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Question

Velocity and acceleration vector of a charged particle moving in a magnetic field at some instant are v=3^i+4^j and a=2^i+x^j. Select the correct options.
  1. x=3
  2. Magnetic field has a component along the z-direction
  3. Kinetic energy of the particle is constant
  4. x=1.5

A
x=3
B
Kinetic energy of the particle is constant
C
x=1.5
D
Magnetic field has a component along the z-direction
Solution
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Under the influence of magnetic field only, speed and hence, kinetic energy remains constant.

As Fv

So Fv=0mav=0

m(2^i+x^j)(3^i+4^j)=0

6+4x=0x=1.5

Let the magnetic field is B=a^i+b^j+c^k, then from F=qv×B

m(2^i+x^j)=q(3^i+4^j)×(a^i+b^j+c^k)

2m^i1.5m^j=q(3b^k3c^j4a^k+4c^i)

4c=2mc0

Hence, magnetic field has a component along z-direction.

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