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A large straight current carrying conductor is bent in the form of L shape. Find vec(B) at P. |
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Answer» <P> Solution :Let us divide the conductor into two semi infinite SEGMENTS 1 and 2. Then, induction at `P` is `vec(B)=vec(B_(1))+vec(B_(2))...(i)` `vec(B_(1))=(mu_(0)i)/(4pia)(sin(90^(@)-theta_(1))+sin90^(@))hat(k)...(ii)` `vec(B_(2))=(mu_(0)i)/(4pia)(sin(90^(@)-theta_(2))+sin90^(@))hat(k)....(iii)` then `vec(B)=(mu_(0)i)/(4pia)(costheta_(1)+costheta_(2)+2)hat(k)`,where `costheta_(1)=costheta_(2)=(1)/(sqrt(2))`Hence, ` vec(B)=(2+sqrt(2))(mu_(0)ihat(k))/(4pia)` |
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