A positively charged particle has a velocity in the negative z direction at poin

Question

A positively charged particle has a velocity in the negative z direction at point P. The magnetic force on the particle at this point is in the negative y direction. Which one of the following statements about the magnetic field at point P can be determined from this data? B_x is positive. B_z is positive. B_y is negative. B_y is positive. B_x is negative. ANS: A A charged particle (mass = 4.0 mu g, charge = 5.0 mu C) moves in a region where the only force on it is magnetic. What is the magnitude of the acceleration of the particle at a point where the speed of the particle is 5.0 km/s, the magnitude of the magnetic field is 8.0 mT, and the angle between the direction of the magnetic field and the velocity of the particle is 60 degree? Ans: 43 km/s^2 A 2.0-m wire carries a current of 15 A directed along the positive x axis in a region where the magnetic field is uniform and given by B (30I - 40J) mT. What is the resulting magnetic force on the wire? Ans: (-1.2 k) N

Explanation / Answer

5. from lorentz law,

magnetic force on a charged particle is

F=qv x B

=q (-vk) x (Bxi+Byj+Bzk)

=-qv(Bxj-Byi)

FB in -j ----->By=0 and Bx>0

so, ans. is A

Bx is positive.

6. mass = 4 micro g

q=5micro C

v=5 km/s

B=8 mT

angle=60deg

from lorentz law

F= q v x B

F= qvB sin(theta)

we know that F=ma

a=F/m

a= qvB sin(theta) /m

substitute all the values

a=(5*10^-6*5*10^3*8*10^-3*sin(60))/(4*10^-6)

a=43 m/s^2

7.given data

l=2m
I=15 i A
B=(30i-40j )mT
F=?

F=l i x B

from the cross product

i x B= (15 i) x (30*10^-3 i - 40*10^-3j) = -0.6 k

F=l i x B = 2 * -0.6 k =-1.2 k N

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