Find `B` at centre `C` in the following cases:

Answer

(i)`(mu_(0)l)/(4R)ox`, (ii)`(mu_(0)l)/(4R)(1+1/pi)ox` , (iii)`(mu_(0)l)/(4R)(1/2+1/pi)ox` , (iv) `(mu_(0)l)/4(1/a+1/b)ox`
(v) `(mu_(0)l)/4(1/a-1/b)ox` , (vi) `(mu_(0)l)/(2R)(1-1/pi)o.`
(vii) `(mu_(0)l)/(2R)(1+1/pi)ox` , (viii) `(mu_(10))/(4pi)(1/a-1/b)o.`

Explanation

(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)`