Phy 1 (7)

Q.1. The induced emf in the rod causes a current to flow counter clockwise in the circuit. Because of this current in the rod, it experiences a force to the left due to the magnetic field. In order to pull the rod to the right with constant speed, the force must be balanced by the puller.

The induced emf in the rod is

| = BLv = (0.25) (0.5)(4) = 0. 5 V I = /R = 0. 5/3A from which,

F = iLBsin90o = (0.05)(0.5/3)(0.25) = (0.0625 /3)N Q.2. I = I0 (1 – e-Rt/L) …(i) U = LI2 2 1 , Umax = LI20 2 1 U = U_max 4 1 2 0 2 LI 8 1 LI 2 1   I = I0 / 2 From (i) 2 I0 _{= I} 0 (1 – e -Rt/L )  t = ln2 R L = 5 ln2 = 3.47 s Q.3. B r2 Q.4. Zero Q.5.  = Bs = (8 + 5t)  10–2  800  10–4 dt d   = 5 10–2  80010–4 = 4  10–3 J/c (a) qMax = C = 10  10–6  4  10–3 = 4  10–8C (b) Max energy stored is ₆

8 8 2 10 10 2 10 4 10 4 c 2 q          = 8  10–11 J (c) Plate B will be positively charged.

A B



Q.6. In accordance with law of potential distribution, for the given network, VA  IR + E  L

dt dI

= VB

And as here I is decreasing (dI/dt) is negative. VB – VA =  5  1 + 15  5  10

-3

VB – VA =  5 + 15 + 5 = 15 V

Q.7. At t = 0 current through inductor will be zero.

Therefore current provided by the source I = E/R1 = 10/10 = 1 amp. At t = , inductor will be shorted

Therefore current provided by the source I =

1 2 1 2 E R R R R = 2 amp. Q.8. 0 0 avrage T I _I 4 I T 2 2   

Q.9. The torque applied by the magnetic force Fm about O is given as dm = r(dFm) = r {i(dr) B}  m =



d_m



r



idrB



; Putting i = R 2 B R E 2   We obtain, m = _



         0 2 2 rdr R 2 B   0 2 2 2 m 2 r R 2 B     R 4 B2 4 m     

 The external torque must be equal and opposite to the magnetic torque  ext = R 4 B24 Power = ext = 2 4 2 B 4R  

Q.10. Writing loop equation





3 3 BA 4_{ }1 10 10 10_{ }  _10 _V AB V 4    volts. A I B R=10 E=4v L=10mH 1amp

Q.11. The current at any time can be given by the expression, i = i0 (1  et/) Where  = L 100mH _{5 10 s}3 R 20      5 ln 2 5 i0.5A 1 e_   _   = io _{ln 2 o} o i 1 1 1 i 1 2 2 e                 Where i0 = 10V 0.5A 20   The energy stored = Li2

2 1 =





2 3 4 1 10 100 2 1          = 320 1 J.

Q.12. Induced emf in the rod  = BL (v cos ) i = /R = cos R BLv  F = BiL = BL        cos R BLv

for uniform velocity force on rod up the plane = force on rod down the plane

 2 2 2 B L v cos mgsin R     Mg cos v cos v Mg Mg sin  v =         2 2 2 cos sin L B Rmg v = tan sec L B Rmg 2 2 Q.13. (b) d = Rd de = Rd . V sin  . B [ = 90 - ] = RVB cos  d e =

_

   2 / 2 / de = RVB

_

    2 / 2 / cos = RVB [sin]_/2_/2 = 2RVB.  d  d Q.14. (a)  =

_

B.d_s = B.S (b) e = - dt dS . B dt d   4b b 2b 3b 0 x BNb  4b b 3b 2b x e e = -BNv e = +BNlv

Q.15. The induced emf in the rod is | = BLv = (0.25) (0.5)(4) = 0. 5 V I = /R = 0.1 A

F = ILBsin90o = 1/80 N

Thus the total force acting on the rod on left is 1/40 N.

Hence 1/40 N of force is to be applied on the rod in the right side to move it with a constant velocity. Q.16. The current at any time can be given by the expression, i = i0 (1  et/)

Where  = L 100mH _{5 10 s}3 R 20      5 ln 2 5 i0.5A 1 e_   _   = io _{ln 2 o} o i 1 1 1 i 1 2 2 e                

Where i0 = 10V

0.5A 20 

 The energy stored = Li2 2 1 =





2 3 4 1 10 100 2 1          = 320 1 J

Q.17. As the current in the spokes would be flowing in radially inward fashion , the force on each spoke due to magnetic field directed into the plane of figure, turns out towards right, the initial torque would be anticlockwise and hence the rotation.

Let i1 , i2 , i3 , i4 , i5 and i6 be the current through each spoke. Consider the spoke through which current I1 flows.

For an element i dx at distance x from the centre, the magnetic force = i dx1 B  

 . Hence the torque of this force =



xi dxB n₁



 (Where n is unit vector in the direction of torque)

As torque for each spoke has the same direction , Total initial torque =



  



_

l xdx 0 6 2 1 i ... i i nˆ B = IB nˆ



0.06Nm



nˆ 2  2 l . Q.18. The loop equation

L iR 0 dt di   At t = 0, i = 0, 12 - 0.008 0 dt di  008 . 0 12 dt di  = 2500 A/s Where dt di

= 500A/s, an equation yields, 12 - (0.008)(500) - 6i = 0

6 i = 12 - 4  i = (4/3) A For the final current

dt di

= 0, and L(0) - 6 I F = 0

IF = 2A

Q.19. The induced emf in the rod causes a current to flow counter clockwise in the circuit . Because of

this current in the rod, it experiences a force to the left due to the magnetic field. In order to pull the rod to the right with constant speed, the force must be balanced by the puller.

The induced emf in the rod is = BLv = (0.25) | (0.5)(4) = 0. 5 V I /R = 0. 5/3A from which, =

F = iLBsin90o = (0.05)(0.5/3)(0.25) = (0.0625 /3)N Q.20. (a) Magnetic induction due to large loop at its centre is

B = b 2 ib 0  and B = _        b 2 i A . B 0b a2 cos 

where ‘’ is angle between loops and  = t  induced emf  = - b 2 i a dt d 0b 2      sin t  current ia =           bR 2 i a R b 0 2 sin t … (i)

(b) Mutual inductance M = cos t b 2 a dt d 0 2        = - (Mi ) dt d dt d a    from (i)  = _              bR 2 t sin i a . t cos b 2 a dt d 2 ₀ 2 _0b = - _                2 t 2 sin dt d b 2 a R i_b 2 ₀ 2 = - 2 t 2 cos 2 b 2 a R i_b 2 ₀ 2              = - cos2 t b 2 a R i_b 2 ₀ 2          

Q.21. A conducting rod 'OA' of mass 'm' and length 'l' is kept rotating in a vertical plane . . . any other resistance. (a) 2 1 Bl2 = e (b) E = iR + L dt di 

_

_

  iR E di L dt   log(E iR) c L Rt     E  iR = EeRT/L  i =



1 e RT/L



R E _  i = Bl2



1 e RT/L



2 1 R 1 _         i = R 2 Bl2 at t   steady state Power = Torque () i2 R = J

J =     2 2 4 2 2 R 4 R l B R i

+ torque due to weight of the rod J =

R 4

l B24

+ torque due to weight of the rod =

R 4

l B24

+ Mg (/2)cos t

Q.22. Let the velocities of the two rods at time t be v1 iˆ and v2 iˆ . Net induced emf in the circuit = B(v2 – v1)  Induced current = R ) v v ( B 2  1 _(clockwise)

As a result, a force F = iˆ R ) v v ( B 2 1 2  

acts on the rod of mass m and a force F = - iˆ R ) v v ( B 2 1 2  

acts on the rod of mass 2m. Equations of motion of m, 2m, and 3m are

m dt dv1 ₌ R ) v v ( B 2 1 2 2   . . .. (i) 2m dt dv2 _{= T -} R ) v v ( B 2 1 2 2   . . .. (ii) 3m dt dv₂ = 3mg – T . . .. (iii) (T being the tension in string)

Adding (ii) and (iii), 5m dt dv2 _{= 3,mg -} R ) v v ( B 2 1 2 2   . . .. (iv)

From equation (i) and (iv) 5m (v v ) dt d 1 2 = 3mg - R ) v v ( B 6 2 2 _{2 1}   Solving this equation, we get v2 – v1 =           5mR t 2 2 B 6 2 2 1 e B 2 mgR   Putting this in (i) m dt dv1 ₌ 2 mg           5mR t 2 2 B 6 e 1  Integrating we get v1 =                    5mR t 2 2 B 6 2 2 1 e B 6 mR 5 t 2 g   Putting values v1 = 5 [t – 5 (1 –e-(t/5) ) ] m/s where t is in seconds.

Q.23.  =

_

         x a x 0 0 x x a n a B ad B    =        vt a 1 n a B₀   =





2 0 B vt a vt a vt a B dt      I =  vt)t a ( 4 a B₀ . d  x _v I Q.24. Q = 0.25dx x ) t 4 3 ( . 2 2 25 . 1 1 0  _  



= (3 4t ) 2 2 0 _    0.25 ln (1.25) e = - 0.25h(1.25) 8t 2 dt dQ 0         L ) 25 . 1 ln( t dt di ₀ i = 0 _t2 10 ) 25 . 1 ln(  

Q.25. (a)The rotation of the ring about point P generates an emf. The ring within P & Q is equivalent to a rod of length PQ.

Now PQ = a2a2 a 2

The emf between P and Q is given by

 = B



a 2



2 2

1

 = Ba2

(b) As the resistance between P & Q is R. Then the current

I = R a B R 2    a  P Q   R

Q.26. From faraday's law of emf (a) e = - (BA) dt d with A =  R2 B = B0sin 2 ft so e = - 22R2 f B0 cos 2 ft. (b) Use R = a L  , (Here R is resistance  R = a R 2 

(c) I =

/

R e

and P = I2R.

Q. 27. (a) (i) F = BId (ii) e = I(R+2x)

(b) If v is the velocity of the rod, the induced emf = Bvd The current is given by,

I = R e  = (R 2 x) Bvd   If I is constant, v = Bd x 2 R  I . . . (1) F - Fm = ma or, F = Fm + ma or, F = Bid + m dt dx Bd I 2 or, F = Bid + 2 2 2 d B mI 2 (R + 2x) . . . (2) (c)                (R 2 x) d B mI 2 BId Fv P dt dw 2 2 2         Bd x 2 R I

= Power dissipated in circuit + Rate of increase of kinetic energy of rod = I2 (R + 2x) + 3 3 3 d B mI 2 (R + 2x)2 The fraction is,

f = 1 3 3 2 d B ) x 2 R ( mI 2 1 P ) x 2 R ( I               . f = 0.961. Q.28.  =

_

             x a x 0 0 x x a ln 2 Ia ady y 2 I =          vt a 1 ln 2 Ia 0  =





2 0 B vt a vt a vt 2 Ia dt        I =



a vt



t 8 Ia 0    ? dy y x _v I

Q.29. At P due to current in (1), magnetic field is in upward direction and due to current in (2), magnetic field is downward direction.

At Q due to current in (1) magnetic field is downward and due to current in (2), magnetic field is in upward direction. Therefore at P, B1 - B2 = 3 . 0 30 2 1 . 0 20 2 o o     

= 2  10-5 N/Ampmeter, along positive z-axis at Q, B1 + B2 = 3 . 0 30 2 1 . 0 20 2 o o     

at R, B2 - B1 = 3 . 0 20 2 1 . 0 30 2 o o     

= 4.7  10-5 N/Ampmeter, along positive z-axis

Q.30. (a) Applying KVL to the circuit at time 't' 2i + dt di 10 2 . 0 = 20

solving this differential equation : i = 10 (1- 0.1e-100t ) (b) EMF induced = - dt d current induced = R 1 dt d = i  idt = R d ? 5A 3m 1m P Q Integrating,

_



_

 R d

idt = net change in flux /R Hence charge flown =

R flux in change net Now net change in flux = i - f =

  2 ) 5 )( 1 ( 0                    3 4 ln 2 3 ln Weber

charge flown = 1.17  10-7 coulomb. Q.31. The loop equation

L iR 0 dt di   At t = 0, i = 0, 60 - 0.008 0 dt di  008 . 0 60 dt di  = 7500 A/s Where dt di

= 500A/s, an equation yields, 60 - (0.008)(500) - 30i = 0 30 i = 60 - 4  i = 1.867 A For the final current

dt di

= 0, and L(0) - 30 I F = 0 IF = 2A

Q.32. (a) Given L = 2.0 mH, C = 5.0 F and Qmax = 200C

Let q be the charge on the capacitor at any time t and I be the current flowing in te circuit.

(a)  0 dt dI L C q   ) 10 5 )( 10 2 ( 10 100 6 LC q dt dI 6 3 6          = 104 amp/sec.

(b) When Q = 200 C, then the total energy of the circuit is stored in the capacitor i.e., no current flows through the circuit. Hence current I in the circuit is zero.

I = 0

(c) Let I0 be the maximum value of the current. Then C Q 2 1 LI 2 1 2 _max2 0 or LC Q I 2 max 2 0  or LC Q I 2 max 0  I0 = 2 amp By conservation of energy C q 2 1 4 Li 2 1 C Q 2 1 20 2 2 max _{ }  Q = 2 3 Q_max = 100 3 C L dt dI C q +  i Q.33. X = a + vt,  = x 0 a Id dx 2 x  



= 0I x d ln 2 a       _{ } emf = d 0Idv0 dt 2 (a vt)      Current I = 0Idv0 2 R(a vt)    F = ILB = 2 2 2 0 0 2 2 I d v 4 (a vt) R    . Q.34. 2₀ Cv2₀ 2 1 Li 2 1  i0 = v0 L C v0 = ₄ 2 6 0 _{1.25 10} 10 0 . 4 10 0 . 5 C q         volt i0 = 1.25  10 -2 4 4 10 33 . 8 09 . 0 10 0 . 4      A umax = Cv 3.125 10 J 2 1 Li 2 1 2 8 0 2 0     3.125  10-8 = 2 4 2 4 1 8.33 10 1 q (0.09) 2 2 2 (4.0 10 )            q = 4.33  10-6 C.

Q.35. Vel. Comp along x-axis v cos will remain constant m x v cos . 2qB    

due to vel comp along y – axis v sin it will move in a circular path of radius R mv sin

qB   y-Co-ordinate yR sin R sin t y - Z-axis  R  Y-axis z mv sin qB m sin . qB m 2qB      _{ }   mv sin mv sin .sin / 2 qB qB      Z - coordinate





z  RR cos





R 1 cos     mv sin 1 cos qB 2     _  _   mv sin qB    Q.36. VB sin 180

_

 

_

VB sin   VB sin I R R     2 2 mag B V sin F I B R     Now 2 2 2 B V sin mg sin 0 R      or B ₂mgR V sin    B  v F mg sin F sin 90-

Q.37. (a) In one complete cycle Iav = 0 Irms = 2 / 2 0 0 / 2 0 2I t dt dt       _      





Irms = 3 I₀ . (b) U = 0.5 2 0.5 2 3 0 0 v (10 sin 2 t) dt dt R 10  





= 1 1 104 = 0.025 J.

Q.38. (i) Impedance Z = 2 2

L C

R (X X )  6436 10 (ii) Current = v 220 22A

z  10  (iii) tan  = | XL X |C 6 3 R 8 4      = 370 (current leads)

(iv) Power factor, cos  = R/Z = 0.8 Q.39. Emf induced in the coil, E =

-dt d

= -nA dt dB

where n = number of turns, A = area, B = magnetic field. Current in the coil i1 = 

     ) 2 . 0 ( 6 . 1 ) 0 8 . 0 ( 5 . 0 dt dB R nA R E - 1.25 A

Current in the coil when magnetic field is decreased uniformly from 0.8 T to 0 in 0.2 sec. i2 = -2 . 0 ) 8 . 0 0 ( 6 . 1 5 . 0 dt dB R nA    = 1.25 A Power dissipated P = i2R = (1.25)2 (1.6) = 2.5 W

Graph showing variation of current Graph showing variation of power

t Current (i) 1.25 A -1.25 A 0.2 s. 0.4 s. 0.8 s. t Power 2.5 W 0.8 s.

Heat dissipated in the time interval from 0 to 0.4 sec. = 2 1

i Rt + 2 2

i Rt = 2  (1.25)2 (1.6) (0.2) = 1 J

Total heat dissipated when the cycle is repeated 12000 times H = (12000) (1) J = 12000 J

Let T = increase in temperature

Therefore H = (m1c1 + m2c2)T  T = 5.63 0C Therefore final temperature T = 35. 63 0C

Q.40. Area (PQRS) = (a x)dx 3 2  Flux over PQRS = d = x d I 2 4 dx ) x a ( 3 2 ₀    

Flux over whole ,  =  d =

_

  a 2 3 0 0 3 I         x d x a S R PQ x _dx d A B I

=                 





a 2 3 0 a 2 3 0 0 _{dx dx} x d d a 3 I I =  R e   R dt d              a 2 3 ) d 2 a 3 1 log( ) d a ( R 3 I 0 _{cos t} Q.41. At t = t Va – Vb = Vc – Vd  L dt dx B dt di    Ldi = Bdx  li = Bx i = x L B       

net force on the connector Fnet = mg – Fm  m mg iB dt dv   x = 0 x t = 0 v = 0 L a b c d mg i Fm t = t X B v = mg - L B22 x x mL B g dt dv 2 2             dt dx . mL B dt v d 2 2 2 2     v dt v d 2 2 2      = mL B  v = v0 sin t, v0 dt dv cos t at t = 0, dt dv = g  g = v0   v0 =  g =  B mL g so v = v0 sin t [where v0 =  B mL g and  = mL B ]  v sin t dt dx 0  



_



_

 t 0 0 x 0 dt t sin u dx  x =  0 v [ - cos t] t 0 x =  0 v (1 – cos t) x = v0



1cost



 Q.42. L Bv dt di  L dt dx B dt di   i = x L B

Magnetic force on the rod, Fm = iB = x L B22



m v C B A D  m x L B dt x d 2 2 2 2  

 (Force is in opposite direction of v)

2 2 2 2 2 d x B dt  m L  x  = mL B . Q.43.  = 1m, mbar = 1kg , B = 0.6 T Let v = terminal velocity of the bar L.

The induced emf developed across the bar is  = Bv

Current in R1 and R2 are I1 =  1 P ; I2 =  2 P

Total current through bar is I = I1 + i2 = (P1 + P2)/ … (i) Force experienced by bar is F = IB

When the bar attains a constant velocity we have mg = IB  I =

 B mg

… (ii) from (i) and (ii)

  Bv P P B mg ₁ ₂   v = 8 . 9 2 . 0 2 . 1 76 . 0 mg P P_{1 2}     = 1.0 m/s  = Bv = 0.6  1  1 = 0.6 v R1 = 0.474 76 . 0 ) 6 . 0 ( P I 2 1 2 1       R2 = 2 . 1 ) 6 . 0 ( P 2 2 2   = 0.3 

Q.44. (a) Equivalent Inductance = L L L . L  + L = 2L 3  f = LC ) 2 / 3 ( 2 1  m

(b) (I) Applying kirchooff's second law to meshes (1) and (2)

2I2 + 2I1 = 12 - 3 = 9 and 2I1 + 2(I1 -I2) = 12 solving I1 = 3.5 A, I2 = 1A

P.D.between AB = 2I2 + 3 = 5 volt

Rate of production of heat iR 24.5 dt dQ 1 2 1         (J) R2 _B 2 R3 R1 2 12V E1 E2 A S 3V I1R1 (I1-I2) (1) (2) I1 I2 I2 (ii) i =



1 e Rt/L



R E _  = i0



1eRt/L



i = i0/2  L Rt = loge2 t = 0.0014 sec. Energy stored = 2 1 Li2 = 0.00045 (J) R2 = 2 R4 = 3 10 mH E2 = 3V Q.45. (a) de = B (x) dx  e = 2 r B 2 (b) i = i0 (1 - eRT/L)  i0 = e/R  i =



Rt/L



2 e 1 R 2 r B    R L i e O A B  x dx (c) In steady state, i = i0

Bet torque  = mg (r/2) cos  +  Fx  = r 0 mgr cos iB.xdx 2  

_

= 

_

 r 0 2 Bxdx . R 2 r B 2 cos mgr  = mgr cos t 1 2 4 B r 2 4R     mg x F dx

Q.46. v  Yiˆ ; ddyjˆ 0Ikˆ B L 2 [d Y] 2        ; d v Ydykˆ 0I Ydy B.(d v) L 2 [d Y] 2         _{ }  vB – v0 = L / 2 0 0 I Ydy L [d y] 2   



_{ } = 0I _d L _{ln 1} L L 2 2 2d 2                        O B A L d I z x y dy Y

Q.47. The repulsive force on the side ps of the current-carrying loop, due to current i1 is N 10 4 . 2 04 . 0 15 . 0 16 20 ) 10 2 ( d L i i 2 F o 12 7 4 1            

This force will be towards RHS and  to the current-carrying wire ps.

Similarly, the attractive force acting on the side qr of

the loop, due to current I1 is(Here R = d+b = 10 cm = 0.1 meter) 10 . 0 15 . 0 16 20 ) 10 2 ( F₂   7      N 10 96 . 0  4  . a q p s i2 i1 b r d

Direction of this force will be towards LHS and  to current-carrying wire qr. The forces acting on the sides pq and rs of the loop will be equal and opposite. Thus net force on the loop = F1  F2 = (2.4  0.96)  10

4

= 1.44  104 N (Acting away from the current-carrying wire)

When the direction of current in the loop becomes clockwise, the net force on the loop remain same, but its direction now becomes towards the current-carrying wire.