A long, straight, horizontal wire carries a left-to-right current of 20 A. If the wire is placed in a uniform magnetic field of magnitude 4.0\times 10^{-5}~\text{T}4.0×10 −5 T that is directed vertically downward, what is the resultant magnitude of the magnetic field 20 cm above the wire?

Explanation:

vertical magnetic field B_v = 4 x 10⁻⁵ T.

Magnetic field due to horizontal current at point 20 cm above

= (μ₀/4π ) x (2i / R)

= 10⁻⁷ x 2 x 20/ 20 x 10⁻²

= 2 x 10⁻⁵ T

It will act  coming out of paper. Hence it will be normal to magnetic field given .

So resultant magnetic field

= √ (4² + 2²) x 10⁻⁵ T

= 4.47 X 10⁻⁵ T .

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures