Calculate the magnitude of the magnetic force on an electron moving at speed 7.0

Question

Calculate the magnitude of the magnetic force on an electron moving at speed 7.09 x 105 m/s for the direction shown in part a of the figure below . The magnetic field has magnitude B = 0.355 T.

1. Magnitude?

What is its direction?

a) The direction is into the page.

b) The direction is out of the page.

2. Now calculate the magnitude of the magnetic force on the same electron for direction b. As shown, 1 = 20.6°?

What is its direction?

a)The direction is out of the page.
b) The direction is into the page.

3. And now calculate the magnitude of the magnetic force on the same electron for direction c. As shown, 2 = 35.0°.

What is its direction?
a) The direction is out of the page.
b) The direction is into the page

What is its direction?

Explanation / Answer

(a)

magnetic force Fb = q*v*B*sintheta

theta = angle between velocity vector and magnetic field = 90

Fb = q*v*B

Fb = 1.6*10^-19*7.09*10^5*0.355

Fb = 4.027*10^-14 N

Fb = q*(v x B)

direction = OUT OF PAGE

==============================

(b)

magnetic force Fb = q*v*B*sintheta

theta = angle between velocity vector and magnetic field = 110

Fb = q*v*B*sin110

Fb = 1.6*10^-19*7.09*10^5*0.355*sin110

Fb = 3.784*10^-14 N

Fb = q*(v x B)

direction = INTO THE PAGE

===============================

(c)

magnetic force Fb = q*v*B*sintheta

theta = angle between velocity vector and magnetic field = 120

Fb = q*v*B*sin120

Fb = 1.6*10^-19*7.09*10^5*0.355*sin120

Fb = 3.487*10^-14 N

Fb = q*(v x B)

direction = OUT OF PAGE

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