Answer:
[tex]5.48\cdot 10^{-14} N[/tex]
Explanation:
When a charged particle is moving in a region with a magnetic field, the particle experiences a force perpendicular to its direction of motion. The magnitude of this force is given by
[tex]F=qvB sin \theta[/tex]
where
q is the charge of the particle
v is its velocity
B is the strength of the magnetic field
[tex]\theta[/tex] is the angle between the direction of v and B
In this problem we have:
[tex]q=1.6\cdot 10^{-19}C[/tex] is the charge of the proton
[tex]v=0.261 c[/tex] is the speed of the proton, where
[tex]c=2.998\cdot 10^8 m/s[/tex] is the speed of light
[tex]B=0.00667 T[/tex] is the strength of the magnetic field
[tex]\theta=139^{\circ}[/tex] is the angle between the direction of the proton and the magnetic field
Substituting, we find the magnitude of the force:
[tex]F=(1.6\cdot 10^{-19})(0.261\cdot 2.998\cdot 10^8)(0.00667)(sin 139^{\circ})=5.48\cdot 10^{-14} N[/tex]
The magnitude of the magnetic force acting on the proton is [tex]5.46*10^{-14} N[/tex]
When a charge particle is moving in magnetic field.
Then force is given as, [tex]F=qvBsin(\theta)[/tex]
where
Given that, [tex]q=1.6*10^{-19}C,v=0.261*3*10^{8}=7.8*10^{7} ,B=0.00667T,\theta=139[/tex]
Substitute all values in above relation.
[tex]F=1.6*10^{-19}*7.8*10^{7} *0.00667T*sin(139)\\\\F=5.46*10^{-14} N[/tex]
Learn more about the magnetic force here:
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