Answer:
The magnetic field strength and the electrons' energy are 0.077 T and 0.8906 eV.
Explanation:
Given that,
Diameter = 2.62 mm
Frequency = 2.15 GHz
(A). We need to calculate the magnetic field strength
Using formula of the magnetic field strength
Where, f = frequency
e = charge of electron
Put the value into the formula
(B). We need to calculate the energy of electron
Using formula of energy
The energy in eV
Hence, The magnetic field strength and the electrons' energy are 0.077 T and 0.8906 eV.
Let F = required force, N
Given:
d = 12 m, distance
W = 280 J, work done
By definition,
W = F*d,
therefore
(F N)*(12 m) = (280 J)
F = 280/12 = 23.33 N
Answer: The force is 23.3 N (nearest tenth)
<span>5.98 x 10^-2 ohms.
Resistance is defined as:
R = rl/A
where
R = resistance in ohms
r = resistivity (given as 1.59x10^-8)
l = length of wire.
A = Cross sectional area of wire.
So plugging into the formula, the known values, including the area of a circle being pi*r^2, gives:
R = 1.59x10^-8 * 3.00 / (pi * (5.04 x 10^-4)^2)
R = (4.77 x 10^-8) / (pi * 2.54016 x 10 ^-7)
R = (4.77 x 10^-8) / (7.98015 x 10^-7)
R = 5.98 x 10^-2 ohms
So that wire has a resistance of 5.98 x 10^-2 ohms.</span>
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.
Work= force*distance
Work= x*12
Force= mass*acceleration
Force= 5 kg*6
Force= 40 N
Work= 40×12
Work= 480 J (joules)
I think this is it
