JEE Mains · Physics · STD 12 - 5. Magnetism and matter
Two short bar magnets oflength \(1\ cm\) each have magnetic moments \(1.20\ Am^2\) and \(1.00\ Am^2\) respectively. They are placed on a horizontal table parallel to each other with their \(N\) poles pointing towards the South. They have a common magnetic equator and are separated by a distance of \(20.0\ cm\). The value of the resultand horizontal magnetic induction at the mid-point \(O\) of the line joining their centres is close to (Horizontal component of earth.s magnetic induction is \(3.6 \times 10^{-5}\) \(Wbm^{-2}\) )
- A \(3.6 \times 10^{-5} \) \(Wbm^{-2}\)
- B \(2.56 \times 10^{-4}\) \( Wbm^{-2}\)
- C \(3.50 \times 10^{-4} \) \(Wbm^{-2}\)
- D \(5.80 \times 10^{-4}\) \(Wbm^{-2}\)
Answer & Solution
Correct Answer
(B) \(2.56 \times 10^{-4}\) \( Wbm^{-2}\)
Step-by-step Solution
Detailed explanation
Given \(: M_{1}=1.20 \,A m^{2}\) and \(M_{2}=1.00 \,A m^{2}\) \(r=\frac{20}{2}\, c m=0.1\, \mathrm{m}\) \(\mathrm{B}_{\mathrm{net}}=\mathrm{B}_{1}+\mathrm{B}_{2}+\mathrm{B}_{\mathrm{H}}\) \(B_{n e t}=\frac{\mu_{0}}{4 \pi} \frac{\left(M_{1}+M_{2}\right)}{r^{3}}+B_{H}\)…
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