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# 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.x=3Magnetic field has a component along the z-directionKinetic energy of the particle is constantx=−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 →F⊥→vSo →F⋅→v=0⇒m→a⋅→v=0⇒m(2^i+x^j)⋅(3^i+4^j)=0⇒6+4x=0⇒x=−1.5Let the magnetic field is →B=a^i+b^j+c^k, then from →F=q→v×→B⇒m(2^i+x^j)=q(3^i+4^j)×(a^i+b^j+c^k)⇒2m^i−1.5m^j=q(3b^k−3c^j−4a^k+4c^i)4c=2m⇒c≠0Hence, magnetic field has a component along z-direction.

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