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 Dx=−1.5
Under the influence of magnetic field only, speed and hence, kinetic energy remains constant.
As →F⊥→v
So →F⋅→v=0⇒m→a⋅→v=0
⇒m(2^i+x^j)⋅(3^i+4^j)=0
⇒6+4x=0⇒x=−1.5
Let 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≠0
Hence, magnetic field has a component along z-direction.