đưa nam chân gần vào khung=> Từ thông tăng=> \(B_0\) ngược chiều \(B_c\)=> theo quy tắc nắm tay phải chiều ADCB

I=\(\dfrac{P}{U}\)= \(\dfrac{50000}{500}\)=100(A)

\(P_{hp}\)=\(I^2\)R= \(100^2\).4= 40,000(W)

\(U_{giam}\)= IR= 100.4= 400(V)

H= (\(P_{tp}\)-\(P_{hp}\))/\(P_{tp}\)= 20%

\(U_{tieuthu}\)= U- \(U_{giam}\)= 500 - 400 = 100(V)

b) \(U_{thucap}\)= 10. \(U_{socap}\)= 10.500= 5000(V)

\(I_{moi}\)=\(\dfrac{P}{U_{thu}}\)=\(\dfrac{50000}{5000}\)= 10(A)

P hao phi moi=\(I^2_{moi}\).R= \(10^2\).4=400(W)

U giam moi= \(I_{moi}\).R=10.4=40(V)

H=(\(P_{tp}\)-\(P_{hpmoi}\))/\(P_{tp}\)=92%

\(U_{moi}\)=\(U_{thu}\)- \(U_{giam}\)moi=5000-40=4960(V)

A) I= \(\dfrac{q}{t}\)=\(\dfrac{N.e}{t}\)=\(\dfrac{10^8.1,6.10^{-19}}{1}\)=0,6(A)

B) F=BIL\(\sin\)α=\(5.10^{-3}\).0,5.0,16.\(\sin90\)= 4.\(10^{-4}\)(N)