consider the loop shown in the figure which is partially embedded in a magnetic
ID: 776795 • Letter: C
Question
consider the loop shown in the figure which is partially embedded in a magnetic field. the mass m is connected to the loop "pulls" the loop out of the wire by virtue its weight.
calculate the terminal fall speed v(magnitude) assuming mass m = 50gram which L =50cm. field strength B = 2.0 T and resistance R = 100.
2.
a real wall transformer in my office has the following rating.
Vp =120VAC Vs = 6VAC Ip = 0.3A Is = 1.8A
calculate the following
a. calculate the turns ratio
b. calculate thr promary power Pp.
c. calculate the secondary power Ps.
d. Calculate the efficiency e.
Explanation / Answer
1. The magnetic flux throught he loop is decreasing, so , by lenz law, current is induced in such a direction so as to oppose the decrease in flux. So induced current increases the flux and so the direction of current is clockwise. So, Now, by right hand thumb rule, forces on the vertical wires are opposite in direction and on the horizontal wire it is upwards.
Now, induced emf = B*A/t = B*L*v = 2*0.5*v = v
so, induced current = I= emf/ resistance = v/ 100
S, Force F on vertical wire moving in magnetic field is = I*L*B= v*0.5*2/ 100 = v/100
Now, this force is same as the weight of mass = 0.050*9.81= 0.4905
So, 0.4905= v/100
So, v= 49.05 m/s
2. the turns ratio = ratio of the number of primary turns to the number of secondary turns = 120/6= 20
primary power Pp = Ip*Vp = 36W
secondary power Ps= Is*Vs= 10.8 W
efficiency e = power output/ power input= Ps/Pp = 10.8/36= 0.3 = 30%
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