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TS EAMCET · Physics · Magnetic Properties of Matter

Two short bar magnets have their magnetic moments \(1.2 \mathrm{Am}^2\) and \(1.0 \mathrm{Am}^2\). They are placed on a horizontal table parallel to each other at a distance of \(20 \mathrm{~cm}\) between their centres, such that their north poles pointing towards geographic south. They have common magnetic equatorial line. Horizontal component of earth's field is \(3.6 \times 10^{-5} \mathrm{~T}\). Then, the resultant horizontal magnetic induction at mid point of the line joining their centers is \(\left(\frac{\mu_0}{4 \pi}=10^{-7} \mathrm{~N} / \mathrm{m}\right)\)

  1. A \(3.6 \times 10^{-5} \mathrm{~T}\)
  2. B \(1.84 \times 10^{-4} \mathrm{~T}\)
  3. C \(2.56 \times 10^{-4} \mathrm{~T}\)
  4. D \(5.8 \times 10^{-5} \mathrm{~T}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(2.56 \times 10^{-4} \mathrm{~T}\)

Step-by-step Solution

Detailed explanation

We knows, \(B=\frac{\mu_0}{4 \pi} \frac{M}{r^3}\) Hence the resultant horizontal magnetic induction point of the line joining their conters is…
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