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The wire loop formed by joining two semicircular sections of radii R_(1) and R_(2) and centre C, carries a current I as shown. The magnetic field at C has a magnitude |
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Answer» `(mu_(0)I)/(2)((1)/(R_(1))-(1)/(R_(2)))`  Magnetic field at the CENTRE C of a semicircular conductor is GIVEN by `B=sumdB` `=sum(mu_(0))/(4pi)(Idlsin90^(@))/(r^(2))` `=(mu_(0))/(4pi)(I)/(r^(2))sumdl=(mu_(0))/(4pi)*(I)/(r^(2))(pir)=(mu_(0)I)/(4r)` Here, magnetic field at C due to the current in the CIRCULAR loop of smaller radius, `vecB_(1)=(mu_(0)I)/(4R_(1))` upwards perpendicular to the plane of the PAPER. Magnetic to the plane of the paper. Magnetic field at C due to the current (I) in the circular loop of bigger radius, `vecB_(2)=(mu_(0)I)/(4R_(2))` downwards perpendicular to the plane of the paper. Also magnetic field due to straight portions is zero. Hence, net magnetic field at C `=vecB_(1)+vecB_(2)=|vecB_(1)|-|vecB_(2)|` `=(mu_(0)I)/(4R_(1))-(mu_(0)I)/(4R_(2))=(mu_(0)I)/(4)[(1)/(R_(1))-(1)/(R_(2))]` upwards. |
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