All categories

3 years ago

Which equation is used to calculate the magnetic force on a charge moving in a magnetic field?

2 answers:

7 0

The force is perpendicular to the plane formed by v and B. Since the force is zero if v is parallel to B, charged particles often follow magnetic field lines rather than cross them.

B) F = |q|vBsin

katrin2010 [14]3 years ago

6 0

Answer: The correct option is (B).

Explanation:

The force on a charge moving in a magnetic field is as follows;

$F= qvBsin\theta$

Here, q is the charge, v is the velocity, B is the magnetic field and $\theta$ is the angle between the velocity and the magnetic field.

The magnetic force is perpendicular to both magnetic field and velocity. Its direction is determined by the right hand rule.

If the angle between the velocity and the magnetic field is 90 degree then the expression of the magnetic force becomes as;

$F= qvB$

Therefore, the correct option is (B).

You might be interested in

Now review the difference between the temperatures on Earth and Mars. Also look at their distances from the sun.

aksik [14]

Answer: Use Question cove you can get it faster you can get the answer faster! ;) hope this helps ;) but yeah use that and answer is done right away

Explanation: HOPE THIS HELPSS!! ;))

7 0

3 years ago

Read 2 more answers

While driving down his street one evening, Jonah notices that his neighbor has laid out an electric fan for the garbage to pick

Kryger [21]

Answer: I have no idea either i need help aswell

Explanation:

4 0

3 years ago

Read 2 more answers

How is the q of an rlc parallel resonant circuit calculated?

notka56 [123]

Answer:

It is calculated by dividing Resistance, R, by Inductive reactance, XL.

Explanation:

Q is called the Q factor of a resonance circuit. In a parallel resonance circuit, it is calculated by finding the ratio of the power stored in the circuit to the power distributed in the circuit. It is a way of measuring the quality of a circuit or how effective the circuit is.

Q factor is the inverse in the resonance series circuit.

Q factor of a resonance parallel circuit,

R = Resistance

XL = Inductive reactance

3 0

3 years ago

A 6.72 particle moves through a region of space where an electric field of magnitude 1270 N/C points in the positive direction,

Romashka [77]

Answer:

$v_{y} = -104 m/s$

Explanation:

Using:

Force = electric field * charge

$F=e*q$

Force = magnitude of charge * velocity * magnetic field * sin tither

$F_{x2}= |q|*v*B*sin \alpha$

Force on particle due to electric field:

$F_{x1}= E*q = (1270N/C)*(-6.72*10^{-6} ) = -8.53*10^{-3}$

Force on particle due to magnetic field:

$F_{x2}= |q|*v*B*sin \alpha = (6.72*10^{-6} )*(1.15)*(sin90)*v = (7.728*10^{-6})*(v)$

$F_{x2}$ is in the positive x direction as $F_{x1}$ is in the negative x direction while net force is in the positive x direction.

Magnetic field is in the positive Z direction, net force is in the positive x direction.

According to right hand rule, Force acting on particle is perpendicular to the direction of magnetic field and velocity of particle. This would mean the force is along the y-axis. As this is a negatively charged particle, the direction of the velocity of the particle is reversed. Therefore velocity of particle, v, has to be in the negative y direction.

Now,

$F_{xnet}- F_{x1 } = F_{x2 }$