Determine the magnitude of the force between two parallel wires 25 m long and 4-0 cm apart- each carrying 25 A in the same direction-

Question

Toppr Admin · Accepted Answer

Correct option is A- -0-078- NThe magnitude of the force between two parallel wires-F-2-dfrac-mu-0-2-pi-dfrac-I-1I-2-d-l-2-F-dfrac-4-pi-times 10-7-T-m-A-2-pi-dfrac-25A-25A-4-0-times 10-2-m-25m-0-078N-

Determine the magnitude of the force between two parallel wires 25 m long and 4.0 cm apart, each carrying 25 A in the same direction.

Correct option is A. $$0.078$$ N
The magnitude of the force between two parallel wires,
$$F_2=\dfrac{\mu_0}{2\pi}\dfrac{I_1I_2}{d}l_2$$
$$F=\dfrac{(4\pi\times 10^{-7}T.m/A)}{2\pi}\dfrac{(25A)(25A)}{(4.0\times 10^{-2}m)}(25m)$$
$$=0.078N$$

Determine the magnitude of the force between two parallel wires 25 m long and 4.0 cm apart, each carrying 25 A in the same direction.

Correct option is A. $$0.078$$ NThe magnitude of the force between two parallel wires,$$F_2=\dfrac{\mu_0}{2\pi}\dfrac{I_1I_2}{d}l_2$$$$F=\dfrac{(4\pi\times 10^{-7}T.m/A)}{2\pi}\dfrac{(25A)(25A)}{(4.0\times 10^{-2}m)}(25m)$$$$=0.078N$$

Correct option is A. $$0.078$$ N
The magnitude of the force between two parallel wires,
$$F_2=\dfrac{\mu_0}{2\pi}\dfrac{I_1I_2}{d}l_2$$
$$F=\dfrac{(4\pi\times 10^{-7}T.m/A)}{2\pi}\dfrac{(25A)(25A)}{(4.0\times 10^{-2}m)}(25m)$$
$$=0.078N$$