# An astronomer detects a signal of 2480 MHz from an electron travelling in a distant nebula. What is the strength of the magnetic field in the nebula?

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When a charged particle q enters with a speed v, in a uniform magnetic field B, experiences a magnetic force given by:

Fm = q (v x B)

Assuming that the velocity is perpendicular to the field; the magnetic force acts as a centripetal force and the electric charge makes a circular motion. Thus we can write:

Fc = Fm = mac

where:

m → is the mass of the charged particle.

ac → is the centripetal acceleration.

Substituting the expressions for the magnetic force and centripetal acceleration, we have:

qvB = mv^2/r

where:

r → is the radius of the circular path.

From the above equation, the speed v is:

v = rqB/m

The angular frequency of the circular motion is expressed as:

w = 2πf = v/r

In this case, this frequency is called, "cyclotron frequency".

So that:

2πf = qB/m → B = 2πfm/q

Substituting the values, we have:

B = (6.28)(2.48 x 10^9)(9.109 x 10^-31)/(1.602 x 10^-19)

B = 88.55 x 10^-3 T