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Oduvanchick [21]

2 years ago

3

A particle (q = –4.0 µC, m = 5.0 mg) moves in a uniform magnetic field with a velocity having a magnitude of 2.0 km/s and a dire

ction that is 50° away from that of the magnetic field. The particle is observed to have an acceleration with a magnitude of 5.8 m/s2. What is the magnitude of the magnetic field?

1 answer:

krok68 [10]2 years ago

5 0

Answer:

The magnitude of the magnetic field is 4.73 T.

Explanation:

Given that,

Charge of the particle, $q=-4\ \mu C=-4\times 10^{-6}\ C$

It is moving with a velocity of 2 km/s and a direction that is 50° away from that of the magnetic field. The particle is observed to have an acceleration with a magnitude of 5.8 m/s². We need to find the magnitude of magnetic field.

When it enters in magnetic field,

$qvB\sin \theta=ma$

B is magnetic field

$B=\dfrac{ma}{qv\sin \theta}\\\\B=\dfrac{5\times 10^{-3}\times 5.8}{4\times 10^{-6}\times 2\times 10^3\sin (50)}\\\\B=4.73\ T$

So, the magnitude of the magnetic field is 4.73 T.

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