A charged particle with velocity v=xi^+yj^ moves in a magnetic field B=yi^+xj^. The force acting on the particle has magnitude F. Which one of the following statements is/are correct?
This question has multiple correct options
A
No force will act on charged particle if x=y
B
If x>y,F∝(x2−y2)
C
If x>y, the force will act along z-axis
D
If y>x, the force will act along y-axis
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Updated on : 2022-09-05
Solution
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Correct options are A) , B) and C)
v=xi^+yj^
B=yi^+xj^
f=q(v×B)=q((xi^+yj^)×(yi^+xj^))=q(x2−y2)k^
If x=y, f=q(x2−y2)=q(0−0)=0
If x>y, f=q(x2−y2)k^>0
If y>x, f=q(x2−y2)k^<0 and it points in the −vez− axis.
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