Question

A square loop of wire with 10 turns and a side length of 1 m is placed in a changing magnetic field. If the magnetic field changes from 2 T to 4 T with 8 seconds, what is the average induced EMF? A) 2.5 V B) 1.25 V C) Zero D) 5 V

Answer 1

Answer:

The average induced emf is 2.5 V.

(A) is correct option.

Explanation:

Given that,

Number of turns = 10

Length of side = 1 m

change in magnetic field = 4-2=2 T

Time t = 8 sec

We need to calculate the average induced emf

Using formula of induced emf

$\epsilon=NA\times\dfrac{\Delta B}{\Delta t}$

Where,

N = number of turns

A = area of cross section

B = magnetic field

t = time

Put the value into the formula

$\epsilon=10\times1\times\dfrac{2}{8}$

$\epsilon = 2.5\ V$

Hence, The average induced emf is 2.5 V.

Answer 2

This question involves the concepts of induced emf and magnetic field.

The average induced emf will be "2.5 volts".

Induced EMF

The induced emf in the coil can be given by the following formula:

$EMF= \frac{NA\Delta B}{\Delta t}$

where,

• EMF = Induced EMF = ?
• N = No. of turns = 10 \
• A = Area of coil = (1 m)(1 m) = 1 m²
• ΔB = change in magnetic field = 4 T - 2 T = 2 T
• Δt = time interval = 8 s

Therefore,

$EMF=\frac{(10)(1\ m^2)(2\ T)}{8\ s}$

EMF = 2.5 volts

Learn more about induced emf here:

https://brainly.com/question/4721624