Question

1.) How much heat is trasnferred in a process where a 2.35-mole sample of an ideal gas maintained at a temperature of 402K is compressed from 1.38 m³ to 0.206 m³, in kilojoules?

2.) What is the magnitude of the magnetic force (in N) on a charged particle (Q = 5.0 ?C) moving with a speed of 80x10³ m/s in the positive x-direction at a point where B_x = 5.0 T, B_y = -4.0 T, and B_z = 3.0 T?

3.) The electric force on a positive point charge of charge +q is 4.0 ?N (-i) (in the negative x-direction) due to anotehr positive point charge of charge +4q, located a distance of 1.0 cm away. What is the electric force on the +4q charge due to the +q charge?

4.) A very large surface of charge of uniform surface charge density 67 nC/m² is coincident with the xy-plane. What is the electric flux through the surface of a cube (side length d = 4.0 cm) that is centered on the origin and therefore encloses a section of this charge surface?

5.) When a positive electric charge is accelerated by an electric field, does it move to a position of higher or lower electric potential? Does it move to aposition of higher or lower electric potential energy?

6.) A flat, circular conducting coil with 100 turns (N = 100) of radius r = 8.00x10^-2 m is rotated in a uniform magnetic field of magnitude B = 10.0 T, sot hat the magnetic flux through one turn/loop of the coil as a function of time is given by ?_B = ?r²Bcos?t, where ? = 20? radians/second. If the total resistance of the coil is R = 10.0 ?, what is the current unduced in the coil at time t = 6.0x10^-3 s?

Please show all work, thank you!

Answer 1

1.

n = number of moles = 2.35

T = temperature = 402 K

V₁ = 1.38 m³

V₂ = 0.206 m³

work done is given as

W = n RT log(V₂/V₁)

W = (2.35) (8.314) (402) log(0.206/1.38)

W = - 1.5 x 10⁴ J

Q = heat transferred = W = - 1.5 x 10⁴ J

2)

Q = 5 x 10^-6 C

= velocity = 80000 i

= magnetic field = 5i - 4j + 3k

magnetic force is given as

= Q (x )

= (5 x 10^-6) ((80000 i) x (5i - 4j + 3k))

= (5 x 10^-6) (- 32 x 10⁴ k - 24 x 10⁴ j)

= - 1.60 k - 1.20 j

= 0 i - 12.0 j - 1.60 k