Question

unI Fald Ines th a veloc 8-fel ㄨㄧㄧㄨㄧㄨ x XxX e the

Answer 1

Part A)

The positive X axis is to right, positive Y axis is upwards and positive Z axis is out of the page.

Unit vectors along Positive X,positive Y and positive Z axes are $\hat{i},\hat{j},\hat{k}$ respectively.

a) In the given figure, magnetic field is directed to right,
B= 1.5 i T
and velocity vector is making an angle 30⁰ with the field direction, that is
3 * 10"(cos 30% + sin 30%) m/s

Magnetic force

F=qv × B

F(1.6 1019)3 10(cos 30+sin 30°j) x (1.5i)

F (1.6*1019) 3 105 sin 301.5(-) *3*10"sin 30° * 1.

F (3.6 10-4)-k)

Magnitude of magnetic force is and direction of magnetic force is in to the page.

F 3.6* 10-14 N

b) Acceleration of proton is $a=\frac{F}{m}=\frac{3.6*10^{-14}}{1.67*10^{-27}}=2.16*10^{13}\,m/s^2$

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Part B)

In the given figure, magnetic field is directed into the page , and velocity vector is to right , that is

B = 1.5(-1) T

2 10 i m/s

1) Path of proton is as shown below.

2) Magnetic force

F=qv × B

F= (1.6 * 10-1912 * 10%) × 1.5(-k)

$\vec{F}=4.8*10^{-14}\,\hat{j}\,\,N$

Magnitude of initial magnetic force on the proton is and its direction is upwards along positive Y direction.

F-4.8* 10-14 AN

3) Radius of path followed by proton is $r=\frac{mv}{qB}=\frac{1.67*10^{-27}*2*10^5}{1.6*10^{-27}*1.5}=0.00139\,m=0.139\,cm$

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