Question

2 identical magnets are arranged with like poles facing each other on a frictionless 30 degree slope. The magnet on the lower position is held in place and can't move. the magnet in the higher position on the track is held in place by the balance of the upslope force due to the magnetic repolsion between the magnets and the downslope force due to the weight of the magnet. The mass of the magnets is 50 grams each. The distance between the magnets is 5 cm.

(1) what is the magnetic force between the magnets? answer in newtons.

(2) If the strength of the magnetic force between the magnets is of the form: F= C*(1/r^3), what is the separation between the magnets when the angle of the slope is only 20 degrees? answer in cm.

(3) what would the angle of the slope be to get a separation of 7.11 cm? answer in degrees.

(4) what would the separation be at a slope of 5 degrees? answer in cm.

(5) what would the force between the magnets be if the slope was 5 degrees? answer in Newtons.

Answer 1

1) given
theta = 30 degrees
m = 50 g = 0.05 kg
r = 5 cm

In the equilibrium, net force acting on the magnet along the inclined surface must be zero.

F_magnetic - m*g*sin(theta) = 0

F_magnetic = m*g*sin(theta)

= 0.05*9.8*sin(30)

= 0.245 N <<<<<<<<--------------Answer

2) Let theta1 = 30 degrees
r1 = 5 cm
theta2 = 20 degrees
r2 = ?

use, sin(theta1)/sin(theta2) = (r2/r1)^3

(r2/r1) = (sin(theta1)/sin(theta2))^(1/3)

r2 = r1*(sin(theta1)/sin(theta2))^(1/3)

= 5*(sin(30)/sin(20))^(1/3)

= 5.67 cm <<<<<<<-----------------------Answer

3) Let theta1 = 30 degrees
r1 = 5 cm
r2 = 7.11 cm
theta2 = ?

use, sin(theta1)/sin(theta2) = (r2/r1)^3

sin(theta2) = sin(theta1)*(r1/r2)^3

= sin(30)*(5/7.11)^3

= 0.1739

theta2 = sin^-1(0.1739)

= 10.0 degrees <<<<<<<<<-------------------Answer

4) Let theta1 = 30 degrees
r1 = 5 cm
theta2 = 5 degrees
r2 = ?

use, sin(theta1)/sin(theta2) = (r2/r1)^3

(r2/r1) = (sin(theta1)/sin(theta2))^(1/3)

r2 = r1*(sin(theta1)/sin(theta2))^(1/3)

= 5*(sin(30)/sin(5))^(1/3)

= 8.95 cm <<<<<<<<<<-------------------Answer

5) In the equilibrium, net force acting on the magnet along the inclined surface must be zero.

F_magnetic - m*g*sin(theta) = 0

F_magnetic = m*g*sin(theta)

= 0.05*9.8*sin(5)

= 0.0427 N <<<<<<<<--------------Answer