1.

Find B at centre C in the following cases:

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Solution :(i) `B=(mu_(0)l)/(2R)xx1/2=(mu_(0)l)/(4R)`
(ii) `B=B_(1)+B_(2)+(mu_(0)l)/(2R)xx1/2=(mu_(0)l)/(4R.)1/R=(mu_(0)l)/(4R)(1+1/pi)`
(III) `B=B_(1)+B_(2)+B_(3)=2B_(1)+B_(2)=(2xx(mu_(0)l)/(4pi)I/R)+((mu_(0)I)/(2R)xx1/2)=(mu_(0)I)/(2R)(1/2+1/pi)`
(iv) `B=B_(1)+B_(2)=(mu_(0))/(2A)xx1/2+(mu_(0))/(2b)xx1/2=(mu_(0)I)/4(1/a+1/b)`
(v) `B=B_(1)-B_(2)=((mu_(0))/(2a)xx1/2+(mu_(0))/(2b)xx1/2)=(mu_(0)I)/4(1/a-1/b)`
(vi)`B=B_(1)-B_(2)=(mu_(0)I)/(2R)-(mu_(0)I)/(2piR)=(mu_(0)I)/(2R)(1-1/pi)`
(VII)`B=B_(1)+B_(2)=(mu_(0)I)/(2R)-(mu_(0)I)/(2piR)=(mu_(0)I)/(2R)(1+1/pi)`
(viii)`B=B_(1)-B_(2)=(mu_(0)I)/(2a)-(mu_(0)I)/(2b)xxtheta/(2pi)=(mu_(0)Itheta)/(4pi)(1/a-1/b)`


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