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Three long straight parallel wires are kept as shown in the fig. The wire (3) carries a current I. (i) The direction of flow of current I in wire (3) is such that the net force on wire (1) due to the other two wires, becomes zero. What will be the direction of current I in the two cases ? Also obtain the relation between the magnitude of currents I_1, I_2 and I_3. |
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Answer» Solution :(i) Since net force per unit length on wire (1) due to other two wires is zero, the direction of `I_2 and I_3` must be mutually opposite and then `|vec(F_(12))| = |vec(F_(13))|` `:. (mu_0 I_1I_2)/(2 pi d) = (mu_0 l_1 I)/(2 pi (2d)) implies I = 2I_(2)` (II) When direction of I of reversed then net force per unit length on wire (2) due to other two wires becomes zero. HENCE, now `vec(F_21) + vec(F_23) = 0` `:. |vecF_(21)| = |vecF_(23)| implies (mu_0 I_1 I_2)/(2 pi d) = (mu_0 I_2I)/(2 pi d) implies I_2 = I` |
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