A region of space contains a magnetic field of 500 G and an electric field of 3 $\times$ 10⁶ N/C.The electrical energy density in a cubical box of side = 20 cm in this region is A)19.9 J/m³ B)39.8 J/m³ C)54.8 J/m³ D)66.9 J/m³ E)74.6 J/m³

Question

Quizplus · Accepted Answer

The electrical energy density formula is given by:
𝑈 = 1/(2 * ε0 ) * E^2

where ε0 is the permittivity of free space and E is the electric field strength.

Here, the given electric field strength is E = 3 * 10^10 N/C.

So, 𝑈 = 1/(2 * ε0) * (3 * 10^10)^2

Now, the magnetic field does not contribute to the electrical energy density calculation.

The side of the cubical box is given as 20 cm. So, the volume of the cubical box is (20*10^-2)^3 m^3.

Thus, the total electrical energy in the cubical box is given by:

𝐸 = 𝑈 * volume = (1/(2 * ε0) * (3 * 10^10)^2) * (20*10^-2)^3

On substituting the values, we get 𝐸 = 3.98 J.

Hence, the electrical energy density is 𝐸 / volume = 39.8 J/m^3.

Therefore, the correct answer is option B.

A Region of Space Contains a Magnetic Field of 500 ×\times×

A Region of Space Contains a Magnetic Field of 500 $\times$