Answer:
Magnetic field, B = 1.84 T
Explanation:
It is given that,
Charge on alpha particle, q = +2e = [tex]3.2\times 10^{-19}\ C[/tex]
Mass of alpha particle, [tex]m=6.6\times 10^{-27}\ kg[/tex]
Speed of alpha particles, [tex]v=1.6\times 10^7\ m/s[/tex]
We need to find the magnetic field strength required to bend them into a circular path of radius, r = 0.18 m
So, [tex]F_m=F_c[/tex]
[tex]F_m\ and\ F_c[/tex] are magnetic force and centripetal force respectively
[tex]qvB=\dfrac{mv^2}{r}[/tex]
[tex]B=\dfrac{mv}{qr}[/tex]
[tex]B=\dfrac{6.6\times 10^{-27}\ kg\times 1.6\times 10^7\ m/s}{3.2\times 10^{-19}\ C\times 0.18\ m}[/tex]
B = 1.84 T
So, the value of magnetic field is 1.84 T. Hence, this is the required solution.
The magnetic field strength required to bend them into a circular path is 1.83 T.
The magnetic force on the emitted charge is given as;
F = qvB
The centripetal force of the emitted charge is given as;
F = mv²/r
The magnetic field strength required to bend them into a circular path is calculated as follows;
qvB = mv²/r
[tex]B = \frac{mv}{rq}[/tex]
[tex]B = \frac{6.6 \times 10^{-27} \times 1.6 \times 10^7}{(2\times 1.6 \times 10^{-19} ) \times 0.18} \\\\B = 1.83 \ T[/tex]
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