InterviewSolution
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InterviewSolution
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(a) Find the magnetic field at the point O if the wire carrying a current i has the shape shown. The radius of the current part of the wire is R, the linear parts of the wire are very long. (i) (ii) (iii) (b) A long straight wire along the z-axis carriers a current i in negative z-direction. Find magnetic field in vector from at a point having co-ordinates (x,y) on z=0 plane. (c) A non-coplaner loop of conducting wire carrying a current I is placed as shown in the figure. Each of the straight sections of the loop of length 2a. find unit vector along magnetic field magnetic filed at the point P(a,0,a). |
Answer» Solution :(a) (i)  At `O:` Wire `①` is in `x-y` plane. Magnetic field DUE to this wire be `bot^(ar)` to plane `x-y` at `O`. `B_(1)=(mu_(0)i)/(4piR)`, along `z`-axis Wire `②` is in `y-z` plane. `B_(2)=(mu_(0)i)/(4R)`, along `-x`-axis `B_(3)=(mu_(0)i)/(4piR)`, along `-z`-axis `vec(B)_(O)=-(B_(1)+B_(3)) hatk-B_(2)hati` `=-(mu_(0)i)/(4R) hati-(mu_(0)i)/(2 pi R) hat(k)` `|vec(B)|= B_(o) = (mu_(0)i)/(4R) SQRT(1 + ((2)/(pi))^(2))` (II)  `B_(1)=(mu_(0)i)/(4piR)`, along `-z` axis `B_(2)=(mu_(0)i)/(4R)`, along `-x` axis `B_(3)=(mu_(0)i)/(4piR)`, along `-x` axis `vec(B)_(O)=-(B_(2)+B_(3)) hati-B_(3)hatk` `=-((mu_(0)i)/(4R)+(mu_(0)i)/(4piR)) hati-(mu_(0))/(4piR) hatk` `|vec(B)_(O)|=B_(O)=(mu_(0)i)/(4R) sqrt((1+1/(pi))^(2)+(1/(pi))^(2))` (iii)  `B_(1)=(mu_(0)i)/(4piR)`, along `-z`direction `B_(2)=0`, current is divided in two PARTS and magnetic field due to these parts is zero. `B_(3)=(mu_(0)i)/(4piR)`, along `-y` direction `vec(B)_(O)=-B_(3)hatj-B_(1)hatk` `=-(mu_(0)i)/(4piR) hatj-(mu_(0)i)/(4piR) hat k` `|vec(B)_(O)|=B_(O)=(sqrt(2)mu_(0)i)/(4piR)=(mu_(0)i)/(2sqrt(2)piR)`  Due to long wire, magnetic field at `P` is `B=(mu_(0)I)/(2piR)` `vec(B)=B_(x)hati-B_(y) hatj` `=B sin thetahati-B cos theta hatj` `=B(y/r hati-x/r hatj)` `=(mu_(0)I)/(2pir)(y/r hati-x/r hatj)` `=(mu_(0)I)/(2pir^(2))(y hati-x hatj)` `=(mu_(0)I)/(2pi(x^(2)+y^(2)))(yhati-xhatj)` (c) There can be two closed loops.  Magnetic field at `P` due to loop, since is anticlockwise, HENCE magnetic field will be towards the observer i.e. along `=x` axis `vec(B)_(1)=Bhatk` where `B`: magnitude of magnetic field due to each loop. `vec(B)_(P)=vec(B)_(1)+vec(B)_(2)=Bhati+Bhatk` Unit vector along magnetic field at `P` `hat(B)_(P)=(vec(B)_(P))/(|vec(B)_(P)|)=(Bhati+Bhatk)/(sqrt(2)B)=1/(sqrt(2))(hati+hatk)` (##CPS_V02_C05_S01_008_S06##) |
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