Question

(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 4.21×107 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?

Answer 1

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

Magnetic Field Force

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

Acceleration of the Moun

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

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