A circular loop of wire of area 25 cm2 lies in the plane of the paper. A decreasing magnetic field B is coming out of the paper. What is the direction of the induced current in the loop?

Answer:

t = T / 2 all energy is stored in the inductor

Explanation:

The circuit described is an oscillating circuit where the charge of the condensation stops the inductor and vice versa, in this system the angular velocity of the oscillation is

          w = √1/LC

          2π / T =√1 / LC

          T = 2π  √LC

The energy is constant and for the initial instant it is completely stored in the capacitor

         Uc = Q₀² / 2C

In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor

        U = L I² / 2

in the intermediate instant the energy is stored in the two elements.

Since the period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor

After t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation  (t = T/2).

The given problem is based on the charging and discharging concepts of capacitor. An oscillating circuit is a circuit where the charge of the capacitor stops the inductor and vice versa, in this system the angular frequency of the oscillation is given as,

[tex]\omega =\dfrac{1}{\sqrt{LC}}\\\\\\\dfrac{2 \pi}{T} =\dfrac{1}{\sqrt{LC}}\\\\\\T = 2\pi \times \sqrt{LC}[/tex]

here, T is the period of oscillation.

Also, the energy stored in the capacitor is constant and for the initial instant it is completely stored in the capacitor. So, the energy stored is given as,

[tex]U =\dfrac{Q^{2}}{2C}[/tex]

here, C is the capacitance.

In the process, the capacitor is discharging and the energy is stored in the inductor until when the charge in the capacitors zero, all the energy is stored in the inductor. So, the expression for the energy stored in the inductor is,

[tex]U'=\dfrac{L I^{2}}{2}[/tex]

here, L is the inductance and I is the current.

Note :- The period of the system is T for time t = 0 all energy is stored in the capacitor and for t = T / 2 all energy is stored in the inductor.

Thus, we conclude that after t = 0 the maximum energy stored in the magnetic field of the inductor is equal to [tex]U'=\dfrac{L I^{2}}{2}[/tex] for the time period, half of period of oscillation  (t = T/2).

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Answer:

e. It is neither attracted nor repelled.

Explanation:

Electrostatic attraction or repulsion occurs between two or more charged particles or conductors. In this case, if the negatively charged ball is brought close to the neutral end A of the rod, there would be no attraction or repulsion between the rod end A and the negatively charged ball. This is because a charged particle or conductor has no attraction or repulsion to a neutral particle or conductor.

Answer:

The electric field  can either oscillates in the z-direction, or the y-direction, but must oscillate in a direction perpendicular to the direction of propagation, and the direction of oscillation of the magnetic field.

Explanation:

Electromagnetic waves are waves that have an oscillating magnetic and electric field, that oscillates perpendicularly to one another. Electromagnetic waves are propagated in a direction perpendicular to both the electric and the magnetic field. If the wave is propagated in the x-direction, then the electric field can either oscillate in the y-direction, or the z-direction but must oscillate perpendicularly to both the the direction of oscillation of the magnetic field, and the direction of propagation of the wave.

Answer:

  t = 1.81 min ,     the correct answer is c

Explanation:

This is a missile throwing exercise

The object is thrown horizontally, so its vertical speed is zero (voy = 0), let's use the equation

             y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

the final height is y = 0 and the initial height is y₀ = 22000 m

            0 = y₀ + 0 - ½ g t²

            t = √y 2y₀ / g

let's calculate

           t = √(2  22000 / 3.72)

           t = 108.76 s

let's reduce to minutes

           t = 108.76 s (1 min / 60 s)

           t = 1.81 min

The correct answer is c

Answer:

The self-inductance in henries for the solenoid is 0.0274 H.

Explanation:

Given;

number of turns, N = 1500 turns

length of the solenoid, L = 13 cm = 0.13 m

radius of the wire, r = 2 cm = 0.02 m

The self-inductance in henries for a solenoid is given by;

[tex]L = \frac{\mu_oN^2A}{l}[/tex]

where;

[tex]\mu_o[/tex] is permeability of free space = [tex]4\pi*10^{-7} \ H/m[/tex]

A is the area of the solenoid = πr² = π(0.02)² = 0.00126 m²

[tex]L = \frac{4\pi *10^{-7}(1500)^2*(0.00126)}{0.13} \\\\L = 0.0274 \ H[/tex]

Therefore, the self-inductance in henries for the solenoid is 0.0274 H.

Answer:

A thermos bottle works well because:

A) Its glass walls are thin

Answer:

A thermos bottle works well because:

C

Vacuum reduces heat radiation

Answer:

L = 0.8 m

Explanation:

Since, the distance between first and third dark fringes is 4 mm. Therefore, the fringe spacing between consecutive dark fringes will be:

Δx = 4 mm/2 = 2 mm = 2 x 10⁻³ m

but,

Δx = λL/d

λ = wavelength of the light = 500 nm = 5 x 10⁻⁷ m

d = slit spacing = 0.2 mm = 0.2 x 10⁻³ m

L = Distance between slits and screen = ?

Therefore, using the values, we get:

2 x 10⁻³ m = (5 x 10⁻⁷ m)(L)/(0.2 x 10⁻³)

L = (2 x 10⁻³ m)(0.2 x 10⁻³ m)/(5 x 10⁻⁷ m)

L = 0.8 m

Answer: Speed = [tex]3.10^{-31}[/tex] m/s

Explanation: Like in classical physics, when external net force is zero, relativistic momentum is conserved, i.e.:

[tex]p_{f} = p_{i}[/tex]

Relativistic momentum is calculated as:

p = [tex]\frac{mu}{\sqrt{1-\frac{u^{2}}{c^{2}} } }[/tex]

where:

m is rest mass

u is velocity relative to an observer

c is light speed, which is constant (c=[tex]3.10^{8}[/tex]m/s)

Initial momentum is zero, then:

[tex]p_{f}[/tex] = 0

[tex]p_{1}-p_{2}[/tex] = 0

[tex]p_{1} = p_{2}[/tex]

To find speed of the heavier fragment:

[tex]\frac{mu_{1}}{\sqrt{1-\frac{u^{2}_{1}}{c^{2}} } }=\frac{mu_{2}}{\sqrt{1-\frac{u^{2}_{2}}{c^{2}} } }[/tex]

[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=\frac{3.10^{-28}.0.844.3.10^{8}}{\sqrt{1-\frac{(0.844c)^{2}}{c^{2}} } }[/tex]

[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=1.42.10^{-19}[/tex]

[tex]1.86.10^{-27}u_{1} = 1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }[/tex]

[tex](1.86.10^{-27}u_{1})^{2} = (1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } })^{2}[/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38}.(1-\frac{u_{1}^{2}}{9.10^{16}} )[/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -[2.02.10^{-38}(\frac{u_{1}^{2}}{9.10^{16}} )][/tex]

[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -2.24.10^{-23}.u^{2}_{1}[/tex]

[tex]3.46.10^{-54}.u_{1}^{2}+2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]

[tex]2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]

[tex]u^{2}_{1} = \frac{2.02.10^{-38}}{2.24.10^{-23}}[/tex]

[tex]u_{1} = \sqrt{9.02.10^{-62}}[/tex]

[tex]u_{1} = 3.10^{-31}[/tex]

The speed of the heavier fragment is [tex]u_{1} = 3.10^{-31}[/tex]m/s.

Answer:

This means that mercury has a higher or faster expansion rate than glass

Explanation:

This is because When a container expands, the reservoir in the glass expands at the same rate as the glass. Thus, if there is something in a glass and both expand at the same rate, they have no change - but if the contents expand faster, they will fill the container to a higher level, and if the contents expand slower, they will fill the container to a lower level (relative to the new size of the container).

Answer:

Maximum altitude to see(L) =  1.47 × 10⁶ m (Approx)

Explanation:

Given:

wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m

Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m

Separation distance (D) = 5.4 cm = 0.054 m

Find:

Maximum altitude to see(L)

Computation:

Resolving power = 1.22(λ / d)

D / L = 1.22(λ / d)

0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]

0.054 / L = 1.22 [0.03 × 10⁻⁶]

L = 0.054 / 1.22 [0.03 × 10⁻⁶]

L = 0.054 / [0.0366 × 10⁻⁶]

L = 1.47 × 10⁶

Maximum altitude to see(L) =  1.47 × 10⁶ m (Approx)

Answer:

a. 365; b. 3; c. 78; d. 1.343 rad; e. 12; f. 10.66

Explanation:

Assume that the function is

[tex]D(x) = 3 \sin \left (\dfrac{2\pi}{365}(x - 78) \right ) + 12[/tex]

The general formula for a sinusoidal function is

      y = A sin(B(x - C))+ D

   |A| = amplitude

     B = frequency

2π/B = period, P

     C = horizontal shift (phase shift)

     D = vertical shift

By comparing the two formulas, we find

|A| = 3

 B = 2π/365

 C = 78

 D = 12

a. Period

P = 2π/B = 2π/(2π/365) = 2π × 365/2π = 365

The period is 365.

b. Amplitude

|A| = 3

The amplitude is 3.  

c. Horizontal shift

C= 78

The horizontal shift is 78.

d. Phase shift  (φ)

Ths phase shift is the horizontal shift expressed in radians.

φ = C × 2π/365 = 78 × 2π/365 ≈ 1.343

The phase shift is 1.343 rad.

e. Vertical shift

D = 12

The vertical shift is 12.

f. Hours of sunlight on Feb 21

Feb 21 is the 52nd day of the year, so x = 51 (the number of days after Jan 1),

[tex]\begin{array}{rcl}D(x) &=& 3 \sin \left (\dfrac{2\pi}{365}(x - 78) \right ) + 12\\\\&=& 3 \sin (0.01721(51 - 78) ) + 12\\&=& 3\sin(-0.4648) + 12\\&=& 3(-0.4482) + 12\\\&=& -1.345 + 12\\& = & \textbf{10.66 h}\\\end{array}[/tex]

There will be 10.66 h of sunlight on Feb 21 of any given year.

The figure below shows the graph of the function from 0 ≤ x ≤ 365.

Answer:

λmax = 372 nm

Explanation:

First we find the energy of photon:

E = hc/λ

where,

E = Energy of Photon = ?

λ = Wavelength of Light = 233 nm = 2.33 x 10⁻⁷ m

c = speed of light = 3 x 10⁸ m/s

h = Planks Constant = 6.626 x 10⁻³⁴ J.s

Therefore,

E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(2.33 x 10⁻⁷ m)

E = 8.5 x 10⁻¹⁹ J

Now, from Einstein's Photoelectric Equation:

E = Work Function + Kinetic Energy

8.5 x 10⁻¹⁹ J = Work Function + (1.98 eV)(1.6 x 10⁻¹⁹ J/1 eV)

Work Function = 8.5 x 10⁻¹⁹ J - 3.168 x 10⁻¹⁹ J

Work Function = 5.332 x 10⁻¹⁹ J

Since, work function is the minimum amount of energy required to emit electron. Therefore:

Work Function = hc/λmax

λmax = hc/Work Function

where,

λmax = maximum wavelength of light that will produce photoelectrons = ?

Therefore,

λmax = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(5.332 x 10⁻¹⁹ J)

λmax = 3.72 x 10⁻⁷ m

λmax = 372 nm

Answer:

The hoop

Explanation:

Because it has a smaller calculated inertia of 2/3mr² compares to the disc

Answer:

The wave is travelling in the ±z-axis direction.

Explanation:

An electromagnetic wave has an oscillating magnetic and electric field. The electric and magnetic field both oscillate perpendicularly one to the other, and the wave travels perpendicularly to the direction of oscillation of the  electric and magnetic field.

In this case, if the magnetic field is in the +x-axis direction, and the electric field is in the +y-axis direction, we can say with all assurance that the wave will be travelling in the ±z-axis direction.

Answer:

 Δy = 1 10⁻⁴ m

Explanation:

In double-slit experiments the constructive interference pattern is described by the equation

           d sin θ = m λ

In this case we have two wavelengths, so two separate patterns are observed, let's use trigonometry to find the angle

         tan θ = y / L

as the angles are small,

         tan θ = sin θ / cos θ = sin θ

substituting

         sin θ = y / L

         d y / L = m λ

         y = m λ / d L

let's apply this formula for each wavelength

λ = 450 nm = 450 10⁻⁹ m

m = 5

d = 5.0 mm = 5.0 10⁻³ m

      y₁ = 5 450 10⁻⁹ / (5 10⁻³  1.4)

      y₁ = 3.21 10⁻⁴ m

we repeat the calculation for lam = 590 nm = 590 10⁻⁹ m

      y₂ = 5 590 10⁻⁹ / (5 10⁻³  1.4)

      y₂=  4.21 10⁻⁴ m

the separation of these two lines is

        Δy = y₂ - y₁

        Δy = (4.21 - 3.21) 10⁻⁴ m

        Δy = 1 10⁻⁴ m

Explanation:

In order of Increasing Wavelength of the Electromagnetic Spectrum :

B) X rays

D) Visible light

A) Microwave

C) Radio Waves

Electromagnetic waves in order of decreasing wavelength  is X-rays,visible light,microwaves and radio waves.

The electromagnetic radiation consists of waves made up of electromagnetic field which are capable of propogating through space and carry the radiant electromagnetic energy.

The radiation are composed of electromagnetic waves which are synchronized oscillations of electric and magnetic fields . They are created due to change which is periodic in electric as well as magnetic fields.

In vacuum ,all the electromagnetic waves travel at the same speed that is with the speed of air.The position of an electromagnetic wave in an electromagnetic spectrum is characterized by it's frequency or wavelength.They are emitted by electrically charged particles which undergo acceleration and subsequently interact with other charged particles.

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Answer: The correct answer for this question is letter (B) The electromagnetic waves reach Earth, while the mechanical waves do not. The sun generates both mechanical and electromagnetic waves. Space, between the sun and the earth is a nearly vacuum. So mechanical wave can not spread out in the vacuum.

Hope this helps!

Answer:

The electromagnetic waves reach Earth, while the mechanical waves do not

Answer:

Explanation:

From the question we are told that

    The radius is  [tex]r = 1.4 \ mm = 1.4 *10^{-3} \ m[/tex]

     The  current is  [tex]I = 4.5 \ A[/tex]

Generally the electric field is mathematically represented as

         [tex]E = \frac{J}{\sigma }[/tex]

Where [tex]\sigma[/tex] is the conductivity of  aluminum with value [tex]\sigma = 3.5 *10^{7} \ s/m[/tex]

J is the current density which mathematically represented as  

      [tex]J = \frac{I}{A}[/tex]

Here A is the cross-sectional area which is mathematically represented as  

       [tex]A = \pi r^2[/tex]

       [tex]A = 3.142 * (1.4*10^{-3})^2[/tex]

       [tex]A = 6.158*10^{-6} \ m^2[/tex]

So

    [tex]J = \frac{ 4.5 }{6.158*10^{-6}}[/tex]

    [tex]J = 730757 A/m^2[/tex]

So

       [tex]E = \frac{ 730757}{3.5*10^{7} }[/tex]

       [tex]E = 0.021 \ N/C[/tex]

Answer:

Both Ellen and Mary are correct.

Explanation:

Both are correct, it's just different ways of saying the same thing.

When the acceleration is always opposite in direction to the displacement, then, the acceleration is directly proportional to the displacement of an object from its equilibrium position

Answer:

5.465 × 10^10 revolutions

Explanation:

Formula for Magnetic Field = m. v/ q . r

M = mass of electron = mass of positron = 9.11 x 10^-31 kg,

radius of the positron = 5.03 mm

We convert to meters.

1000mm = 1m

5.03mm = xm

Cross multiply

x = 5.03/1000mm

x = 0.00503m

q = Electric charge = -1.6 x 10^-19 C

Magnetic field (B) = 0.85 T

Speed of the positron is unknown

0.85 = 9.11 x 10^-31 kg × v/ -1.6 x 10^-19 C × 0.00503

0.85 × 1.6 x 10^-19 C × 0.00503 = 9.11 x 10^-31 kg × v

v = 0.85 × -1.6 x 10^-19 C × 0.00503/9.11 x 10^-31 kg

v = 6.8408 ×10-22/ 9.11 x 10^-31 kg

v = 750911086.72m/s

Formula for complete revolutions =

Speed × time / Circumference

Time = 2.30s

Circumference of the circular path = 2πr

r =0.00503

Circumference = 2 × π × 0.00503

= 0.0316044221

Revolution = 750911086.72 × 2.30/0.0316044221

= 1727095499.5/0.0316044221

= 546541562294 revolutions

Approximately = 5.465 × 10^10 revolutions

Complete Question

A 590-turn solenoid is 12 cm long. The  current in it is 36 A . A 2 cm straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).

What is the magnitude of the force on this wire assuming the solenoid's field points due east?

Answer:

The force is  [tex]F = 0.1602 \ N[/tex]

Explanation:

From the question we are told that

   The number of turns is  [tex]N = 590 \ turns[/tex]

   The  length of the solenoid is  [tex]L = 12 \ cm = 0.12 \ m[/tex]

   The current is  [tex]I = 36 \ A[/tex]

   The  diameter is  [tex]D = 4.5 \ cm = 0.045 \ m[/tex]

   The  current carried by the wire is  [tex]I = 27 \ A[/tex]

    The  length of the wire is  [tex]l = 2 cm = 0.02 \ m[/tex]

Generally the magnitude of the force  on this wire assuming the solenoid's field points due east is mathematically represented as

           [tex]F = B * I * l[/tex]

Here  B  is the magnetic field which is mathematically represented as

          [tex]B = \frac{\mu_o * N * I }{L}[/tex]

Here   [tex]\mu _o[/tex] is permeability of free space with value  [tex]\mu_ o = 4\pi *10^{-7} \ N/A^2[/tex]

substituting values

         [tex]B = \frac{4 \pi *10^{-7} * 590 * 36 }{ 0.12}[/tex]

           [tex]B = 0.2225 \ T[/tex]

So

      [tex]F = 0.2225 * 36 * 0.02[/tex]

      [tex]F = 0.1602 \ N[/tex]

Answer:

d) (CATA + CBTB) / (CA + CB)

Explanation:

According to the given situation, the final temperature of both objects is shown below:-

We assume T be the final temperature

while m be the mass

So it will be represent

m CA (TA - T) = m CB (T - TB)

or we can say that

CATA - CA T = CB T - CBTB

or

(CA + CB) T = CATA + CBTB

or

T = (CA TA + CBTB) ÷ (CA + CB)

Therefore the right answer is d

The final temperature of both objects is [tex]T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex].

The given parameters;

The final temperature of both objects is calculated as follows;

heat lost by object A is equal to heat gained by object B

[tex]mC_A (T_A - T) = mC_B(T- T_B)\\\\C_AT_A-C_AT = C_BT - C_BT_B\\\\C_BT+C_AT = C_AT_A+ C_BT_B\\\\T(C_B + C_A) = C_AT_A+ C_BT_B \\\\T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex]

Thus, the final temperature of both objects is [tex]T = \frac{C_AT_A+ C_BT_B}{C_B + C_A} \\\\[/tex].

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Answer:

Torque = 0.012 N.m

Explanation:

We are given;

Mass of wheel;m = 750 g = 0.75 kg

Radius of wheel;r = 25 cm = 0.25 m

Final angular velocity; ω_f = 0

Initial angular velocity; ω_i = 220 rpm

Time taken;t = 45 seconds

Converting 220 rpm to rad/s we have;

220 × 2π/60 = 22π/3 rad/s

Equation of rotational motion is;

ω_f = ω_i + αt

Where α is angular acceleration

Making α the subject, we have;

α = (ω_f - ω_i)/t

α = (0 - 22π/3)/45

α = -0.512 rad/s²

The formula for the Moment of inertia is given as;

I = ½mr²

I = (1/2) × 0.75 × 0.25²

I = 0.0234375 kg.m²

Formula for torque is;

Torque = Iα

For α, we will take the absolute value as the negative sign denotes decrease in acceleration.

Thus;

Torque = 0.0234375 × 0.512

Torque = 0.012 N.m

Answer:

8.1 m

Explanation:

Convert km/h to m/s.

45 km/h × (1000 m/km) × (1 h / 3600 s) = 12.5 m/s

Distance = speed × time

d = (12.5 m/s) (0.65 s)

d = 8.125 m

Answer:

Explanation:

We shall apply formula of Doppler's effect

Here source is fixed and observer is approaching the source

f = f₀ x [(V + v ) / V ]

f₀ is original and f is apparent frequency , V is velocity of sound and v is velocity of motorcyclist .

f = 1000 x [(331 + 27.27 ) / 331 ]

= 1082 .4 Hz

This is the frequency heard by motorcyclist .

When police car is approaching him when he is stopped

f = f₀ x [V /(V - v ) ]

v is velocity of police car .

= 1000  x 331 / (331 - 27.27)

= 1090 Hz  

Explanation:

u = 95 m/sec ( Initial speed)

t = 7 sec ( Time of ascent)

According to Equations of Motion :

[tex]s = ut - \frac{1}{2} g {t}^{2} [/tex]

Max. Height = 95 * 7 - 4.9 * 49 = 424. 9 = 425 m

Answer:

332.5 m

Explanation:

At the maximum height, the velocity is 0.

Given:

v₀ = 95 m/s

v = 0 m/s

t = 7 s

Find: Δy

Δy = ½ (v + v₀) t

Δy = ½ (0 m/s + 95 m/s) (7 s)

Δy = 332.5 m

Answer: Ф = 17.2657 ≈ 17°

Explanation:

we simply apply ET =0 about the ending of the rod

so In.g.L/2sinФ - In.a.L/2cosФ = 0

g.sinФ - a.cosФ = 0

g.sinФ = a.cosФ

∴ tanФ = a/g

Ф =  tan⁻¹ a / g

Ф = tan⁻¹ ( 10 / 32.17405)

Ф = tan⁻¹ 0.31080948777

Ф = 17.2657 ≈ 17°

Therefore the angle of rotation of the rod during this acceleration is 17.2657 ≈ 17°

Answer:

The temperature difference [tex]\Delta T = 258.6 \ ^ o\ C[/tex]

Explanation:

From the question we are told that

   The  thickness is [tex]\Delta x = 0.630 cm = 0.0063 m[/tex]

    The  area is  [tex]A = 1.45 *10^{-2 } \ m^2[/tex]

     The rate is  [tex]P = 500 J/s[/tex]

       The  thermal conductivity is  [tex]\sigma = 0.84J[\cdot s \cdot m \cdot ^oC ][/tex]

Generally the rate heat conduction mathematically represented as

       [tex]P = \sigma * A * \frac{\Delta T}{\Delta x }[/tex]

=>    [tex]\Delta T = \frac{P * \Delta x }{\sigma * A }[/tex]

=>     [tex]\Delta T = \frac{ 500 * 0.00630 }{ 0.84 * 1.45 *10^{-2} }[/tex]

=>    [tex]\Delta T = 258.6 \ ^ o\ C[/tex]

Answer:gamma ray

Explanation:

Answer:

The induced current is clockwise