what must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a 0.46 t magnetic field?



Answer :

The magnitude of a uniform electric field is 1.47 × [tex]10^{8}[/tex] V/m.

What is an electric field?

When charge is present in any form, a point in space has an electric field that is connected to it. The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field. Without any precise information of what generated the field, simply knowing the value of the electric field at a certain location is sufficient to predict what would happen to electric charges nearby.

Given,

B = 0.49 T

Energy density due to the magnetic field

[tex]u_{b}[/tex] = [tex]\frac{1}{2mu_{0}}[/tex]B²

Energy density due to the electric field

[tex]u_{e}[/tex] = [tex]\frac{1}{2}[/tex]ε₀E²

We know, [tex]u_{b}[/tex] = [tex]u_{e}[/tex]

[tex]\frac{1}{2mu_{0} }[/tex]B² = [tex]\frac{1}{2}[/tex]ε₀E²

[tex]\frac{1}{mu_{0} }[/tex]B² = ε₀E²

E² = [tex]\frac{1}{mu_{0}E_{0} }[/tex]B²

E =  [tex]\frac{1}{\sqrt{mu_{0}E_{0}}}[/tex]B

[tex]\frac{1}{\sqrt{4\pi X 10^{-7} X 8.852 X 10^{-12}}}[/tex] × 0.49

3 × [tex]10^{8}[/tex] × 0.49

= 1.47 × [tex]10^{8}[/tex] V/m

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