Question

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 oriented 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.

Answer 1

(a). The emf induced in the loop is 0.005467 V.

(b). The direction of the current is clock wise direction.

Step-by-step explanation:

Given that,

Circumference = 165.0 cm

Constant rate = 12.0 cm/s

Magnetic field = 0.500 T

Time = 9.0 s

We need to calculate the emf induced

Using formula of emf

`\epsilon=-(d\phi)/(dt)`

`\epsilon=-(d(BA\cos\theta))/(dt)`

`\epsilon=B(dA)/(dt)`...(I)

Area of the coil is

`(dA)/(dt)=2\pi r(dr)/(dt)`

Put the value of area in equation (I)

`\epsilon=B*2\pi r(dr)/(dt)`...(II)

The circumference of the coil is

`C=2\pi r`

`r = (C)/(2\pi)`

Put the value into the formula

`r=(165.0)/(2\pi)`

`r=26.26\ cm`

The rate of change of radius is

`(dr)/(dt)=(12)/(2\pi)`

`(dr)/(dt)=1.90\ cm/s`

The radius of the coil in 9 sec

`r=26.26-1.90*9`

`r=9.16\ cm`

Put the value in the equation (I)

`\epsilon=-0.500*2\pi*9.16*10^(-2)*1.90*10^(-2)`

`\epsilon=-0.005467\ V`

The magnitude of the emf

`|\epsilon|=0.005467\ V`

(b). Right hand rule :

According to right hand rule,

The thumb shows the direction of force, the index finger shows the direction of magnetic field and the middle finger shows the direction of current.

So, The direction of the current is clock wise direction.

Hence, (a). The emf induced in the loop is 0.005467 V.

(b). The direction of the current is clock wise direction.