A copper wire having resistance 0.01 ohm in each metre 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 centre of the solenoid.
Solution
Given:
Resistance per unit length of the wire, `R/l`= 0.01 Ω/m
Radius of the wire, r = 1.0 cm = 0.01 m
Total no. of turns, N = 400
Magnetic field intensity, B = 1.0 × 10−2 T
Now,
Let E be the emf of the battery and R0 be the total resistance of the wire.
`therefore i = E / (R_0) = E /(0.01 xx 2 pi r xx 400)`
`= E/(0.01 xx 2 xx pi xx0.01 xx400)`
The magnetic field near the centre of the solenoid is given by
`B = mu_0 ni `
`= 1 xx 10^-2 = 4 pi xx 10^-7 xx 400/(20xx10^-2) xx E/(2pi xx 4 xx10^-2)`
`E = (10^-2 xx 20 xx 10^-2 xx 2 xx 10^-2)/(10^-7 xx 4 xx 10^2)`
= 1 V
