A copper wire having resistance 0.01 ohm in each meter is used to wind a 400-turn solenoid of radius 1.0 cm and length 20 cm. Find the emf of a battery which when connected across the solenoid will cause a magnetic field of 1.0×10–2 T near the center of the solenoid.
Given:
Resistance per unit length(R0) = 0.01 ohm
Total number of turns(N) = 400
Radius of the wire(r) = 1 cm = 0.01 m
Length of the wire(l) = 20 cm = 0.2 m
Magnetic field near the center(B) = 1.0×10–2 T
Formula used:
Magnetic field at the center of a solenoid, B = μ0ni,
Where
μ0 = magnetic permeability of vacuum = 4π x 10-7 T m A-1,
n = number of turns per unit length,
i = current carried by the wire
Emf(E) = iR,
where i = current, R = resistance
Now, n(number of turns per unit length) = Total number of turns(N)/Length of wire(l) = 
= 2000
Total resistance (R of all turns) = R0 x 2πr x 400 = (0.01 x 2π x 0.01 x 400) = 0.25 ohm
Now, substituting the given values:
1.0×10-2 = 4π x 10-7 x 2000 x i => i = 3.98 A
Therefore, Emf(E) = iR = (3.98 X 0.25) = 0.995 V which is almost equal to 1 V.
Emf of the battery = 1 V (Ans)