Question

A 9.6 cm diameter circular loop of wire is in a 1.10 T magnetic field perpendicular to the plane of the loop. The loop is removed from the field in 0.15 s What is the average induced emf? Draw the loop and field and indicate the direction of the induced current

Answer 1

Answer:
`emf=0.531 V`

Step-by-step explanation:

Induced EMF is equals to the change in the Magnetic flux per unit change in time.

It is calculated by the Faraday's law which is given mathematically as:

`emf=N * (\Delta (B.A) )/(\Delta t)` ..................................(1)

where:

• N= number of turns in looped coil of wire

• B= magnetic field

• A= area of the loop formed by the wire

• t= change in time

• 
  `\Delta`= change in the quantity from initial to final state.

Given:

N=1 (since it is a loop not a coil of loops)

Diameter of loop, d = 9.6 cm

`B = 1.10 T`

`{\Delta t}`= 0.15 s

Firstly, we find the area of loop:

`A= \pi (d^(2))/(4)`

`A= \pi ((9.6 * 10^(-2) )^(2))/(4)`

`A= 7.2382 * 10^(-3) m^(2)`

Putting respective values in eq. (1)

`emf=1* (\Delta (1.10* 7.2382 * 10^(-3)) )/(0.15)`

`emf=0.531 V`

THe direction of the induced emf is in accordance with the Lenz Law which states that the direction of an induced current is always such as to oppose the cause by which it is produced.

Answer 2

Thus induced emf is 0.0531 V

Solution:

As per the question:

Diameter of the loop,
`d = 9.6\ cm = 0.096\ m`

Thus the radius of the loop, R = 0.048 m

Time in which the loop is removed, t = 0.15 s

Magnetic field, B = 1.10 T

Now,

The average induced emf, e is given by Lenz Law:

`e = - (\Delta \phi_(B))/(\Delta t)`

`e = - (\Delta \phi_(B))/(\Delta t)`

where

`\phi_(B)` = magnetic flux =
`A\Delta B`

where

A = cross sectional area

Also, we know that:

`e = - (A\Delta B)/(\Delta t)`

`e = - (\pi r^(2)* (0 - 1.10))/(0.15)`

`e = - (\pi * 0.048^(2)* (0 - 1.10))/(0.15)`

e = 0.0531 V

The sketch is shown in the figure, where I indicates the direction of the induced current.