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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 :

mavila18

Answer:

y = 0.108m

Explanation:

The magnitude of the magnetic force is given by:

[tex]B=\frac{\mu_oI}{2\pi r}[/tex]    ( 1 )

and the force per unit length is:

[tex]\frac{F}{\Delta L}=\frac{\mu_oI_1I_2}{2\pi d}[/tex]   ( 2 )

you first calculate I2 from (2):

[tex]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[/tex]

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:

[tex]\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[/tex]

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

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