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Q.

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?

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a

No force will act on charged particle if x = y

b

lf x>y,Fx2y2

c

lf x>y and the charge is positive, the force will act along z-axis

d

lf y>x the force will act along y-axis

answer is A.

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Detailed Solution

If x=y, then vB, i.e., Fmag =0

Hence, (a) is correct.

Fmag=q(v×B)=q[(xi^+yj^)×(yi^+xj^)]Fmag=qx2y2k^

If x > y, then the force is along z-axis.

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