(SOLVED)A circular loop of flexible iron wire has an initial circumference of 165.0 cm, but its circumference is decreasing at a constant rate of 12.0 cm>s due to a tangential pull on the wire

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Question

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A circular loop of flexible iron wire has an initial circumference of 165.0 cm, but its circumference is decreasing at a constant rate of 12.0 cm>s due to a tangential pull on the wire. The loop is in a constant, uniform magnetic field ori- ented perpendicular to the plane of the loop and with magnitude 0.500 T. (a) Find the emf induced in the loop at the instant when 9.0 s have passed. (b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.

Expert Answer

Concepts and reason

The concepts required to solve this problem is circumference of the circle, magnetic flux, right hand thumb rule and induced emf.

Initially, derive the expression for the radius by using circumference of the circle. Later, calculate the magnetic flux and then calculate the induced emf in the circular loop. Finally, find the direction of the induced current in the loop.

Fundamentals

The expression for magnetic flux is,

$ϕ = B A$

Here, $ϕ$ is the magnetic flux, B is the magnetic field, and A is the area of the circular loop.

The expression for induced emf in the loop is,

$ε = - \frac{d ϕ}{d t}$

Here, $ε$ is the induced emf.

Right hand thumb rule states that the thumb indicates the direction of the current and other fingers indicates the direction of the field.

Step-by-step

1st step

All steps

Answer only

(a)

The expression for the circumference is,

$c = 2 π r$

The circumference is decreasing at a constant rate is,

$c = 1.65 m - 0.12 t m/s$

Rewrite the above expression for r.

$r = \frac{c}{2 π}$

Substitute $1.65 m - 0.12 t m/s$ for c in the above equation.

$r = \frac{1.65 m - 0.12 t m/s}{2 π} = 0.26 - 0.019 t$

The expression for magnetic flux is,

$ϕ = B A$

Substitute $(3.14) (0.26 - 0.019 t)^{2}$ for A and 0.500 T for B in the above equation.

$ϕ = (0.500 T) (3.14) (0.26 - 0.019 t)^{2}$

The expression for induced emf in the loop is,

$ε = - \frac{d ϕ}{d t}$

Substitute $(0.500 T) (3.14) (0.26 - 0.019 t)^{2}$ for $ϕ$ in the above equation.

$ε = - \frac{d}{d t} (0.500 T) (3.14) (0.26 - 0.019 t)^{2} = (0.500 T) (3.14) 2 (0.19) (0.26 - 0.019 t)$

Substitute 9.0 s for t in the above equation.

$ε = (0.500 T) (3.14) 2 (0.019) (0.26 m - 0.019 m/s (9.0 s)) = 0.0053097 V$

Therefore, the induced emf in the loop is 0.0053097 V.

Part a

Therefore, the induced emf in the loop is 0.0053097 V.

Explanation
Common Mistakes
Hint for Next Steps

The expression for the area of the circular disk is,

$A = π r^{2}$

Substitute $0.26 - 0.019 t$ for r in the above equation.

$A = (3.14) (0.26 - 0.019 t)^{2}$

The derivative of the function $f (t) = t^{2}$ is as follows:

$\frac{df ( t )}{d t} = 2 t$

(b)

The induced current produce a magnetic field in the opposite direction called the induced field.

The direction of the induced current in the loop is clockwise.

Part b

The direction of the induced current in the loop is clockwise.

Explanation
Common Mistakes

If you put the fingers of your right hand inside the coil in the direction of that induced field, the thumbs points in the direction of the current must flow to produce field.