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कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

A loop, made of straight edges has six corners at A(0, 0, 0), B(L, O, 0) C(L, L, 0), D(0, L, 0) E(0, L, L) and F(0, 0, L). A magnetic field B=Bo(i^+k^)T is present in the region. - Physics

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प्रश्न

A loop, made of straight edges has six corners at A(0, 0, 0), B(L, O, 0) C(L, L, 0), D(0, L, 0) E(0, L, L) and F(0, 0, L). A magnetic field `B = B_o(hati + hatk)`T is present in the region. The flux passing through the loop ABCDEFA (in that order) is ______.

विकल्प

  • `B_o L^2  Wb`

  • `2B_o L^2  Wb`

  • `sqrt(2) B_o L^2  Wb`

  • `4B_o L^2  Wb`

MCQ
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उत्तर

A loop, made of straight edges has six corners at A(0, 0, 0), B(L, O, 0) C(L, L, 0), D(0, L, 0) E(0, L, L) and F(0, 0, L). A magnetic field `B = B_o(hati + hatk)`T is present in the region. The flux passing through the loop ABCDEFA (in that order) is `underline(2B_o L^2  Wb`).

Explanation:

In this problem first we have to analyse area vector, loop ABCDA lies in x-y  plane whose area vector `vecA_1 = L^2  hatk` whereas loop ADEFA lies in y-z plane whose area vector `vecA_2 = L^2 hati`

And the magnetic flux is `phi_m = vecB * vecA`

`vecA = vecA_1 + vecA_2 = (L^2  hatk + L^2 hati)`

And `vecB = B_0(hati + hatk)`

Now, `phi_m = vecB * vecA = B_0(hati + hatk)*(L^2  hatk + L^2hati)`

= `2B_0L^2  Wb`

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अध्याय 6: Electromagnetic Induction - MCQ I [पृष्ठ ३३]

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एनसीईआरटी एक्झांप्लर Physics [English] Class 12
अध्याय 6 Electromagnetic Induction
MCQ I | Q 6.02 | पृष्ठ ३३

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