Q. 585.0( 1 Vote )

# 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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