a) We know that the magnetic force on the charge partical
F =qVB sin()
So
sin()=F/(qVB) ...1
here F=1.40*10^(-16) , B=1.25T ,V=4.00*10^3 m/s ,q=-1.6*10^(-19) C
put these values in equation 1
sin()=1.40*10^(-16)/(-1.6*10^(-19)*4.00*10^3*1.25)
sin()=-0.175
means
since sin(180o+) = -sin()
so =180+10.08 =190.08 o
Answer is
b) In a circular path
the magnetic force on the proton is equal to the centripetal force .
qVBsin() =mV^2/R
Here So sin(90) =1 therefore
qVB=mV^2/R
B=mV/Rq
here m=1.673*10^(-27) kg , V=7.5*10^7 ,R=0.80 m ,q=1.6*10^(-19) C
B=1.673*10^(-27)*7.5*10^7/(0.80 *1.6*10^(-19))
B=9.803*10-1
B=0.9803 T answer
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