Question

Two long wires, each 10 m long and parallel, hang from wires 4.00 cm long, as shown in the figure. The wires carry equal currents and each carry a current of 160.0 Amps.

a) (20pts) Find the magnitude of the magnetic field generated by one wire just the distance where the other wire is.

b) (30pts) Based on the previous answer, find the mass of each wire so that a 6o angle is formed between the wire and the vertical as shown in the figure.

4.00 cm In 6.000 6.00°

Answer 1

Refer to the diagram: (end view)

2-4cm - 0 Fm 이지 하고 X fm song ma

the direction of current is opposite.

the separation between the wires is

$r = l\sin6^{\circ}$

a) Magnetic field field due to a current carrying straight long wire at a distance r is $B = \frac{\mu_{o}I}{2\pi r}$ .

As the current in the wires is same, therefore,

the strength magnetic field at the location of a wire due to the other wire is [answer]

B μοΙ 2πι 4π και 10 και 160 2π * 4 * 10-2 και sin69 0.0076T

b) The magnetic force on a wire due to the other wire is $F_{m}=ILB$ [ L = length of the wire carrying current]

We can see the system of wire is in equilibrium. It means we can say the resistant of magnetic force and force due to gravity is balanced by the tension in the string, and the string makes as angle $\theta$ with vertical.

Therefore, applying trigonometry, we can write,

$tan\theta=\frac{F_{m}}{mg}$

$\Rightarrow tan\theta=\frac{ILB}{mg}$

$\Rightarrow m=\frac{ILB}{gtan\theta} = \frac{160*10*0.0076}{9.81*tan6^{\circ}}=11.79kg$

Therefore, the mass of each wire is 11.79 kg. [answer]