Question

The magnetic field in a region of space is given by B = 0.510 j T. The velocity vector of an electron moving at 7.60 ✕ 106 m/s makes an angle of 45.0° with this field. What are the magnitudes of the following?(a) the magnetic force on the electron

(b) the electron's acceleration

Answer 1

(a) Magnetic force will be equal to
`4.384* 10^(-13)N`

(b) Acceleration will be equal to
`=0.4818* 10^(18)m/sec^2`

Step-by-step explanation:

We have given magnetic field in a region of space B = 0.510 T

Velocity of electron moving
`v=7.6* 10^6m/sec`

Angle between velocity and field
`\Theta =45^(\circ)`

Charge on electron
`q=1.6* 10^(-19)C`

(A) Magnetic force will be equal to
`F=qvBsin\Theta`

So magnetic force
`F=1.6* 10^(-19)* 7.6* 10^6* 0.510* sin45^(\circ)=4.384* 10^(-13)N`

(B) Mass of electron
`m=9.1* 10^(-31)kg`

According to newton's law we know that F = ma , here m is mass and a is acceleration

So acceleration
`a=(F)/(m)=(4.384* 10^(-13))/(9.1* 10^(-31))=0.4818* 10^(18)m/sec^2`