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A narrow copper wire of length L and radius b is attached to a wide copper wire of length L and radius 2b, forming one long wire of length 2L. This long wire is attached to a battery, and a current is flowing through it. If the electric field in the narrow wire is E, the electric field in the wide wire is ___

(A) E
(B) 2E
(C) 4E
(D) E/2
(E) E/4

Respuesta :

Answer:

E) E₂=E/4

Explanation:

Given that

Both wire are connects in the series so the current flow will be same in the both wire but voltage difference will be different.

Resistance R

1 - for first wire

2 - for second wire

R₁=ρL/A                

R₂=ρL/A'

Given that radius of second wire is double then the area of second wire will be 4 times of first wire.

A'= 4 A

R₂=ρL/A'

R₂=ρL/4A

R₂ = R₁/4

For wire 1

We know that

V= I R

V₁= I R₁

V₂=I R₂

R₂ = R₁/4

so

V₁=4 V₂

So we can say that

E₂=E₁/4

Given that E₁=E

E₂=E/4

onyebuchinnaji

If the electric field in the narrow wire is E, then electric field in the wide wire is E/4.

Resistance of the copper wire

The initial resistance of the copper wire given as;

R₁ = ρL/A      

where;

  • A is area of the wire = πr²

when the radius of the wire doubles, the area becomes 4 times larger.

The new resistance of the wire

R₂ = ρL/4A

R₂ = R₁ /4

Apply ohms law

E = IR

E₁ = IR₁

E₂ = IR₂

E₁/R₁ = E₂/R₂

E₁/R₁ = E₂/R₁/4

E₁/R₁ = 4E₂/R₁

E₁ = 4E₂

E₂ = E₁/4

Thus, if the electric field in the narrow wire is E, then electric field in the wide wire is E/4.

Learn more about resistance of wire and electric field here: https://brainly.com/question/13197035

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