2. (a) A 6.28 m wire is wrapped around 100 times into a circular coil and then p
ID: 2305059 • Letter: 2
Question
2. (a) A 6.28 m wire is wrapped around 100 times into a circular coil and then placed inside a magnetic field with magnitude B-500 G. If a current of 20A passs through the wire, determine the maximum torque on the coil. A long, straight wire carries a current of 40 A. Calculate the magnetic field produced by this current at a point 10 cm perpendicular to the wire. (b) . (a) A coil with an area of 400 cm is placed in a magnetic field B-0.25 T such that the normal of the area is parallel to the field. If it took 400 ms to rotate the coil such that the normal is now in the opposite direction, find the induced voltage in the coil.Explanation / Answer
Torque on coil T =niAxB
Where n=no of turns=100
Total length =6.28m
We have 2piR*100=6.28
2*3.14*R*100=6.28
R=6.28/628 =0.01 m
So area A=pixR² =3.14*(0.01)² =0.0031 m²
Current=i=20A
Magnetic field, B=500 G
As 1 G=0.0001T.,so 500G =0.05T
Torque =100*20*0.0031*0.05=0.31
2):as the wire is long and point is perpendicular to midpoint we have
B=magnetic field =muo°I/2piR
As mjo°=4*pi*10^-7 T
We have B=4*pi*10^-7*40/2*pi*0.10=8*10^-5 T
C) induced emf = - N (del fi)/del t
Where del fi =fi final - fi initial
And fi =B.S, i.s dot product of B(magnetic field) and S=area
As initially angle between B and S=0,because they are parallel
Fi initial=BScosx=BScos0=BS =0.25*400*10-?
Fi final, as B amd S are anti parallel x=180°
B.S=BS cos180=-BS = - 0.25*400*10^-?
Del T=400ms =400*10—³ s
Put value in above formula we get
Emf= - (-0.25*400*10^-4 - 0.25*400*10—?)/(400*10—³)
Emf = - (-200*10—?)/400*10—3
Emf =5*10—2 v
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