Answer:
the total force vector, on test charge is points from origin to point C( 1, 1 )
Explanation:
Given the data in the question, as illustrated in the image below;
from the Image, OA = 1, OB = AC = 1
so using Pythagoras theorem
a² = b² + c²
a = √( b² + c² )
so
OC = √( OB² + AC² )
we substitute
OC = √( OA² + AC² )
OC = √( 1² + 1² )
OC = √( 1 + 1 )
OC = √2
Coordinate of C( 1, 1 )
Hence, the total force vector, on test charge is points from origin to point C( 1, 1 )
Answer:
a) 4.9*10^-6
b) 5.71*10^-15
Explanation:
Given
current, I = 3.8*10^-10A
Diameter, D = 2.5mm
n = 8.49*10^28
The equation for current density and speed drift is
J = I/A = (ne) Vd
A = πD²/4
A = π*0.0025²/4
A = π*6.25*10^-6/4
A = 4.9*10^-6
Now,
J = I/A
J = 3.8*10^-10/4.9*10^-6
J = 7.76*10^-5
Electron drift speed is
J = (ne) Vd
Vd = J/(ne)
Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)
Vd = 7.76*10^-5/1.3584*10^10
Vd = 5.71*10^-15
Therefore, the current density and speed drift are 4.9*10^-6
And 5.71*10^-15 respectively
Answer:
option D
Explanation:
given,
A conductor is carrying current = 2.0 A is 0.5 mm thick
Hall voltage = 4.5 x 10-6 V
uniform magnetic field = 1.2 T
density of the charge = n =?
hall voltage =
n = 6.67 × 10²⁷ charges/m
hence the correct answer is option D
