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A wire AB is carrying current of 12 A and is lying on the table. Another wire CD carrying 5 A is held directly above AB at a height of 1 mm. Find the mass per unit length of the wire CD so that it remains suspended at its position when left free.Give the direction of the current flowing in CD with respect to that remains suspended at its position when left free. Give the direction of the current flowing in CD with respect to that in AB . [Take the value of g = 10 ms^(-2)] |
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Answer» Solution :The wire CD will remain steady above AB at a height of `d = 1 mm = 10^(-3) m`, if force on it due to magnetic FIELD produced on account of current flowing in wire AB just balances its weight. As weight ACTS VERTICALLY downward, magnetic force must ACT in vertically upward and it is possible when current in wire CD is flowing along a direction opposite to that in AB as shown in Fig. If m be mass of wire CD per unit LENGTH, then `mg = (mu_0)/(4 pi) cdot (2 I_1 I_2)/(d)` `implies m = (mu_0)/(4pi) (2 I_1 I_2)/(dg) = (10^(-7) xx 2 xx 12 xx 5)/(10^(-3) xx 10) = 1.2 xx 10^(-3) kg m^(-1)` 
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