
Magnetic field produced by a magnet on its axis and its equator are given by \(B_{\text {axis }}=\frac{\mu_0}{4 \pi} \cdot \frac{2 M}{r_1^3}\) \(B_{\text {equator }}=\frac{\mu_0}{4 \pi}=\frac{M}{r_2^3}\) \(B_{\text {axis }}=B_{\text {equator }} \times \frac{2 r_2^3}{r_1^3}\)…
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