A tightly-wound, long solenoid carries a current of 2.00 A. An electron is found to execute a uniform circular motion inside the solenoid with a frequency of 1.00 × 108 rev s–1. Find the number of turns per meter in the solenoid.
Given:
Current(i) = 2 A
Frequency(f) = 1.00 × 108 rev s–1.
Formula used:
Magnetic field inside 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
Now, charge of electron(q) = 1.6 x 10-19 C
Mass of electron(m) = 9.1 x 10-31 kg
Now, frequency of the particle in uniform circular motion in a magnetic field is given by
f = 
,
Where
q is the charge,
B is the magnetic field,
m is the mass of particle
Hence, B = 
Substituting the given values, we obtain B as,
Hence, 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
=> Number of turns per meter(n) = 
= 0.0036/(4π x 10-7 x 2) = 1420 turns/meter. (Ans)
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