Option (B) is correct.The magnetic field strength=8.5 x 10² T
Explanation:
the magnetic force Fm is given by
Fm= q V B sinθ
q= charge=1.4 x 10⁻⁷ C
v= velocity= 1.3 x 10⁶ m/s
B= magnetic field strength
Fm= magnetic force= 1.5 x 10² N
θ=75°
so 1.5 x 10²=(1.4 x 10⁻⁷) (1.3 x 10⁶ ) (B) sin75
B=8.5 x 10² T
Answer:
Explanation:
Given that,
Current in loops are
i1 = 12A
i2 = 20A
The loops are 3.4cm apart
The magnetic field at the center is found to be zero, so when want to find the radius of bigger loop
Magnetic Field is given as
B= μoi/2πr
Where,
μo is a constant = 4π×10^-7 Tm/A
r is the distance between the two wires
i is the current in the wires
B is the magnetic field
NOTE
Field due to large loop should be equal to the smaller loop.
B1 = B2
μo•i1 / 2π•r1 = μo•i2 / 2π•r2
Then, μo, 2π cancels out, so we have
i1 / r1 = i2 / r2
Make r2 subject of formula
i1•r2 = i2•r1
r2 = i2•r1 / i2
r2 = 20×3.4/12
r2 = 5.67cm
The radius of the bigger loop is 5.67cm.
D. The osculations show a variable rate of motion. Hope this helps:)
Answer:
The thermal conductivity of the wall = 40W/m.C
h = 10 W/m^2.C
Explanation:
The heat conduction equation is given by:
d^2T/ dx^2 + egen/ K = 0
The thermal conductivity of the wall can be calculated using:
K = egen/ 2a = 800/2×10
K = 800/20 = 40W/m.C
Applying energy balance at the wall surface
"qL = "qconv
-K = (dT/dx)L = h (TL - Tinfinity)
The convention heat transfer coefficient will be:
h = -k × (-2aL)/ (TL - Tinfinty)
h = ( 2× 40 × 10 × 0.05) / (30-26)
h = 40/4 = 10W/m^2.C
From the given temperature distribution
t(x) = 10 (L^2-X^2) + 30 = 30°
T(L) = ( L^2- L^2) + 30 = 30°
dT/ dx = -2aL
d^2T/ dx^2 = - 2a
Answer:
the body has energy due to its constant motion. it means it moves in a uniform acceleration which has zero velocity
Explanation:
Uniform or constant acceleration is a type of motion in which the velocity of an object changes by an equal amount in every equal time period.
