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Two long, parallel wires are separated by a distance of `0.400m` (as shown in Fig.) The current `I_1 and I_2` have the direction shown. (a) Calculate the magnitude of the force exerted by each wire on a 1.20m length of the other. Is the force attractive or repulsive ? (b) Each current is doubled, so that `I_1` becomes 10.0A and `I_2` becomes 4.00A. Now, what is the magnitude of the force that each wire exerts on a 1.20m length of the other? |
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Answer» (a) `F=(mu_0I_1I_2L)/(2pir)` `=(mu_0(5.00A)(2.00A)(1.20m))/(2pi(0.400m))=6.00xx10^-6N` The force is repulsive since the currents are in opposite directions. (d) Doubling the current make the force increase by a factor of 4 to `F=2.40xx10^-5N` |
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