Ahemispherical surface (half of a spherical surface) of radius r is located in a uniform electric field of magnitude e that is parallel to the axis of the hemisphere. what is the magnitude of the electric flux through the hemisphere surface?

π*R²*E

Explanation:

According to the definition of electric flux, it can be calculated integrating the product E*dA, across the surface.

As the electric field E is uniform and parallel to the hemisphere axis,  and no charge is enclosed within it, the net flux will be zero, so, in magnitude, the flux across the opening defining the hemisphere, must be equal to the one across the surface.

The flux across the open surface can be expressed as follows:

As E is constant, and parallel to the surface vector dA at any point, can be taken out of the integral, which is just the area of the surface, π*R².

⇒Flux = E*π*R²

answer: option c.

explanation:

adding a(n) to an electric circuit would increase the magnetic field around a wire;

b. solenoid

a solenoid is a type of electromagnet whose purpose is to generate a controlled magnetic field.

if the purpose of the solenoid is instead to impede changes in the electric current, a solenoid can be more specifically classified as an inductor rather than an electromagnet.