Question

Problem 3. a. An electron moving at 4.00x103 m/s in a 1.25-T magnetic field experiences a magnetic force of 1.40*10-16 N. What angle does the velocity of the electron make with the magnetic field? There are two answers. b. A proton moves at 7.50x107 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius 0.800 m. What is the field strength?

Answer 1

a) We know that the magnetic force on the charge partical

F =qVB sin($\theta$)

So

sin($\theta$)=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($\theta$)=1.40*10^(-16)/(-1.6*10^(-19)*4.00*10^3*1.25)

sin($\theta$)=-0.175

$\theta=sin^{-1}(-0.175)$

$\theta=-10.08^{o}$

means

= 360° – 10.08° = 349.52°

since sin(180^o+$\theta$) = -sin($\theta$)

so $\theta'$ =180+10.08 =190.08 ^o

Answer is

$\theta =190.08^o , 349.52^o$

b) In a circular path

the magnetic force on the proton is equal to the centripetal force .

qVBsin($\theta$) =mV^2/R

Here $\theta=90^o$ 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