JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
A small cube of side \(1\) mm is placed at the centre of a circular loop of radius \(10\) cm carrying a current of \(2\) A. The magnetic energy stored inside the cube is \(\alpha \times 10^{-14}\) J. The value of \(\alpha\) is _______. (\(\mu_o = 4\pi \times 10^{-7}\) Tm/A, \(\pi = 3.14\))
- A \(6.28\)
- B \(6.28 \times 10^{-6}\)
- C \(628\)
- D \(6.28 \times 10^{-4}\)
Answer & Solution
Correct Answer
(A) \(6.28\)
Step-by-step Solution
Detailed explanation
The magnetic field at the centre of the circular loop is given by \(B = \dfrac{\mu_0 I}{2R}\) Substituting the given values, \(B = \dfrac{4\pi \times 10^{-7} \times 2}{2 \times 0.1} = 4\pi \times 10^{-6}\) T The magnetic energy density is \(u = \dfrac{B^2}{2\mu_0}\)…
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