Consider an electron being ejected from the surface of a heated filament at nearly zero specd [a] through what potential difference must it be accelerated to achieve a speed of 8.4x106m/s? any relativistic effects can be neglected [b] what is the minimum b-field required to cause the electron in part [a] above to move in a circular orbit of radius r 5.0 cm?

The potential difference and the magnetic field are 200.655 V and

Explanation:

Given that,

Speed

radius = 5.0 cm

We need to calculate the potential difference

We know that,

The kinetic energy acquired by electron

Where, m = mass of  electron

q = charge

v = velocity

Put the value into the formula

(b). We need to calculate the magnetic field

Using formula of magnetic field

Put the value into the formula

Hence, The potential difference and the magnetic field are 200.655 V and

a) 200 v

b )9.555 x 10⁻⁴ T

Explanation:

Equation for electron acceleration under potential difference

q V = 1/2 mv²

V is potential diff , v is velocity after acceleration, q is charge on electron.

V= .5 x 9.1x 10⁻³¹ x( 8.4 x 10⁶)² / 1.6 x 10⁻₁¹⁹

= 200.655 V

b ) radius of circular orbit in a magnetic field of electron

r =

B )  magnetic field required =

= 9.1 x 10⁻³¹ x 8.4 x 10⁶ / 1.6 x 10⁻¹⁹ x 5 x 10⁻²

9.55 x 10⁻⁴ T

multiply 2kg by latent heat of vap

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards