Anarrow 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

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

magnityde and direction

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation: