Answer:
[tex]3.24\times 10^{-7}\ \text{m}[/tex]
Explanation:
m = Mass of bullet = 4.8 g
v = Velocity of bullet = 170 m/s
B = Magnetic field of Earth = [tex]5\times 10^{-5}\ \text{T}[/tex]
q = Charge of bullet = [tex]1.06\times 10^{-8}\ \text{C}[/tex]
a = Acceleration
Time the bullet will be in the air for is [tex]t=\dfrac{1000}{170}=5.88\ \text{s}[/tex]
Force is given by
[tex]F=ma[/tex]
Magnetic force is given by
[tex]F=qvB[/tex]
So
[tex]ma=qvB\\\Rightarrow a=\dfrac{qvB}{m}\\\Rightarrow a=\dfrac{1.06\times 10^{-8}\times 170\times 5\times 10^{-5}}{4.8\times 10^{-3}}\ \text{m/s}^2[/tex]
From the linear equations of motion we have
[tex]s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0+\dfrac{1}{2}\times \dfrac{1.06\times 10^{-8}\times 170\times 5\times 10^{-5}}{4.8\times 10^{-3}}\times 5.88^2\\\Rightarrow s=3.24\times 10^{-7}\ \text{m}[/tex]
The defelection of the bullet is [tex]3.24\times 10^{-7}\ \text{m}[/tex]
