JEE Mains · Physics · STD 12 - Magnetism and matter
- JEE Mains 2013
- Hard
- MCQ
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 \(3.6 \times 10^{-5}\) \(Wbm^{-2}\) )
Pick an option to check your answer.
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 \(M_{2}=1.00 \,A m^{2}\)
\(r=\frac{20}{2}\, c m=0.1\, \mathrm{m}\)
\(\mathrm{B}_{\mathrm{net}}=\mathrm{B}_{1}+\mathrm{B}_{2}+\mathrm{B}_{\mathrm{H}}\)
\(B_{n e t}=\frac{\mu_{0}}{4 \pi} \frac{\left(M_{1}+M_{2}\right)}{r^{3}}+B_{H}\)
\(=\frac{10^{-7}(1.2+1)}{(0.1)^{3}}+3.6 \times 10^{-5}\)
\(=2.56 \times 10^{-4}\, \mathrm{wb} / \mathrm{m}^{2}\)
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