To solve this problem we must first find the potential on the body which is given as a product between the number of turns, the area and the variation of the magnetic field as a function of time. Once the potential is found, we will apply Ohm's Law with which we can find the induced current on the body. Our values are,
![A = 7.00*10^{-4} m^2](https://tex.z-dn.net/?f=A%20%3D%207.00%2A10%5E%7B-4%7D%20m%5E2)
![\Delta B = 3.30T-0.5T](https://tex.z-dn.net/?f=%5CDelta%20B%20%3D%203.30T-0.5T)
![t = 0.99s](https://tex.z-dn.net/?f=t%20%3D%200.99s)
![R = 1.8\Omega](https://tex.z-dn.net/?f=R%20%3D%201.8%5COmega)
a) The magnitude of average induced emf is given by
![\epsilon_{emf} = NA (\frac{dB}{dt})](https://tex.z-dn.net/?f=%5Cepsilon_%7Bemf%7D%20%3D%20NA%20%28%5Cfrac%7BdB%7D%7Bdt%7D%29)
Here N =1
![\epsilon_{emf} = (1)(7.00 x 10^-4 m2)[\frac{(3.3 T - 0.5 T)}{(0.99s)}]](https://tex.z-dn.net/?f=%5Cepsilon_%7Bemf%7D%20%3D%20%281%29%287.00%20x%2010%5E-4%20m2%29%5B%5Cfrac%7B%283.3%20T%20-%200.5%20T%29%7D%7B%280.99s%29%7D%5D)
![\epsilon_{emf}= 0.00196 V](https://tex.z-dn.net/?f=%5Cepsilon_%7Bemf%7D%3D%200.00196%20V)
b) The magnitude of the induced current is
![I = \frac{V}{R}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BV%7D%7BR%7D)
Here Resistance is
![R = 1.80\Omega](https://tex.z-dn.net/?f=R%20%3D%201.80%5COmega)
![I = \frac{(0.00196V)}{(1.80\Omega)}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B%280.00196V%29%7D%7B%281.80%5COmega%29%7D)
![I = 0.00108 A](https://tex.z-dn.net/?f=I%20%3D%200.00108%20A)
Therefore the induced current is 0.00108A
Answer:
The work done on the canister is 15.34 J.
Explanation:
Given;
mass of canister, m = 1.9 kg
magnitude of force acting on x-y plane, F = 3.9 N
initial velocity of canister in positive x direction,
= 3.9 m/s
final velocity of the canister in positive y direction, ![v_j = 5.6 \ m/s](https://tex.z-dn.net/?f=v_j%20%3D%205.6%20%5C%20m%2Fs)
The change in the kinetic energy of the canister is equal to net work done on the canister by 3.9 N.
ΔK.E = ![W_{net}](https://tex.z-dn.net/?f=W_%7Bnet%7D)
ΔK.E ![= K.E_f -K.E_i](https://tex.z-dn.net/?f=%3D%20K.E_f%20-K.E_i)
The initial kinetic energy of the canister;
![K.E_i = \frac{1}{2} mv_i^2\\\\K.E_i = \frac{1}{2} m(\sqrt{v_i^2 +v_j^2 + v_z^2}\ )^2\\\\K.E_i = \frac{1}{2} *1.9(\sqrt{3.9^2 +0^2 + 0^2}\ )^2 = 14.45 \ J](https://tex.z-dn.net/?f=K.E_i%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20mv_i%5E2%5C%5C%5C%5CK.E_i%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%28%5Csqrt%7Bv_i%5E2%20%2Bv_j%5E2%20%2B%20v_z%5E2%7D%5C%20%20%29%5E2%5C%5C%5C%5CK.E_i%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A1.9%28%5Csqrt%7B3.9%5E2%20%2B0%5E2%20%2B%200%5E2%7D%5C%20%20%29%5E2%20%3D%2014.45%20%5C%20J)
The final kinetic energy of the canister;
![K.E_f =\frac{1}{2} mv_j^2 \\\\K.E_f = \frac{1}{2} m(\sqrt{v_i^2 +v_j^2 + v_z^2}\ )^2\\\\K.E_f = \frac{1}{2} *1.9(\sqrt{0^2 +5.6^2 + 0^2}\ )^2 = 29.79 \ J](https://tex.z-dn.net/?f=K.E_f%20%3D%5Cfrac%7B1%7D%7B2%7D%20mv_j%5E2%20%5C%5C%5C%5CK.E_f%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20m%28%5Csqrt%7Bv_i%5E2%20%2Bv_j%5E2%20%2B%20v_z%5E2%7D%5C%20%20%29%5E2%5C%5C%5C%5CK.E_f%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A1.9%28%5Csqrt%7B0%5E2%20%2B5.6%5E2%20%2B%200%5E2%7D%5C%20%20%29%5E2%20%3D%2029.79%20%5C%20J)
ΔK.E = 29.79 J - 14.45 J
ΔK.E =
= 15.34 J
Therefore, the work done on the canister is 15.34 J.
Answer:
Explanation:
We are given the following formula:
(1)
Where:
is the amount of heat
is the mass of water
is the specific heat of water
is the variation in temperature, which in this case is
Rewriting equation (1) with the known values at the right side, we will prove the result is
:
(2)
This is the result