Question

An electron is accelerated through 2.40×10³V from rest and then enters a uniform 1.70-T magnetic field. What are(b) the minimum values of the magnetic force this particle experiences?

Answer 1

The maximum value of Force is 7.88x$10^{-12}$

The minimum value of magnetic force is 0

Potential V=2.40x$10^{3}$.

Magnetic field = B = 1.70 T.

we need to find

a) The maximum value of Force

Since the particle is moving from rest, the potential energy change is 2.40x$10^{3}$?

Potential Energy= vq= 2,40x$10^{3}$x (-1.6 x 10x$10^{-19}$%).

= -8.84x$10^{-16}$

Now this will be equal to 1/2 mv²

=  1/2 mv²= 8.84x $10^{-16}$.

v=2.903x$10^{7}$m/s :

Wow Fmax = qvb

=1.6x$10^{-19}$x2.903x$10^{7}$x1.70 .

=7.88x$10^{-12}$

b) The minimum value of magnetic force

The minimum Value of magnetic force will be

when sinθ=0

The minimum value is 0

To know more about magnetic field refer to brainly.com/question/13091447

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