Question

2. Two very long wires are oriented parallel to the y axis and carry identical but unknown currents Ia inopposite directions. The wires cross the x-axis at positions x = +2a and +3a. The wires produce amagnetic field at the origin with known magnitude B0 and direction -z (i.e., into the page). Finally, aparticle of charge q passes through the origin, with velocity y in the +y direction.(a) Using the given magnitude and direction of the magnetic field ii at the origin, determine the current Ia. Bear in mind that Ia might be negative, indicating directions opposite to those shown in the figure.(b) What is the force F on the charged particle as it crosses the origin?

Answer 1

Magnetic Field Due to a current carrying wire is given by:

$B=\frac{\mu_oI}{2\pi r}$

a) Net Magnetic field at origin due to Ia in two wires is given by:

$B=\frac{\mu_oI_a}{2\pi (2a)}-\frac{\mu_oI_a}{2 \pi (3a)}$

$\Rightarrow B_o=\frac{\mu_oI_a}{2\pi (6a)}$

$\Rightarrow B_o=\frac{\mu_oI_a}{12\pi a}$

Now Putting the values:

$\Rightarrow 8*10^{-7}=\frac{4\pi *10^{-7}I_a}{12\pi 0.03}$

$\Rightarrow 8=\frac{I_a}{3* 0.03}$

$\Rightarrow I_a=720 mA$

$\Rightarrow I_a=0.72A$

b)

Force is on a charged particle in a magnetic field is given by the vector cross product

$\Rightarrow F=qv\times B$

$\Rightarrow F=(40*10^{-3}*2*10^7)j \times (8*10^{-7})(-k)$

$\Rightarrow F=(40*10^{-3}*2*10^7)(8*10^{-7})(j \times -k)$

$\Rightarrow F=-0.64i$

Hence Magnitude of Force =0.64 N

Direction of force is = Negative x-axis

Answer 2