Question

an electron is moving along x-axis in xy plane with velocity 10 power 2 m per second a magnetic field of magnitude 20 is present in the region along Z Axis if the magnitude of the force on the particle is 1.6 m into 10 power minus 16 Newton find them

Answer 1

Magnetic force is given as F=qvBsinθ it is given that angle between field B and velocity v is 90

Magnetic force is given as F=qvBsinθ it is given that angle between field B and velocity v is 90 0

Magnetic force is given as F=qvBsinθ it is given that angle between field B and velocity v is 90 0 and q=1.6×10

Magnetic force is given as F=qvBsinθ it is given that angle between field B and velocity v is 90 0 and q=1.6×10 −19

Magnetic force is given as F=qvBsinθ it is given that angle between field B and velocity v is 90 0 and q=1.6×10 −19 C so we get the value of force as F=1.6×10

Magnetic force is given as F=qvBsinθ it is given that angle between field B and velocity v is 90 0 and q=1.6×10 −19 C so we get the value of force as F=1.6×10 −16

Magnetic force is given as F=qvBsinθ it is given that angle between field B and velocity v is 90 0 and q=1.6×10 −19 C so we get the value of force as F=1.6×10 −16 N so m=1 is the required answer.

Explanation:

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