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(a) If BCD is in the plane of AB and DE, find the magnetic induction at centre O due to a current i in the conductors. (b) The loop is made of same wire and uniform in cross-section. Find magnetic field at O. |
Answer» Solution :(a)  `d=R sin alpha` `theta_(1)=(pi-alpha),theta_(2) to 0` `B_(1)=(mu_(0)i)/(4pid)(costheta_(1)+cos theta_(2))` `=(mu_(0)i)/(4pirsin alpha)[cos(pi-a)+cos0]` `=(mu_(0)i)/(4pir sin alpha) (1-cos alpha), o.`  `B_(2)=(mu_(0)i)/(4r), o.`  `d=rsin alpha` `theta_(1)=alpha,theta_(2)to0` `B_(3)=(mu_(0)i)/(4pir sin alpha)(cos alpha+cos0)` `=(mu_(0)i)/(4pir sin alpha)(1+cos alpha), o.` `B_(O)=B_(1)+B_(3)+B_(2)=(mu_(0)i)/(2pir sin alpha)+(mu_(0)i)/(4r), o.` (b)  Length of `ABC=2` (length of `ADC`) `R_(ABC)=R_(1)=2R_(0),R_(ADC)=R_(2)=R_(0)` `i_(1)=(R_(2))/(R_(1)+R_(2))i=(R_(0))/(2R_(0)+R_(0))i=i/3` `i_(2)=i-i_(1)=(2i)/3` `B_(1)=(mu_(0)i_(1))/(2R).(240^(@))/(360^(@))=(mu_(0)i_(1))/(3R)=(mu_(0)i)/(9R), ox` `B_(2)=(mu_(0)i_(2))/(2R).(120^(@))/(360^(@))=(mu_(0)i_(1))/(6R) (mu_(0)i)/(9R), o.` `B_(O)=B_(1)-B_(2)=0` |
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