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What is the speed of a high-speed electron traveling at right angles to a 0.80-tesla magnetic field with a force of -7.5 x 10^-12 newtons? The charge of an electron is 1.60 x 10^-19

2 Answers

1 vote

Step-by-step explanation:

Answer:

Magnetic field, B = 0.016 Tesla

Step-by-step explanation:

It is given that,

Velocity of electron, v=9.6\times 10^5\ m/sv=9.6×10

5

m/s

Magnetic force, F=2.6\times 10^{-15}\ NF=2.6×10

−15

N

Charge, q=1.6\times 10^{-19}\ Nq=1.6×10

−19

N

The magnetic force is given by :

F=qvBF=qvB

B=\dfrac{F}{qv}B=

qv

F

B=\dfrac{2.6\times 10^{-15}}{1.6\times 10^{-19}\times 9.6\times 10^5}B=

1.6×10

−19

×9.6×10

5

2.6×10

−15

B=0.016\ TB=0.016 T

So, the magnitude of magnetic field is 0.016 Tesla. Hence, this is the required solution.

User Dhirschl
by
5.8k points
4 votes

Answer:

F = q v B so v = F/q B

v = -7.5 × 10-12/(-1.60 × 10-19) (0.80) = 5.9 × 107 m/s

Step-by-step explanation:

User Nayana Madhu
by
5.6k points