Answer :

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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