Question

A 4.80 g bullet moves with a speed of 170 m/s perpendicular to the Earth's magnetic field of 5.00×10−5T.

Answer 1

Answer:

$3.24\times 10^{-7}\ \text{m}$

Explanation:

m = Mass of bullet = 4.8 g

v = Velocity of bullet = 170 m/s

B = Magnetic field of Earth = $5\times 10^{-5}\ \text{T}$

q = Charge of bullet = $1.06\times 10^{-8}\ \text{C}$

a = Acceleration

Time the bullet will be in the air for is $t=\dfrac{1000}{170}=5.88\ \text{s}$

Force is given by

$F=ma$

Magnetic force is given by

$F=qvB$

So

$ma=qvB\\\Rightarrow a=\dfrac{qvB}{m}\\\Rightarrow a=\dfrac{1.06\times 10^{-8}\times 170\times 5\times 10^{-5}}{4.8\times 10^{-3}}\ \text{m/s}^2$

From the linear equations of motion we have

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0+\dfrac{1}{2}\times \dfrac{1.06\times 10^{-8}\times 170\times 5\times 10^{-5}}{4.8\times 10^{-3}}\times 5.88^2\\\Rightarrow s=3.24\times 10^{-7}\ \text{m}$

The defelection of the bullet is $3.24\times 10^{-7}\ \text{m}$