![]() dl r sin 90q 4S r3 P0 Idl - (10.47) Example 10.5: A wire has 2 straight 4S r 2 I d sections and one arc as shown in the figure. (10.34), the F21 = - F12 Is this consistent with Newton's third law? magnetic field produced at O is: (Consider for example the gravitational pull experienced by the Earth towards the dB P0 I.dl u r Sun and that by the Sun towards the Earth.) 4S r3 dB P0 I. The circular arc AB subtends an angle θ at the centre O of We have seen that in case of parallel conducting wires carrying steady currents, the circle of which the arc is a part, and r is its the Biot-Savart law and the Lorentz force law give the result in Eq. We can first B P0 I - (10.49) obtain the magnetic field produced by one 2r current-length element of the arc and then Use your brain power integrate over the entire arc length. Figure 10.18 depicts a circular arc of a wire (AB), carrying a current I. (10.48), produced by a current in a circular arc of a B P0 I 2S 4S r wire. Magnetic field at the centre of a full circle of 10.12 Magnetic Field Produced by a Current a wire, carrying a current I : in a Circular Arc of a Wire: For a full circular wire carrying a current I, the After considering straight parallel wires magnetic field at the centre of the circle, using let us obtain the magnetic field at a point Eq. ³P0 T r dT P0 I T, - (10.48) r2 4S r 4S I 0 Here it is assumed that the wire diameter where the angle θ is in radians. ![]() The total field at O is therefore, L 2S d ³ ☻ dB P0 I B dl P0 4S A r2 4S =10-7 Wb/m d= 1 m F For I1 = I2 = 1A, L = 2×10-7 N per meter. direction of each of the dB is into the plane of F P0 I1I2 the paper. It is a straight forward evaluation from is indicated by the curling fingers.
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