(a) The maximum speed of the proton is 3.39 x 10⁵ m/s.
(b) The current generated through the looped wire is 28.2 A.
The maximum speed of the proton is calculated as follows;
K.E = eV
¹/₂mv² = eV
v² = 2eV/m
where;
v² = (2 x 1.6 x 10⁻¹⁹ x 600) / (1.67 x 10⁻²⁷)
v² = 1.15 x 10¹¹
v = √((1.15 x 10¹¹)
v = 3.39 x 10⁵ m/s
V(max) = NBAω
where;
A = πr²
A = π(0.04)² = 0.00503 m²
ω = v/r
ω = (3.39 x 10⁵)/(0.04) = 8,475,000 rad/s
B = V/(NAω)
B = (600) / (1000 x 0.00503 x 8,475,000)
B = 1.41 x 10⁻⁵ T
The current in the loop at a given distance from the magnetic field is calculated as follows;
[tex]B = \frac{\mu I}{2\pi d} \\\\I = \frac{2B\pi d}{\mu} \\\\[/tex]
where;
[tex]I = \frac{2(1.41 \times 10^{-5}) \times \pi \times 0.4}{4\pi \times 10^{-7}} \\\\I = 28.2\ A[/tex]
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