Q. 564.7( 6 Votes )

# 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)

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