Question

SHOW ALL STEP PLZZ

Wires A and B are identical, except that the length of wire A is one-half that of wire B. If the same potential difference is applied between the two ends of each wire, how does the electric field within wire A compare to that in wire B?

why is the answer

The electric field in wire A is twice that in wire B.

Answer 1

let RA and RB are the resistance of wires A and B.

we know, resistance of a wire , R = rho*L/A

rho --> rsistivity of the material
L --> length of the wire
A --> cross sectional area of the wire.

let RA = R (rho*L/A)

RB = rho*(2*L)/A

= 2*rho*L/A

= 2*R

if delta_V is the applied potential.

current through wire A, IA = delta_V/RA

= delta_V/R

current through wire B, IB = delta_V/(RB)

= delta_V/(2*R)

= IA/2

we know from Ohm's law, E = J*rho

E = I*rho/A

EA = IA*rho/A

EB = IB*rho/A

EA/EB = IA/IB

= IA/(TA/2)

= 2

EA/EB = 2

so, electric in wire A is twice that in wire B.