Q. 24.4( 12 Votes )

# An electron is projected horizontally with a kinetic energy of 10 keV. A magnetic filed of strength 1.0 × 10^{–7} T exists in the vertically upward direction.

(a) Will the electron deflect toward right or towards left of its motion?

(b) Calculate the sideways deflection of the electron in travelling through 1m. Make appropriate approximations.

Answer :

Given-

The kinetic energy of the electron when projected towards horizontal direction,

*K.E* = 10 keV = 1.6 × 10^{−15} J

Magnetic field, *B* = 1 × 10^{−7} T

Charge on an electron =1.60× 10^{-19}

mass of an electron = 9.1× 10^{-31} kilograms

(a) The direction electron deflection can be found by the right-hand screw rule.

Given the direction of magnetic field is vertically upward.

So, the electron will be deflected towards left.

(b) Kinetic energy-

_{}

Where

m is the mass of the electron

v is the velocity of the electron

Solving for v, we get

We know, Lorentz force F is given by -

where,

q = charge

v = velocity of the charge

B=magnetic field

and

θ= angle between V and B

And Newton's second law of motion

where m = mass

a= acceleration

we know

⇒

From (1) and (2) -

Applying 2^{nd} equation of motion

where,

a= acceleration

u = initial velocity

t= Time taken to cross the magnetic field

Since, the initial velocity is zero,

from (1),(2) and (3)

substituting the values -

From (1)

substituting the values-

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