Item 5 Part A The figure shows a 28-cm-diameter loop in three different magnetic

Question

Item 5 Part A The figure shows a 28-cm-diameter loop in three different magnetic fields. The loop's resistance is 0.30 In (Figure 1, what are the size and direction of the induced current? Express your answer to two significant figures and include the appropriate units. x.10 Value Units Submit Incorrect; Try Again; 8 attempts remaining Part B There is no induced current The loop current direction is cow The lcop current direction is ew. Submit My Answer Give Up Correct Fire 1 13 Part C In (Figure 2. what are the size and direction of the induced current? Express your answer to two significant figures and include the appropriate units. (a) B increasing at 0.50 T/s 1.1 Value Units Submit My Answere Glve Up Incorrect; Try Again; 8 attempts remaining

Explanation / Answer

Given,

Diameter of loop, D = 28 cm = 0.28 m

Radius, R = 0.14 m

Resistance, R = 0.30 ohm

a)Rate of change of magnetic field, B/t = 0.5 T/s

magnetic flux, = A(B cos) = r2(B)

To calculate the magnitude of current, I , we need to calculate average Emf.

So, average EMF induced in the loop during time t is
= -N(/t), where N =1
= -(/t) = -r2(B/t)
the average EMF induced in the loop is
= -r2(B/t)
= - 0.0308 V
so, the magnitude of the induced EMF is
mod () = 0.0308 V

Current, I = mod() / R
Current, I = 0.103 A

direction of the current: clock wise

c) A is perpendicular to B, A.B = AB cos = AB cos90° = 0

magnetic flux, = A(B cos) = r2(B cos90°) = 0

Again, we need to find average Emf to calculate current.
the average EMF induced in the loop during time t is, = -N(/t) = 0
current, I = mod()/R
So, current, I = 0 A

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