An electron moves in the magnetic field B =0.490 i T with a speed of 1.50 *10^7

Question

An electron moves in the magnetic field

B=0.490 i T with a speed of 1.50 *10^7 m/s in the direction shown in the figure. For each, what is magnetic force F on the electron ?

PART A :

Express vector F in the form of F x , F y , F z , where the x, y , and z components are separated by commas.

PART B :

Express vector F in the form of F x , F y , F z , where the x, y , and z components are separated by commas.

An electron moves in the magnetic field B=0.490 i T with a speed of 1.50 *10^7 m/s in the direction shown in the figure. For each, what is magnetic force F on the electron ? Express vector F in the form of F x , F y , F z , where the x, y , and z components are separated by commas. Express vector F in the form of F x , F y , F z , where the x, y , and z components are separated by commas.

Explanation / Answer

Magnetic force F= q*V*B (V*B is cross product of velocity vector and magnetic field vector).

a)V = - 1.50 *10^7 m/s j , B=0.490 i T , q=charge of electron .

--> F= 1.602*10^-19* 1.5*10^7 * 0.49 k N

--> F = 1.177*10^-12 N in the direction of k

Fx=0 N ,Fy=0 N,Fz= -1.177*(10^-12) N

b)V=1.5*10^7/sqrt(2) (k-j) , B = 0.49 i T

--> F = 1.602*10^-19 *1.5*10^7 * 0.49/sqrt(2) (j-k)

Fx= 0 N, Fy = 0.832*10^-12 N,Fz = -0.832*10^-12 N

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