Question

What would be the acceleration voltage that we have to apply in the e/m ratio apparatus, when the current in the Helmholtz coils is equal to 2 amp, to achieve a radius of 5 cm of the circle of the electron path? Turns per Coil, N=132. Coil Radius, a=147.5mm.

Answer 1

For an ideal Helmholtz pair, the magnetic field at the center of a coil is given by -

B = 8 ₀ N I / (5 5) a

where, ₀ = permeability of free space = 4 x 10^-7 mT/A

N = number of turns per coil = 132

a = radius of a coil = 147.5 x 10^-3 m

I = current in the Helmholtz coil = 2 A

then, we get

B = [(8) (4 x 10^-7 mT/A) (132) (2 A)] / [(5 5) (147.5 x 10^-3 m)]

B = [(0.002652672 mT) / (1.64910 m)]

B = 0.001608 T

We know that, F_B = F_C

e v B = m v² / r

(e / m) = v / B r                                                                                             { eq.1 }

From conservation of energy, we have

K.E = P.E

(1/2) m v² = e V_acc

v = 2 e V_acc / m

Inserting the value of 'v' in eq.1 & we get

(e / m) = 2 V_acc / B² r²

where, e = charge on electron = 1.6 x 10^-19 C

m = mass of electron = 9.11 x 10^-31 kg

r = radius of the circle of electron path = 0.05 m

then, we get

[(1.6 x 10^-19 C) / (9.11 x 10^-31 kg)] = 2 V_acc / [(0.001608 T)² (0.05 m)²]

(1.75 x 10¹¹ C/kg) = 2 V_acc / (6.46416 x 10^-9 m².T²)

2 V_acc = [(1.75 x 10¹¹ C/kg) (6.46416 x 10^-9 m².T²)]

V_acc = (1131.2 V) / (2)

V_acc = 565.6 V