A proton experiences an eastward force of 30 × N while traveling in a magnetic field of 5 × T north.
To be given in question
Magnetic force ="30\\hat{i}N"
Magnetic field= "5\\hat{j}T"
To be asked in question
(a)Magnitude of proton velocity (v)=?
(b) velocity direction
(C) diagram
Part(a)
We know that
"\\overrightarrow{F}=q(\\overrightarrow{v}\\times\\overrightarrow{B})"
F=qvBsin"\\theta""(\\hat{n})"
Magnitude of F
F=qvB "\\rightarrow(1)"
"v=\\frac{F}{qB}"
Put value
"v=\\frac{30}{1.6\\times10^{-19}\\times5}" m/sec
"v=3.75\\times10^{19}m\/sec"
Part {b}
State the direction of proton velocity
"\\overrightarrow{F}=q(\\overrightarrow{v}\\times\\overrightarrow{B})"
"F=qvBsin\\theta\\hat{n}"
"\\hat{n}" Direction"\\overrightarrow({q}\\times \\overrightarrow{B} )direction"
F direction is "\\hat{i}" And
"(\\overrightarrow{v}\\times\\overrightarrow {B})"
direction "\\hat{i}" Possible
When
Velocity V direction ("-\\hat{k}" )
Then we can written as
"\\hat{i}=(-\\hat{k}\\times\\hat{j})"
Which means velocity of proton move negetive z -axis
"V=-v\\hat{k}"
Part(c)
According to diagram velocity direction is"-\\hat{k}"
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