Tiêu đề 06. ELECTROMAGNETIC INDUCTION(H).pdf ✅

NEET REVISION 15. () : Explana on The perpendicular length of the triangular area en‐ closed by the structure and bar is the same as the distance travelled by it in me , where . Here the "base" of that triangle (the distance be‐ tween the intersec on points of the bar with the structure) is . Thus, the area of the triangle is Since the field is a uniform , then the magnitude of the flux (in units) is The magnitude of the emf is the (absolute value of) Faraday's law: At , this yields . 16. () : Explana on When the switch has been closed for a long me, ba ery, resistor and coil carry constant current . When the switch is opened, current in ba ery and resistor drops to zero, but the coil car‐ ries this same current for a moment as oscilla ons begin in the loop. We interpret the problem to mean that the voltage amplitude of these oscilla‐ ons is , in . Then, 17. () : Explana on Method - 1 Consider a small element on the rod The emf across the element is Tan Method - II Join the two rods of the semi-infinite rod to the centre. With conduc ng wires the total flux through , where is at the other end of the rod, is As one-fourth of the circular area is present inside the triangle 18. () : Explana on EMF induces across the length of the wire which cuts the magne c field. Length of Length of Length of Length of ). 19. () : Explana on When is closed and is open, when is open and is closed, current as a func on of me ' ' in circuit is t : d = vt v = 5.0 m/s 2d A = ( base )( height ) = (2vt)(vt) = v 2t 2 1 2 1 2 B = 0.350 T SI φB = BA = (0.40)(5.0) 2 t 2 = 10.0t 2 ε = = 10.0 = 20.0 t dΦB dt dt 2 dt t = 4.0 s ε = 80.0 V Ii = ε R LC ΔV C(ΔV ) 2 = LI 2 i 1 2 1 2 L = = C(ΔV ) 2 I 2 i C(ΔV ) 2R2 ε 2 = (1.0×10−6)(100) 2 (200) 2 (50) 2 = 0.16 H = 160 mH dx dε = Ecos θdx cos θ = , Tan θ = ⇒ x = R R r x R θ dx = R sec 2 θdθ E = = R 2 2r dB dt R 2 α 2r dε = ( ) Rsec 2θdθ dε = sec 2 θdθ dε = ( ) 2 dθ ε = ∫ π/2 0 dθ = R 2 α 2r R r R 4 α 2r 2 R 4 α 2r 2 r R R 2α 2 R 2απ 4 OAB B φ = ( ) R2 B π 4 OAB εind = = = dφ dt πR2 4 dB dt πR2α 4 c =( d) >( a = b So (εc = εd) > (εa = εb) K1 K2 I0 = E R K1 K2 t LR I = I0e −R = e −5 = = 0.67 mA t L 1 10 1 1500