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An isosceles right angled current carrying uniform magnetic field vector B pointing along pR. on the arm PQ is F, then the magnetic force will be loop PQR is placed in a If the magnetic force acting which acts on the arm QR(A) F(B) \(\frac{F}{√2}\)(C) √2F (D) –F |
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Answer» The correct option is (D) -F. Force along AB is zero as magnetic field is along AB. The net force on a current carrying loop is always zero. Fnet = F on AB → +F on BC→+F Hence on AC, ⟹0 = 0 + F+ (F on AC) Hence, F on AC is −F. |
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