Question

A 10 g bullet with a charge of 4.00 μC is fired at a speed of 270 m s−1 in a horizontal direction. A vertical magnetic field of 500 µT exists in the space. Find the deflection of the bullet due to the magnetic field as it travels through 100 m. Make appropriate approximations.

Answer 1

Explanation:

Step 1:

In the above question, it is given that

Bullet’s mass, $m=10 \mathrm{g}$

Bullet’s charge, $q=4.00 \mu \mathrm{C}$

In horizontal direction, the bullet speed, $v=270 \mathrm{m} / \mathrm{s}$

Upright magnetic field, $B=500 \mu \mathrm{T}$

Bullet travelled a distance of 100 m (d)

Magnetic force is given as

$F \rightarrow=q v \rightarrow x B$  ….(i)

Step 2:

And,

$F=m a$

Step 3:

We can write using equation (i):

$\mathrm{ma}=\mathrm{qv} \rightarrow \times \mathrm{B} \rightarrow$

$a=q v B m$

Step 4:

Bullet travels at the distance of 100 m horizontally and the time taken for that is $t=d v=100270 s$

In this time interval, the deflection caused because of magnetic field,  

$y=12 a t 2$  

$=12 \times 4.00 \times 10-6 \times 10-610 \times 270 \times 500 \times 10-3 \times 1002702=3.7 \times 10-6 \mathrm{m}$