Question

12. If a Hall voltage of 2.50 x 10^-3V is produced by a 0.300-T field applied across a 1.50-cm-diameter aorta, what is the blood velocity inside?

14. A solenoid has 300 turns and a length of 10.0 cm. When a current of 2.50 A flows through the solenoid, how strong is the magnetic field produced inside it?

15. In a mirror's ray diagram, the distance between the object and the image is 30 cm. The magnification of the image is -0.3. What type of mirror is used and what is the focal length of the mirror?

Answer 1

Answer

Q 12

GIVEN

Hall voltage E = 2.50 x 10^-3V

Magnetic field B = 0.300 T

Aorta Diameter L= 1.50-cm = 1.5 x 10^-2 m

Formula

Hall voltage = BLv where v is the velocity

Velocity v = E/ BL = = 0.56 m/s

Q.14

GIVEN

No of Turns = 300

Length = 10 cm = 0.01 m

Current I = 2.5 A

Formula :-    B = µ_onI   where ‘n’ is the number of turns per unit length

B = 4π x 10^-7 x 300 x 2.5/.01 = 9.42 x 10^-2 T

Magnetic Field inside the solenoid = 9.42 x 10^-2 T

Q.15

GIVEN

The mirror used is a concave mirror , since the magnification given is negative and so the image is real and inverted

distance between the object and the image = 30 cm

u – v = 30

v = u-30………………..1

Magnification = -0.3

Magnification = - (v/u)

- (v/u) = -0.3

v = 0.3 u………….2

From eqn 1 and 2

u- 30 =0.3u

u – 0.3 u = 30

0.7 u = 30

u = 30/0.7 = 42.85 cm

v = 42.85 – 30 = 12.85 cm

By sign convention

U = -42.85

V = - 12.85

Now using mirror formula

1/u + 1/v = 1/f

f = uv/ u+v = ( -42.85 x -12.85)/( -42.85 + -12.85)

f = 550.62/ -55.7 = - 9.89 cm

The focal length = -9.89 cm