(A) 6.32 (B) 9.49 (C) (D) (E) 18.97

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1 Univesity of Bhin Physics 10 Finl Exm Key Fll 004 Deptment of Physics 13/1/005 8:30 10:30 e = C, m e = Kg, m p = Kg k=910 9 Nm /C, ε 0 = C /Nm, µ 0 =4π10 7 T.m/A Pt : 10 MCQ (5 mks ech) Q1 Two identicl chges e plced t distnce = 0.3 m s shown in the figue. f the mgnitude of the electic foce etween them is 4 N, then the vlue of chge in (µc) is: Q Q (A) 6.3 (B) 9.49 (C) 1.65 (D) (E) Q A positive point chge q is plced t the cente of inveted hemispheicl sufce ( hlf spheicl shell) s shown in the figue. The electic flux though the hemispheicl sufce is: q (A) Zeo (B) q/ε 0 (C) q/ε 0 (D) q/3ε 0 (E) q/4ε 0 Q3 A poton is cceleted fom est t point A. f the chge eches speed of v B =110 5 m/s t point B, then the electic potentil diffeence V in (V) etween points A nd B is: Stts fom est (A) (B) 08.8 (C) (D) (E) 51.9 p A p B v B Q4 n the cicuit shown, if C= 8 µf then the equivlent cpcitnce (in µf) etween the two points nd is; C C C (A) 4 (B) 8 (C) 1 (D) 16 (E) 0 C Q5 A coppe wie cuies cuent of 3 A t 0 C. f the tempetue coefficient of coppe is α=0.004 C 1, then the cuent (in A) in the wie when its tempetue inceses to 100 C is: (Assume tht the voltge supplied to the wie emins the sme) (A) 0.76 (B) 1.5 (C).7 (D) 3.03 (E)

2 5Ω Q6 n the cicuit shown, if ε = 1 V, then the vlue of potentil diffeence (in V) etween the two points nd is: 0V ε 10Ω 5Ω (A) (B) 4 (C) 6 (D) 8 (E) 10 6 Q7 A chge q = µc is moving with velocity v / = [ 3 iˆ 4 ˆj ] 10 m/s. f the chge entes mgnetic field B = 0.4kˆ T, then the mgnitude of mgnetic foce (in N) on the chge is: (A) 4 (B) 8 (C) 1 (D) 16 (E) 0 Q8 Two long, stight, pllel wies cies cuents 1 =3 nd =, oth diected out of the pge s shown in the figue. f the wies e septed y distnce then the mgnitude of the net mgnetic field t point P tht is locted t distnce fom is given y; 1 P (A) Zeo (B) µ 0 / (C) µ 0 / (D) µ 0 /4 (E) 3µ 0 / Q9 A cuent entes ing of dius R = 0.m. The cuent splits into unequl potions 1 =/4 nd =3/4 s shown in the figue. f = 5 A, then the mgnitude of the net mgnetic field (in µt) t the cente of the ing ( t point P) is: /4 P R 3/4 (A) 3.93 (B) 7.85 (C) (D) 15.7 (E) Q10 A od (length 0. m) moves on two fictionless hoizontl conducting ils. A 0.5 T mgnetic field is diected into the plne of the loop s shown in the figue. f the od is moving t constnt speed when F pp =1 N, then the vlue of induced voltge (in V) etween points nd is: Ω F pp 0.m (A) 0 (B) 40 (C) 60 (D) 80 (E) 100

3 Solution of the ove questions Q1: F = kq /, Q = 6.3µ c q 1 q φ = ( sphee), φ = ( hemisphee) Q: o o Q3: Consevtion of enegy: TA UA = TB UB, 1 Zeo eva = mυ B evb VB VA = 5.V Q4: Q5: ( C)( C) Ceq = C = C = 16µ F C C V = R 100 0R = = =.7A V R0 1 α ( 80) 1 80α = R0 Q6: 0 = = A 5 5, kε V = R Σ, V = 5 1 = V, V = V k k k / / / / / / / / F = q B = 10 3i 4j k = 0.8 4i 3 j, F = 4N 6 6 m Q7: ( υ ) µ µ µ π π π o 1 o o B = B B = = Upwds p 1 Q8: ( 3 ) µ o µ o 3 /4 µ o /4 Bc = θ, B= B B1 = π π, B= 3.93µ T Q9: 4πR 4πR 4πR Q10: ε ε F = F = B nd = R, = R/ B F = 0V pp mg i i i i pp m 3

4 Pt : Polem 1 (10 mks) A od of length = 0.5 m lies long the xxis nd hs unifom y chge density λ=100 nc/m. Clculte: () The totl chge on the od. de O Q=λ =100*0.5 =50 nc =x dq λ x () The electic field (give its mgnitude nd diection) t the oigin if = 0.1m. de= k dq/, dq=λdl=λdx, E = kdq = kλdx = kλ dx 1 = kλ x x = kλ ( x ) 9 9 kq 9 10 *50 10 E = = = 7500 N / C ( ) 0.1( ) E = 7500 i N / C (c) The electic foce (give its mgnitude nd diection) on n electon plced t the oigin. / 15 Fe = qe= ( e)( 7500) i = in (d) The mgnitude of the electon s initil cceletion t the oigin. 15 F m 1. 10, = = = / m s 15 4

5 Pt : Polem (15 mks) n the cicuit shown, the cuent 1 =1 A. Find: () The mgnitude of the potentil diffeence etween points nd. 1 =1A 6V 10Ω 10V 5Ω 3 ε 5Ω V = = 4V () The unknown emf ε. V = 4 = 5 10, = 1.A = 0. V 3 ε () = 4= 0.5 ε = 1V o, eft loop : = 16 gives = 1. A = gives = 0. A Right loop :5 5 = 10 ε, gives ε = 1V (c) The powe dissipted in the 5Ω esisto. P = 5 = 7.W 5

6 Pt : Polem 3 (10 mks) A coxil cle consists of thin conducting wie concentic with conducting tue with inne dius = 0. m nd oute dius = 0.3 m s shown in the figue. The wie nd the tue cy equl cuents =10A in opposite diections. The cuent in the tue is distiuted unifomly ove its coss section. Clculte the mgnitude of the mgnetic field t the following points: () Point P 1 tht is locted t distnce 0.1 m fom the xis of the conducto. P 3 P P 1 Bdl. = µ o en µ o B π = µ o, B= B P 1 7 4π = = 0µ T 0.1 ( π ) () Point P tht is locted t distnce 0.5 m fom the xis of the conducto. Font view ( ) ( ) [ ] Bdl. = µ = µ JA o enc o in B = µ o J π J = = A π µ o B= π π µ o B = 1 B P 7 ( 0.5) ( 0.) 4π = 1 = 4.4 µ T ( 0.5) ( 0.3) ( 0. ) (c) Point P 3 tht is locted t distnce 0.5 m fom the xis of the conducto.. µ o en µ o 3 [ ] B dl = = = Zeo B P = Zeo 6

7 Pt : Polem 4 (15 mks ) n the figue shown, sque loop of side =0.m nd esistnce R=0.5Ω. The loop lies t distnce ove long, stight wie cying cuent tht is vying with time ccoding to the eltion: = 10 sin(100πt) whee is in (A) nd t in (s). 0.5Ω dy () Show tht the mgnetic flux (Φ) though the sque loop is given y: µ 0 Φ = ln() µ o Φ= BdAcosθ = cos 0 y s µ o dy µ o Φ= = n y ( dy) ( ) y =10 sin(100πt) () Detemine the induced emf (in µv), s function of time, in the loop. ( t) ( t) dφ d µ o sin 100π εind = = n = 4π n cos 100πt dt dt ε = 87.1cos 100π µ V ind 7 (c) Detemine the mximum induced cuent (in µa) in the loop. εind 87.1 ind = = cos ( 100 πt) = 174.cos ( 100 π t) µ A R 0.5 = 174. µ A [ ] ind mx 7

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