Question

Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.320 m and carries a current of 30.0 A in the +x direction. The second wire lies along the x-axis. The wires exert attractive forces on each other, and the force per unit length on each wire is 290 µN/m. What is the y-value (in m) of the line in the xy-plane where the total magnetic field is zero?

Answer 1

Answer:

y = 0.108m

Explanation:

The magnitude of the magnetic force is given by:

$B=\frac{\mu_oI}{2\pi r}$    ( 1 )

and the force per unit length is:

$\frac{F}{\Delta L}=\frac{\mu_oI_1I_2}{2\pi d}$   ( 2 )

you first calculate I2 from (2):

$I_2=(\frac{F}{\Delta L})\frac{2\pi d}{\mu_o I_1}=(290*10^{-6}N/m)\frac{2\pi (0.320m)}{(4\pi*10^{-7}m)(30.0A)}=15.46A$

With this values of I2 you can the position in which the magnitude of the magnetic field is zero, by using (1) for both wires:

$\frac{\mu_o}{2\pi}[\frac{I_1}{0.320-r}-\frac{I_2}{r}]=0\\\\I_1r=I_2(0.320-r)\\\\r(I_1+I_2)=0.320I_2\\\\r=\frac{0.320I_2}{I_1+I_2}=\frac{0.320(15.46)}{15.46+30.0}=0.108m$

hence, for y=0.108m the magnitude of B is zero