An electron moves at 2.60×10 6 m/s through a region in which there is a magnetic

Question

An electron moves at 2.60×106 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.70×102 T .

Part A

What is the largest possible magnitude of the acceleration of the electron due to the magnetic field?

Part B

What is the smallest possible magnitude of the acceleration of the electron due to the magnetic field?

Part C

If the actual acceleration of the electron is 14 of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?

Explanation / Answer

The magnitude of the magnetic force is F = Q v B sin = m a

so that

a = Q v B sin / m

where 0 sin 1

it follows that

a (min) = 0 and a (max) = Q v B / m

Q v B = ( 1.6 × 10 ¹ C ) ( 2.6 × 10 m / s ) ( 7.70 × 10 ² T )

Q v B = 3.203 × 10 ¹

a (max) =3.203 × 10 ¹ / 9.11 × 10 ³¹ = 3.5 × 10 ¹ m / s ²

a) a(max) = 3.5 × 10 ¹ m / s ²

b) a(min) = 0

c) a = ¼ a (max) = ¼ ( Q v B / m ) = Q v B sin / m

sin = ¼

= sin ¹ ( ¼ ) = 14.5°

= 14.5°

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