Two straight parallel wires carry currents in opposite directions as shown in th

Question

Two straight parallel wires carry currents in opposite directions as shown in the figure. One of the wires carries a current of I2 = 10.1 A. Point A is the midpoint between the wires. The total distance between the wires is d = 10.8 cm. Point C is 5.33 cm to the right of the wire carrying current I2. Current I1 is adjusted so that the magnetic field at C is zero.

1. Calculate the value of the current I1. 3.06×10^1

2. Calculate the magnitude of the magnetic field at point A.

3.What is the force between two 1.39 m long segments of the wires?

Need help with 2 and 3!

Consider the mass spectrometer shown schematically in the figure. The electric field between the plates of the velocity selector is 957 V/m, and the magnetic fields in both the velocity selector and the deflection chamber have magnitudes of 0.147 T. Calculate the radius of the path in the system for a singly charged ion with mass 2.18×10-26 kg.

Explanation / Answer

2) magnetic field at A due to current I1 is B1 = mu_o*I1/(2*pi*r1) = (4*3.142*10^-7*30.6)/(2*3.142*0.054) = 1.13*10^-4 T out of the page

magnetic field at A due to current I2 is B2 = mu_o*I2/(2*pi*r1) = (4*3.142*10^-7*10.1)/(2*3.142*0.054) = 37.41*10^-6 T = 0.374*10^-4 T out of the page

then

B = B1 + B2

B = (1.13+0.374)*10^-4 = 1.5*10^-4 T

3) F = mu_o*i1*i2*l/(2*pi*r) = (4*3.142*10^-7*30.6*10.1*1.39)/(2*3.142*0.108) = 7.955.43*10^-4 N

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