Two long parallel wires carry currents of 200 A and 100 A in opposite directions

Question

Two long parallel wires carry currents of 200 A and 100 A in opposite directions. The distance between the wires is 3 cm. A.) What is the magnitude of the force per unit length exerted iby one wire on the other? B.) What are the directions of the forces on each wire? C.) what is the total force exerted on a 10 cm length of the 100 A wire? D.) From this force, compute the strength of the magnetic field produced by the 200 A wire, at the position of the 100 A wire(Hint: F =I1B where I is the current,1 is the length from c. and B is the magnetic field)? E.)What is the direction of the magnetic felid produced by the 200 A wire at the position of the 100 A wire? The wires are traveling in opposite directions. 100A is going south 200A is going north

Explanation / Answer

Magnetic field due to a wire is proportional to I/R. [B = (M0)I / 2(pi)R] So, setting B1 = B2 and canceling constants, we get I1 / R1 = I2 / R2 20(R2) = 5(R1) R1 = 4(R2) So, it should be 4 times further from the wire carrying 20A, or 4/5 of the distance separating the wires. r = .8 * .20m = .16m (from the 20A wire)

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