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.
Answer:
<h2>0.056 W</h2>
Explanation:
From ohms law we know that
Given data
P1 = 0.5 Watt
P2 = ?
V1= 3 Volts
V2= 1 Volt
Thus we can solve for the power dissipated as follows
<em>The resistor will dissipate 0.056 Watt</em>
Answer:
The equation used to calculate the work done is: work done = force × distance. W = F × d. This is when: work done (W) is measured in joules (J)
Explanation:
Given:
v₀ = 250 mph
v = 0 mph
t = 25 s
Find: a
v = at + v₀
(0 mph) = a (25 s) + (250 mph)
a = -10 mph/s
