Velocity and acceleration vectors of a charged particle moving in a magnetic field at some instant are $\to v = 3^i + 4^j$ and $\to a = 2^i + x^j$ . Select the correct alternative(s) :
$x = 1.5$
$x = 3$
kinetic energy of the particle is constant
magnetic field is along z-direction

Question

Velocity and acceleration vectors of a charged particle moving in a magnetic field at some instant are $\to v = 3^i + 4^j$ and $\to a = 2^i + x^j$ . Select the correct alternative(s) :
$x = 1.5$
$x = 3$
kinetic energy of the particle is constant
magnetic field is along z-direction

Toppr Admin · Accepted Answer

Force -amp- velocity are perpendicular to each other in magnetic fieldSo-x2192-V-x22C5-x2192-a-0-3-i-4-j-x22C5-2-i-x-j-06-4x-0x-x2212-1-5No work is done by magnetic field- So- KE of particle is constant

Velocity and acceleration vectors 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 alternative(s) :x=1.5x=3kinetic energy of the particle is constantmagnetic field is along z-direction

Force & velocity are perpendicular to each other in magnetic fieldSo→V⋅→a=0(3^i+4^j)⋅(2^i+x^j)=06+4x=0x=−1.5No work is done by magnetic field. So, KE of particle is constant

Velocity and acceleration vectors of a charged particle moving in a magnetic field at some instant are $\to v = 3^i + 4^j$ and $\to a = 2^i + x^j$ . Select the correct alternative(s) :
$x = 1.5$
$x = 3$
kinetic energy of the particle is constant
magnetic field is along z-direction

Force & velocity are perpendicular to each other in magnetic field
So
$\to V \cdot \to a = 0$
$(3^i + 4^j) \cdot (2^i + x^j) = 0$
$6 + 4 x = 0$
$x = - 1.5$
No work is done by magnetic field. So, KE of particle is constant