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mihalych1998 [28]

3 years ago

9

(PLEASE HELP) : A muon (a particle with the same charge as an electron but with a mass of 1.88×10−28 kg) is traveling at 4.21×10

7 m/s at right angles to a magnetic field. The muon experiences a force of −5.00×10−12 N.
a. How strong is the magnetic field?

b. What acceleration does the muon experience?

1 answer:

Georgia [21]3 years ago

8 0

The magnetic force is 0.74T and acceleration of the Moun is calculated as 2.66*10^16 m/s^2

Data;

Force (F) = -5.00*10^-12N
Velocity = 4.21*10^7 m/s
Mass(m) = 1.88 * 10 ^-28 kg

<h3>Magnetic Field Force</h3>

The formula of magnetic field force is given by

$F =BqV$

Let's make the magnetic field force the subject of formula

$F=BqV\\B = \frac{F}{qV}\\B = \frac{5.00*10^-^1^2}{1.60*10^-^1^9 * 4.21*10^7} \\B = 0.74T$

The magnetic force is 0.74T

<h3>Acceleration of the Moun</h3>

The acceleration of moun can be calculated as

$F=ma\\a = F/m\\a = \frac{5.0*10^-^1^2}{1.88*10^-^2^8} \\a = 2.66*10^1^6m/s^2$

The acceleration of the Moun is calculated as 2.66*10^16 m/s^2

Learn more on magnetic force here;

brainly.com/question/7802337

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