Question

a particle with a charge of 0.030 C experiences a magnetic force of 1.5 N while moving at right angles to a uniform magnetic field. If the speed of the charge is 620 m/s, what is the magnitude of the magnetc field the particle passes through?

Answer 1

The magnetic force experienced by a charged particle moving at right angle in a magnetic field is given by:
$F=qvB$
where
q is the charge
v is the speed of the particle
B is the intensity of the magnetic field

In our problem, q=0.030 C, v=620 m/s and F=1.5 N, therefore if we re-arrange the equation and we plug these data into it, we find the intensity of the magnetic field:
$B= \frac{F}{qv}= \frac{1.5 N}{(0.030 C)(620 m/s)}=0.08 T$

Answer 2

The value of the magnitude of the magnetic field, the particle passes through is 0.08 T.

What is the magnetic field?

The magnetic field is the field in the space and around the magnet in which the magnetic field can be filled.

The magnetic field experienced by a charged particle can be given as,

$B=\dfrac{F}{qv}$

Here, (q) is the charge of the particle, (v) is the speed of the particle, and (F) is the magnitude of the magnetic force.

The charge on the particle is 0.030 C and the magnetic force experienced by the particle is 1.5 N.

As the speed of the charge is 620 m/s. Thus put the values of the given data in the above formula, to find the value of the magnetic field as,

$B=\dfrac{1.5}{0.030\times620}\\B=0.08\rm T$

Thus, the value of the magnitude of the magnetic field, the particle passes through is 0.08 T.

Learn more about the magnetic fields here;

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