Answer:
Explanation:
Magnetic field near current carrying wire
= [tex]\frac{\mu_0}{4\pi} \frac{2i}{r}[/tex]
i is current , r is distance from wire
B = 10⁻⁷ x [tex]\frac{2\times49}{r}[/tex]
force on second wire per unit length
B I L , I is current in second wire , L is length of wire
= 10⁻⁷ x [tex]\frac{2\times49}{r}[/tex] x 33 x 1
= 3234 x [tex]\frac{10^{-7}}{r}[/tex]
This should balance weight of second wire per unit length
3234 x [tex]\frac{10^{-7}}{r}[/tex] = .075
r = [tex]\frac{3234}{.075}[/tex] x 10⁻⁷
= .0043 m
= .43 cm .
