Question

A charge of 8.4 × 10–4 C moves at an angle of 35° to a magnetic field that has a field strength of 6.7 × 10–3 T.

If the magnetic force is 3.5 × 10–2 N, how fast is the charge moving?

9.1 × 10–5 m/s
1.3 × 10–4 m/s
7.6 × 103 m/s
1.1 × 104 m/s


ive taken this test twice and gotten a 68 both times lol

Answer 1

The Lorentz force exerted on a charge moving in a magnetic field is given by:
$F=qvB\sin \theta$
where
q is the charge
v is the speed of the charge
B is the intensity of the magnetic field
$\theta$ is the angle between the directions of the field and of the velocity

The data we have in our problem are:
$q=8.4 \cdot 10^{-4}C$
$B=6.7 \cdot 10^{-3}T$
$\theta=35^{\circ}$
$F=3.5 \cdot 10^{-2} N$

If we substitute these data and we rearrange the initial equation, we can calculate the speed of the particle:
$v= \frac{F}{qB\sin \theta}= \frac{3.5 \cdot 10^{-2}N}{(8.4\cdot 10^{-4}C)(6.7 \cdot 10^{-3}T)(\sin 35^{\circ})} =1.1 \cdot 10^4 m/s$

so, the correct answer is
1.1 × 104 m/s

Answer 2

The velocity of a point charge moving under the influence of electric and magnetic fields is the velocity of that moving charge . The velocity of the moving charge will be 1.1 × 104 m/s.

What is Lorentz force?

Lorentz force is defined as the force acting on point charge when it is moving in an electric and magnetic field. Under the electric and magnetic field, a force is act on the charge known as the Lorentz force.

The given data in the problem is ;

q is the value of charge = 8.4 × 10–4 C

θ is the angle to the magnetic field= 35°

B is the magnetic field strength = 6.7 × 10–3 T.

v is the velocity if moving charge =?

Mathematically the Lorentz force is given by

$\rm F = qvBsin\theta$

$\rm v = \frac{F}{qBsin\theta} \\\\ \rm v = \frac{3.5\times 10^{-2}}{8.4\times10^{-4}sin23^0} \\\\ \rm v =1.1\times10^{-4}$

To learn more about the Lorentz force refer to the link;

https://brainly.com/question/13791875