A particle of mass 0.2g carries a charge 10^-8 C. It was given an initial horizo

Question

A particle of mass 0.2g carries a charge 10^-8 C. It was given an initial horizontal velocity v vector = (5 times 10^4m/s) i. vector What are the magnitude and direction of the minimum magnetic field that will keep the particle moving in the same direction? Neglect Earth's magnetic field. An electron at the origin has a velocity v vector_0(5 times 10^6 m/s)j vector a)Find the magnitude and direction of the magnetic field that will cause the electron to follow the circular path in xy-plane of radius r=10cm. b) What is the period of rotation?

Explanation / Answer

3) m =0.2 g, q =10^-8C, v = 5*10^4 i m/s

weight of particle = magnetic force

mg = qvB

0.2*0.001*9.8 = 10^-8*5*10^4*B

B =3.92 T

F =q(vxB)

-j = (i xk)

B is along out of page

4) vo =5*10^6 m/s j , mass of electron m =9.1*10^-31 kg, charge q =-1.6*10^-19 C , r = 10 cm

mv^2/r = qvB

mv/r = qB

9.1*10^-31*5*10^6/(0.1) = 1.6*10^-19*B

B =2.844*10^4 T

F is perpendicular to xy plane

F =q(vxB)

k = - (jxi)

B is along x direction (i)

(b) v = 2pi*r/T

5*10^6 =2*3.14*0.1/T

T =1.256*10^-7 s

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