1.

A rectangular coil 0.065 m by 0.080 m is positioned
so that its cross-sectional area is perpendicular
to the direction of a magnetic field.
The coil has 66 turns and a total resistance of
7.6 Ω and the field decreases at a rate of 2.5
T/s.
What is the magnitude of the induced current
in the coil?
Answer in units of A.

2.
A student attempts to make a simple generator by passing a single loop of wire between the poles of a horseshoe magnet with a magnetic field of 2.0 ×10−2 T. The area of the loop is 4.87 ×10−3 m2 and is moved perpendicular to the magnetic field lines.
In what time interval will the student have
to move the loop out of the magnetic field in
order to induce an EMF of 1.9 V?
Answer in units of s.

3.
A student attempts to make a simple generator
by wrapping a long piece of wire across
a cylinder with a cross-sectional area of 1.076×10−3 m2. She then passes the coil betweenthe poles of a horseshoe magnet with a magnetic field of 2.2 ×10−2 T. The student finds that by removing the coil perpendicular to the magnetic field lines during 0.29 s, a voltage drop of 145 mV can be induced across the wire.
How many turns of wire are wrapped
around the coil?

4.
A bolt of lightning, such as the one shown on
the left in the figure, behaves like a vertical
wire conducting electric current. As a result,
it produces a magnetic field whose strength
varies with the distance from the lightning. A
139-turn circular coil with a radius of 0.842
m is oriented perpendicular to the magnetic
field as shown. The magnetic field at the coil
drops from 4.99×10−3 T to 0.00 T in 10.7 µs.
What is the average emf induced in the
coil?
Answer in units of V.

1. Ah, the induced current in the coil! Let's calculate this positively shocking answer. We know the cross-sectional area, the number of turns, the total resistance, and the rate at which the magnetic field decreases. But first, let's find the magnetic flux through the coil. The magnetic flux is the product of the magnetic field, the area, and the cosine of the angle between them. Since the coil is perpendicular to the field, the angle is 90 degrees, so the cosine is 0. Now multiply this delightful flux by the rate of change of the magnetic field, divide it by the resistance, and you'll have the magnitude of the induced current in the coil, expressed in amps! Ta-da!

2. Ah, the time interval needed to induce an EMF! Let me grab my stopwatch and calculate this electrifying answer for you. We know the magnetic field, the area of the loop, and the desired EMF. To find the time interval, we divide the EMF by the rate of change of the magnetic flux, which is the product of the magnetic field and the area of the loop. It's like racing against time to induce that EMF. And the answer, my friend, is in seconds!

3. Ah, the number of turns in the wire! Let me untangle this problem for you. We know the magnetic field, the time interval, and the induced voltage across the wire. To find the number of turns, we divide the induced voltage by the product of the magnetic field and the cross-sectional area of the coil. It's like unwrapping the secrets of this generator. And the answer, my friend, is the number of turns!

4. Ah, the average emf induced in the coil by the lightning bolt! Let's calculate this electrifying answer. We know the magnetic field at the coil and how quickly it drops to zero. To find the average emf, we multiply the rate of change of the magnetic field by the number of turns in the coil and the area of the coil. It's as if the lightning bolt is singing a magnetic symphony for our coil. And the answer to this electrifying question is in volts!

1. To find the magnitude of the induced current in the coil, we can use Faraday's law of electromagnetic induction.

Faraday's law states that the magnitude of the induced electromotive force (EMF) in a coil is given by the rate of change of magnetic flux through the coil.

The magnetic flux is given by the product of the magnetic field strength, the area of the coil, and the cosine of the angle between the magnetic field and the normal to the coil.

In this case, the area of the coil is 0.065 m x 0.080 m = 0.0052 m^2.

The angle between the magnetic field and the normal to the coil is 90 degrees, so the cosine of the angle is 1.

The rate of change of magnetic field is given as -2.5 T/s (negative because it is decreasing).

Therefore, the magnitude of the induced EMF is:

EMF = (magnetic field strength) * (area of coil) * (cosine of angle) * (rate of change of magnetic field)
= (2.5 T/s) * (0.0052 m^2) * (1) * (-2.5 T/s)
= -0.0325 V

Since the total resistance of the coil is given as 7.6 Ω, we can use Ohm's law to find the magnitude of the induced current in the coil:

Current = EMF / Resistance
= (-0.0325 V) / (7.6 Ω)
= -4.28 mA

The magnitude of the induced current in the coil is 4.28 mA.

2. To find the time interval in which the student needs to move the loop out of the magnetic field in order to induce an EMF of 1.9 V, we can use Faraday's law of electromagnetic induction.

Faraday's law states that the magnitude of the induced EMF in a coil is given by the rate of change of magnetic flux through the coil.

The magnetic flux is given by the product of the magnetic field strength, the area of the loop, and the cosine of the angle between the magnetic field and the normal to the loop.

In this case, the area of the loop is 4.87 × 10^-3 m^2.

The angle between the magnetic field and the normal to the loop is 90 degrees, so the cosine of the angle is 1.

The induced EMF is given as 1.9 V.

Therefore, we can rearrange the equation for magnetic flux to solve for the rate of change of magnetic field:

Rate of change of magnetic field = EMF / ((magnetic field strength) * (area of loop) * (cosine of angle))
= (1.9 V) / ((2.0 × 10^-2 T) * (4.87 × 10^-3 m^2) * (1))
= 196.5 T/s

To find the time interval, we can rearrange the equation for magnetic flux to solve for time:

Time = (change in magnetic flux) / (rate of change of magnetic field)
= (0 - (2.0 × 10^-2 T) * (4.87 × 10^-3 m^2)) / (196.5 T/s)
= -0.0000974 s

Note that the negative sign indicates that the loop needs to be moved out of the magnetic field in the opposite direction of the initial magnetic field.

The time interval in which the student needs to move the loop out of the magnetic field is 0.0000974 s (or approximately 97.4 µs).

3. To find the number of turns of wire wrapped around the coil, we can use Faraday's law of electromagnetic induction.

Faraday's law states that the magnitude of the induced electromotive force (EMF) in a coil is given by the rate of change of magnetic flux through the coil.

The magnetic flux is given by the product of the magnetic field strength, the area of the coil, and the cosine of the angle between the magnetic field and the normal to the coil.

In this case, the area of the coil is 1.076 × 10^-3 m^2.

The angle between the magnetic field and the normal to the coil is 90 degrees, so the cosine of the angle is 1.

The rate of change of magnetic field is given as -2.2 × 10^-2 T.

The voltage drop across the wire is given as 145 mV, which is equivalent to 0.145 V.

Therefore, we can rearrange the equation for magnetic flux to solve for the number of turns of wire:

Number of turns = Voltage drop / ((magnetic field strength) * (area of coil) * (cosine of angle) * (rate of change of magnetic field))
= (0.145 V) / ((2.2 × 10^-2 T) * (1.076 × 10^-3 m^2) * (1) * (-0.29 s))
= -18031 turns

The negative sign may indicate an error in the calculation or an incorrect interpretation of the problem. Please recheck the values and calculations provided.

4. To find the average emf induced in the coil, we can use Faraday's law of electromagnetic induction.

Faraday's law states that the magnitude of the induced electromotive force (EMF) in a coil is given by the rate of change of magnetic flux through the coil.

The magnetic flux is given by the product of the magnetic field strength, the area of the coil, and the cosine of the angle between the magnetic field and the normal to the coil.

In this case, the area of the coil is π * (0.842 m)^2 = 2.227 m^2 (approx).

The angle between the magnetic field and the normal to the coil is 90 degrees, so the cosine of the angle is 1.

The rate of change of magnetic field is given as (0.00 T - 4.99 × 10^-3 T) / (10.7 × 10^-6 s).

Therefore, we can calculate the average emf induced in the coil as:

Average emf = ((magnetic field strength) * (area of coil) * (cosine of angle) * (rate of change of magnetic field)) / (time)
= ((4.99 × 10^-3 T) * (2.227 m^2) * (1) * ((0.00 T - 4.99 × 10^-3 T) / (10.7 × 10^-6 s))) / (10.7 × 10^-6 s)
= 4.99 V

The average emf induced in the coil is 4.99 V.

To find the answers to these questions, we will use Faraday's Law of Electromagnetic Induction. According to Faraday's Law, the induced electromotive force (emf) is equal to the rate of change of magnetic flux through the coil. The magnetic flux is given by the product of the magnetic field strength (B) and the area (A) of the coil.

1. To find the magnitude of the induced current in the coil:
- Given: Area of the coil (A) = 0.065 m × 0.080 m = 0.0052 m^2, Magnetic field decrease rate (dB/dt) = 2.5 T/s
- The change in magnetic flux (deltaΦ) = B × deltaA, where deltaA is the change in area.
- The change in area is equal to the initial area because the field decrease is perpendicular to the coil, so deltaA = A.
- The induced emf (E) is equal to the rate of change of magnetic flux, E = -deltaΦ/dt = -(B × A)/dt.
- To find the magnitude of the induced current (I), divide the induced emf by the total resistance (R) of the coil, I = E/R.
- Plugging in the values: I = -(B × A)/R × dt = -(2.5 T/s × 0.0052 m^2)/(7.6 Ω) = -0.011 A (Taking the magnitude, the answer is 0.011 A).

2. To find the time interval required to induce an EMF of 1.9 V:
- Given: Magnetic field strength (B) = 2.0 × 10^(-2) T, Area of the loop (A) = 4.87 × 10^(-3) m^2, Induced EMF (E) = 1.9 V.
- From Faraday's law, E = -(B × A)/dt. Rearranging the equation gives -dt = (B × A)/E.
- Using the given values, we have dt = -(2.0 × 10^(-2) T × 4.87 × 10^(-3) m^2)/(1.9 V) = -0.05179 s (Taking the magnitude, the answer is 0.05179 s).

3. To find the number of turns of wire:
- Given: Magnetic field strength (B) = 2.2 × 10^(-2) T, Cross-sectional area of the coil (A) = 1.076 × 10^(-3) m^2, Voltage drop (V) = 145 mV = 0.145 V, Time interval (dt) = 0.29 s.
- From Faraday's law, E = -(B × A)/dt. Rearranging the equation gives E × dt = -B × A.
- Rearranging again, we get N = -(E × dt)/(B × A), where N is the number of turns.
- Substituting the values, N = -(0.145 V × 0.29 s)/(2.2 × 10^(-2) T × 1.076 × 10^(-3) m^2) = -73.144.
- Since the number of turns cannot be negative, the answer is 73 turns of wire.

4. To find the average emf induced in the coil:
- Given: Number of turns in the coil (N) = 139, Radius (r) = 0.842 m, Magnetic field decrease rate (dB/dt) = (4.99 × 10^(-3) T - 0 T)/(10.7 × 10^(-6) s) = -466.36 T/s.
- The change in magnetic flux (deltaΦ) is given by deltaΦ = B × deltaA = (B1 - B2) × A, where B1 and B2 are the initial and final magnetic field strengths, respectively.
- The average emf (Eavg) is equal to the rate of change of magnetic flux multiplied by the number of turns, Eavg = N × deltaΦ/dt.
- Plugging in the values, Eavg = 139 × [(4.99 × 10^(-3) T - 0 T) × πr^2]/(10.7 × 10^(-6) s) = 4.65 V.

Therefore, the answers are:
1. The magnitude of the induced current in the coil is 0.011 A.
2. The time interval required to induce an EMF of 1.9 V is 0.05179 s.
3. The number of turns of wire wrapped around the coil is 73.
4. The average EMF induced in the coil is 4.65 V.

a. EMF=N*area*dB/dt

then current= EMF/resistance

b. EMF= above, solve or dt.