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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.

A
x=1.5
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B
x=3
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C
Magnetic field has a component along the z-direction
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D
Kinetic energy of the particle is constant
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Solution

The correct options are
B Magnetic field has a component along the z-direction
C Kinetic energy of the particle is constant
D x=1.5
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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