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A long wire having a semi-circular loop of radius r carries a current I, as shown in Fig. Find the magnetic field due to entire wire. A. `(mu_(0)I)/(4r)`B. `(mu_(0)I^(2))/(4r)`C. `(mu_(0)I)/(4r^(2))`D. None of these |
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Answer» Correct Answer - A According to Biot savart Law, the magnetic induction at a point to a current carrying elemnet i `delta` l is given by `deltaB=(mu_(0))/(4pi)(ideltalxxr)/(r^(3))or B=(mu_(0))/(4pi)(ideltalsintheta)/(r^(2))` Directed normal to plane containing `deltal` and r theta being angle between `delta`l and r r `theta` being angle between `delta` l and r Magnetic field due to linear portions of wire If we consider anuy element of lenth `deltal` of either linear portion ab or de, the angle between `delta` l and r is 0 or `pi` therefore. `sintheta=0` Hence, magnetic induction due to both linear portins of wire is `(mu_(0))/(4pi)(ideltalsintheta)/(r)=0` Field due to semi circular are Now angle between a current element `delta` l of semi circular arc and the radius vector of the elment to point c is `pi//2` Therefoe the magnitude of magnetic induction B at O due to this element is `deltaB=(mu_(0))/(4pi)(ideltalsinpi//2)/(r^(2))=(mu_(0)ideltal)/(4pir^(2))` Hence magnetic indution due to whole semi circular loop is `B=Sigma(mu_(0))/(4pi)(ideltal)/(r^(2))=(mu_(0))/(4pir^(2))=(mu_(0))/(4pi)(i)/(i^(2))Sigmadeltai=(mu_(0)i)/(4piR^(2))(pir)=(mu_(0)i)/(4r)` |
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