The map below shows major ocean currents in the North Atlantic and North Pacific Oceans. In general, currents flowing toward the
Equator bring cooler waters to some regions, while currents flowing away from the Equator bring warmer waters to other regions.
North
British
Isles
Askan
North Atlantic
Azor
U.S.A
California
Gulf Stream
Loop
n
Canbean
North Equatorial
North Equatorial CC
North Fuatorial
Equator
South Equatorial
Not
South Equatorial
Image courtesy of NOAA
Judging from the map, which region probably has cooler summers than it would without the effect of a nearby ocean current?
A the Central U.S.
B. the British Isles
C. the U.S. East Coast
D. the US West Coast

Answer:

0.09kg of air

Explanation:

The dimensions of the room are given

change the height to meters by dividing it by thousand.

For the volume multiplying the length,width and height (all should be in the same unit most suitable being meters).

Volume refers to the amount of space inside a box or a object.

The amount of air is equal to the volume.

Answer:

90 m^3

Explanation:

Volume of the room:

    6 m * 5 m * 3 m         =  90 m^3   <=====( I changed 3mm to 3 m)

if  3mm is not a typo mistake

 volume becomes     ( 3 mm = .003 m)

      6 m * 5 m * .003 m   = .09 m^3   ( though unlikely )

The volume of the helium balloon in order to lift the weight is 17,760m³.

To find the answer, we need to know about the buoyant force.

     => V= 3179/0.179 = 17,760m³

Thus, we can conclude that the volume of the helium balloon in order to lift the weight is 17,760m³.

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The kinetic energy of the projectile at the highest point of its motion will be E/4.

When a projectile will be thrown obliquely near the surface of the earth, it travels a curved path under uniform acceleration directed toward the center of the Earth. The path of a particle is called a projectile while the motion of a projectile is projectile motion.

Given, the angle of projection  with horizontal, [tex]\theta = 60 ^o[/tex]

Consider that 'E' is the initial value of the kinetic energy of the projectile.

The equation for the initial kinetic energy is : [tex]E =\frac{1}{2}mu^2[/tex]

where m is the mass of the given projectile.

The component of the velocity of the projectile in the horizontal direction:

uₓ = u cosθ

uₓ = u cos 60°

uₓ = u/2

From the equation of motion: v = u +at

v = (u/2) + (0) t

v = u/2

The final kinetic energy of the projectile:

[tex]E_f = \frac{1}{2}mv^2[/tex]

[tex]E_f = \frac{1}{2}m(\frac{u}{2} )^2[/tex]

[tex]E_f = \frac{1}{4} (\frac{1}{2}mu^2 )[/tex]

[tex]E_f = \frac{E}{4}[/tex]

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Based on the weight of the objects in the water, the water level in the given scenario is as follows;

The rise or fall of a fluid when an object is placed in it is determined by the density, mass, and volume of the object.

According to Archimedes' principle, the upthrust or upward force acting on a body fully or partially  immersed in a fluid, is equal to the weight of the fluid displaced.

Considering the given situations:

In conclusion, the rise in water level depends on the weight of the objects.

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The distance from A to B is 39.375 Km.

The total distance travelled by the cyclist is given by the formula below:

Total distance = (15 + 20)/2 * 4.5

Total distance = 78.75 Km

Thus, distance from A to B = 78.75/2 = 39.375 Km

In conclusion, the total distance travelled is determined from the average speed and total time taken.

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The voltage between the plates is 3.9 × [tex]10^-^3[/tex] V.

The work-energy theorem states that the delta in the equation equals the change in kinetic energy plus the change in potential energy. Here, a charge's potential energy is expressed as qV, where V is the position's electric potential.  The greater the change in voltage per unit distance, the greater the electric field.

The kinetic energy of the electrons = 4.1 × [tex]10^-^1^5[/tex] J

Charge of the electron = 1.602 × [tex]10^-^1^9[/tex] coulomb

Using,

     ΔU = q × ΔV

4.1 × [tex]10^-^1^5[/tex] = 1.602 × [tex]10^-^1^9[/tex] × ΔV

      ΔV  = 3.9 × [tex]10^-^3[/tex] V

Therefore, the voltage between the plates is 3.9 × [tex]10^-^3[/tex] V.

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Answer:

4 C

Explanation:

The strength of an electric field can be defined as: [tex]E = \frac{f}{q}[/tex] where f=force and q=charge, and e=strength

Since we're given the strength and the force we can simply rearrange the equation so that we solve for Q:

Original Equation:

[tex]E=\frac{f}{q}[/tex]

Multiply both sides by q

[tex]E*q = f[/tex]

Divide both sides by E

[tex]q=\frac{f}{e}[/tex]

So now plug the known values into the equation

[tex]q=\frac{1000 N}{250 N/C}[/tex]

Simplify:

[tex]q = 4 c[/tex]

The initial momentum of the yellow and the orange train is 1000kgm/s.

Momentum is the product of the mass and velocity of any object.

Momentum is denoted by P.

Momentum P = mv , where m = mass and v = velocity.

Mass of the orange train = 200kg

Velocity of the orange train = 1m/s

So, the momentum of the orange train will be,

                            ∴    P = mv

                                  P = 200 x 1

                                  P = 200 kgm/s

∴   The initial momentum of the orange train is 200kgm/s.

Mass of the yellow train = 100kg

Velocity of the yellow train = 8m/s

So, the momentum of the yellow train will be,

                            ∴    P = mv

                                  P = 100 x 8

                                  P = 800 kgm/s

∴ The initial momentum of the yellow train is 800kgm/s.

Therefore, the initial momentum of the yellow and the orange train is 1000kgm/s.

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Answer: the index of refraction of material X will be 1.09 and the angle the light makes with the normal in the air is 81.25°.

Explanation: To find the answer, we need to know the Snell's law.

                      [tex]\frac{sin (i)}{sin(r)} =\frac{n_r}{n_i}[/tex]

where, i is the incident angle, r is the refracted angle, n is the refractive index.

                           [tex]\frac{sin 65}{sin48} =\frac{n_w}{n_X} \\n_X=\frac{1.33*sin48}{sin65} =1.09\\[/tex]

                         [tex]\frac{sin 48}{sin r}=\frac{1}{1.33} \\\\ sin(r)=\frac{1.33*sin 48}{1}=0.988\\\\r=sin^{-1}(0.988)=81.25 degrees.[/tex]

Thus, we can conclude that, the index of refraction of material X will be 1.09 and the angle the light makes with the normal in the air is 81.25°.

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The light makes an angle of 81.25° with the normal in the air and the index of refraction of material X will be 1.09.

In order to determine the solution, we must understand Snell's law.

                            [tex]\frac{sin(i)}{sin(r)}=\frac{n_r}{n_i}[/tex]

where the refractive index is n and the incidence angle is i. The refracted angle is r.

                      [tex]n_X=\frac{n_w*sin 48}{sin65} =1.09[/tex]

                  [tex]sin(r)=n_w*sin48=0.988\\r=sin^{-1}(0.988)=81.25 degrees.[/tex]

As a result, we can infer that the material X will have an index of refraction of 1.09 and that the angle the light makes with the normal in the air is 81.25°.

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The double-slit experiment shows that both matter and light can exhibit properties of conventionally defined waves and particles.

The double-slit experiment  is a part of a class of "double path" experiments in which a wave is split into two separate waves that later combine to form a single wave (the wave is typically composed of many photons and is better known as a wave front, which should not be confused with the wave properties of the individual photon).

Isaac Newton's corpuscular theory of light, which had previously prevailed as the accepted explanation of light transmission in the 17th and 18th centuries, was defeated by double-slit experiment , which was conducted in the early 1800s.

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(a) The magnitude of the magnetic field at point P1 , midway between the wires is  1.005 x 10⁻⁴ T and the direction will be out of the page.

(b) The magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 is 2.67 x 10⁻⁶ T and the direction is into the page.

(c) The magnitude of the magnetic field at point P3 , 20.0 cm directly above P1  is 3.88 x 10⁻⁶ T and the direction is downwards.

B = μ/2π[I₁/0.5r + I₂/0.5r]

B = (μ/2π) x (I/0.5r + I/0.5r)

B = (μ/2π) x (2I/0.5r)

B = μI/0.5r

B = 2μI/r

where;

B = (2 x 4π x 10⁻⁷ x 4)/(0.1)

B = 1.005 x 10⁻⁴ T

The direction of the magnetic field is out of the page.

B = μI/2πd

d = 5 cm + 25 cm = 30 cm

B =  (4π x 10⁻⁷ x 4)/(2π x 0.3)

B = 2.67 x 10⁻⁶ T

The direction of the magnetic field is into the page towards P1.

B = μI/2πd

d = √(20² + 5²)

d = 20.62 cm

B =  (4π x 10⁻⁷ x 4)/(2π x 0.2062)

B = 3.88 x 10⁻⁶ T

The direction of the magnetic field is downwards towards P1.

Thus, the magnitude of the magnetic field at point P1 , midway between the wires is  1.005 x 10⁻⁴ T and the direction will be out of the page.

The magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 is 2.67 x 10⁻⁶ T and the direction is into the page.

The magnitude of the magnetic field at point P3 , 20.0 cm directly above P1  is 3.88 x 10⁻⁶ T and the direction is downwards.

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The mutual inductance of the pair of coils is 6.67 x 10⁻³ H.

The the flux through each turn is  2.78 x 10⁻⁴ Tm².

The magnitude of the induced emf in the first coil is 2.435 x 10⁻³ V.

M = -NΦ/I

M = emf/I

M = -(1.6 x 10⁻³)/(-0.24)

M = 6.67 x 10⁻³ H

M = NΦ/I

MI = NΦ

Φ = MI/N

Φ = (6.67 x 10⁻³  x 1.25)/(30)

Φ = 2.78 x 10⁻⁴ Tm²

emf = MI

emf = 6.67 x 10⁻³ x 0.365

emf = 2.435 x 10⁻³ V

Thus, the mutual inductance of the pair of coils is 6.67 x 10⁻³ H.

The the flux through each turn is  2.78 x 10⁻⁴ Tm².

The magnitude of the induced emf in the first coil is 2.435 x 10⁻³ V.

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(a) The magnitude of the net magnetic force per unit length on the top wire is μI²/πd.

(b) The magnitude of the net magnetic force per unit length on the middle wire is zero.

(c) The magnitude of the net magnetic force per unit length on the bottom wire is 3μI²/4πd.

The magnitude of the net magnetic force per unit length on the top wire is calculated as follows;

F₁/L = (μI₁/2π) x (I₂/d + I₃/d)

F₁/L = (μI/2π) x (I/d + I/d)

F₁/L = (μI/2π) x (2I/d)

F₁/L = μI²/πd

The magnitude of the net magnetic force per unit length on the middle wire is calculated as follows;

F₂/L = (μI₂/2π) x (I₃/d - I₁/d)

F₂/L = (μI/2π) x (I/d -  I/d) = 0

The magnitude of the net magnetic force per unit length on the middle bottom is calculated as follows;

F₃/L = (μI₂/2π) x (I₁/d + I₂/d)

F₃/L =  (μI/2π) x (I/2d + I/d)

F₃/L =  (μI/2π) x (3I/2d)

F₃/L =  3μI²/4πd

Thus, the magnitude of the net magnetic force per unit length on the top wire is μI²/πd.

The magnitude of the net magnetic force per unit length on the middle wire is zero.

The magnitude of the net magnetic force per unit length on the bottom wire is 3μI²/4πd.

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Hi there!

a)
We can use the equation for the magnetic dipole moment to solve for the number of turns:
[tex]\mu_m = NIA\vec{n}[/tex]

[tex]\mu_m[/tex] = Magnetic dipole moment (0.194 Am²)

N = Number of loops (?)

A = Area of loop (4.0 cm²)

[tex]\vec{n}[/tex] denotes the area vector, or the normal line perpendicular to the area.

We first need to convert cm² to m² using dimensional analysis.

[tex]4.0 cm^2 * \frac{0.01m}{1 cm} * \frac{0.01 m}{1cm} = 0.0004 m^2[/tex]

Rearranging the equation to solve for 'N':

[tex]N = \frac{\mu_m}{IA}\\\\N = \frac{0.194}{(8.8)(0.0004)} = \boxed{55.11 \text{ turns}}[/tex]

**Since we cannot have part of a turn, the coil has about 55 turns.

b)

For this, we can use the Right-Hand-Rule for current. Looking at the coil from the left with your curled fingers going around the coil with the fingertips pointing through and to the left in the direction of the magnetic moment, your thumb points in the COUNTERCLOCKWISE direction.

c)
Now, let's use the equation for the torque produced by a magnetic field:
[tex]\tau = \mu_m \times B[/tex]

This is a cross-product, but since our magnetic field is perpendicular to the magnetic moment, we can disregard it.

Plugging in the values for the magnetic moment and the magnetic field:

[tex]\tau = 0.194 * 45 = \boxed{8.73 Nm}[/tex]

d)

Using the other RHR (current, field, force), the coil will spin about its vertical axis in the field. In more detail, if you look at the coil from the left-hand side with its opening towards you, from this perspective, the left of the coil will come towards you, and the right side of the coil will move away.

The point where m3 experiences a zero net gravitational force due to M1 and m2 is 57.42 m.

m1<------(x)------> m3 <-----------(94.8 m - x)-------->m2

a point, x, where m3 experiences a zero net gravitational force due to M1 and m2;

Force on m3 due to m1 = Force on m3 due to m2

Gm1m3/d² = Gm2m3/r²

m1/d² = m2/r²

where;

m1/(x²) = m2/(94.8 - x)²

m1(94.8 - x)² = m2x²

(94.8 - x)² = (m2/m1)x²

(94.8 - x)² = (10.6/25)x²

(94.8 - x)² = 0.424x²

(94.8 - x)² = (0.651)²x²

94.8 - x = 0.651x

94.8 = 1.651x

x = 94.8/1.651

x = 57.42 m

Thus, the point where m3 experiences a zero net gravitational force due to M1 and m2 is 57.42 m.

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5.48 V is the potential drop across the cable for a copper transmission cable of 50.0 km long and 10.0 cm in diameter carries a current of 105 A

Ohm's Law states that the potential drop is determined by the equation: V = IR, where I is the current and R is the wire resistance.
R=PL/A
Under the assumption that all physical parameters and temperatures remain constant, Ohm's law asserts that the voltage across a conductor is directly proportional to the current flowing through it.

Only when the given temperature and the other physical variables remain constant does Ohm's law apply. Increasing the current causes the temperature to rise in some components. The filament of a light bulb serves as an illustration of this, where the temperature increases as the current increases. Ohm's law cannot be applied in this situation. The filament of the lightbulb defies Ohm's Law.

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(a) The induced current in the loop will be counterclockwise.

(b) The rate at which electrical energy is being dissipated by the resistance of the loop is 0.012 W.

The induced current in the loop will be counterclockwise to the direction of magnetic field.

emf = -NdФ/dt

emf = -NBA/dt

where;

A is area of the loop

A = πr² = π(0.041)² = 5.28 x 10⁻³ m²

emf = -(-0.605 - 7.78) x 5.28 x 10⁻³

emf = 8.385 x 5.28 x 10⁻³

emf = 0.0442 V

P = emf²/R

P = (0.0442)²/0.169

P = 0.012 W

Thus, the rate at which electrical energy is being dissipated by the resistance of the loop is 0.012 W.

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As, per the buoyancy force,the volume that the balloon should have is 2863[tex]m^{3}[/tex]

What is buoyancy force?

Air buoyancy is the upward force exerted on an object by the air that is displaced by object. Air buoyancy is responsible for the buoyancy created by the displaced air.

[tex]F_{b}[/tex] = -Vρg ,       where V= volume of the object

                                   ρ = density of the object

                                   g = acceleration due to gravity

                                   [tex]F_{b}[/tex] = buoyant force

The buoyancy force must be equal to the total load lifted

ρ[tex]_{He}[/tex] × V × g + 269 + 2910 = ρ[tex]_{air}[/tex] × V × g

0.179 × V + 3179 = 1.29 V

0.179V + 3179 = 1.29V

0.179V- 1.29V = - 3179

1.11V = 3179

On solving , we get

V = 2863 [tex]m^{3[/tex]

Therefore, the volume that the balloon should have is 2863[tex]m^{3}[/tex].

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The output current for the given step down transformer is 30A.

What is the step down transformer?

A step down transformer is a passive device that converts high voltage power to low voltage power, while the output current is higher than the  input current. They are used in power adaptors and rectifiers to decrease the voltage to the desired level. It works according to Faraday's law of Electromagnetic induction.

The current in the windings of a step down transformer is inversely proportional to the voltage in windings as:

Input voltage / Output voltage = Output current / Input current

220 / 110 = Outout current / 15

Output current = 30A

Hence, the output current (30A) obtained is higher than the given input current (15A).

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The angle of projection given that the range is 256√3 m and the maximum height reached is 64 m is 30°

R = u²Sine(2θ) / g

256√3 = u²Sine(2θ) / 9.8

Cross multiply

256√3 × 9.8 = u²Sine(2θ)

Divide both sides by Sine(2θ)

u² = 256√3 × 9.8 / Sine(2θ)

H = u²Sine²θ / 2g

64 = [256√3 × 9.8 / Sine(2θ)] × [Sine²θ / 2 × 9.8]

64 = [256√3 / Sine(2θ)] × [Sine²θ / 2]

Recall

Sine²θ = SineθSineθ

Sine2θ = 2SineθCosθ

Thus,

64 = [256√3 / 2SineθCosθ] × [SineθSineθ / 2]

64 = 256√3 × Sineθ / 4Cosθ

Recall

Sineθ / Cosθ = Tanθ

Thus,

64 = 256√3 / 4 × Tanθ

Divide both side by 256√3 / 4

Tanθ = 64 ÷ 256√3 / 4

Tanθ = 0.5774

Take the inverse of Tan

θ = Tan⁻¹ 0.5774

θ = 30°

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The horizontal force applied is  160 N while the velocity is  2.03 m/s.

The work done by the car is obtained as the product of the force and the distance;

W = F x

F = ?

x = 30.0 m

W = 4,800 J

F = 4,800 J/30.0 m

F = 160 N

But F = ma

a = F/m

a = 160 N/2.30 ✕ 10^3-kg

a= 0.069 m/s

Now;

v^2 = u^2 + 2as

u = 0/ms because the car started from rest

v = √2as

v = √2 * 0.069 * 30

v = 2.03 m/s

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Answer:

To give light yo people and the neutron are to give you health care

The deceleration of the bullet over 0.2 second, given the data from the question is –2500 m/s²

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

a is the acceleration

v is the final velocity

u is the initial velocity

t is the time

NOTE: Deceleration is the opposite of acceleration

With the above equation for acceleration, we can obtain the deceleration of the bullet. Details below:

a = (v – u) / t

a = (0 – 500) / 0.2

a = –500 / 0.2

a = –2500 m/s²

Thus, the deceleration of the bullet is –2500 m/s²

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Answer:

See below

Explanation:

T = period of orbit = sqrt ( 4  pi^2  r^3  /  (G  Me)  )

   G = gravitational constant   6.6743 x 10^-11  m^3 / (kg-s^2)

Me = mass of earth      =  6 x 10^24 kg

 r = radius = 6600 km = 6,600,000 m

plug in the values to find T = 5323.75 seconds

(check my math)

The amount of heat needed to raise the temperature of 6.00 g of copper by 15.0°C is 34.65J (option A).

The amount of heat absorbed or released by a substance can be calculated using the following formula:

Q = mc∆T

Where;

Q = 6 × 0.385 × 15

Q = 90 × 0.385

Q = 34.65J

Therefore, the amount of heat needed to raise the temperature of 6.00 g of copper by 15.0°C is 34.65J.

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Answer:

36 cm

Explanation:

Mass of stick; m1 = 0.20kg at midpoint.

Total length; L=1.0 m

Pivot at 0.40m

Atached mass m2 = 0.50kg

Applying rotational equilibrium we have;

Ʈnet = 0

(m1g) • r1 = (m2g) • r2

(0.2) (0.1m) = (0.5)(x)

x = 0.04m =4cm

measured away from 40cm mark gives a position on the stick of; 40cm - 4cm = 36 cm

An object is considered to be in a condition of equilibrium when it is balanced with regard to all external forces.

Equilibrium:

An object is considered to be in equilibrium if both its angular acceleration and the acceleration of its center of mass are equal to zero. In layman's terms: The item must either be at rest or moving at a constant speed if it is not accelerating because F = ma (force = mass x acceleration). Even in motion, a body can be in equilibrium. This kind of equilibrium is referred to as a dynamic equilibrium.

A weight suspended by a spring or a brick laying on a flat surface is an example. The equilibrium is unstable if the force with the smallest deviation tends to increase the displacement. As an example, imagine a ball bearing on the edge of a razor blade.

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The force required to pull the two hemispheres apart is 4.2×10⁴ N and 29 number of horses are needed to pull these hemispheres apart.

=955×100 N/m²= 9.55×10⁴ N/m²

No. of horses required to separate the hemispheres = 4.2×10⁴/1450 = 29

Thus, we can conclude that the 29 horses are needed to pull the two hemispheres with a force of 4.2×10⁴ N.

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The magnitude of the magnetic field and the direction of the magnetic field at point P is mathematically given as

B=1.9*10^{-5}T

To determine the magnetic field direction, use the right-hand -rule on the page magnetic field is going.

Generally, the equation for Biot savant law is  mathematically given as

[tex]B=(\frac{u}{4\pi}*{\frac{Ilsin\theta}{r^2})[/tex]

Their net field is

Bn=2B

[tex]Bn=2* B=(\frac{u}{4\pi}*{\frac{Ilsin\theta}{r^2})[/tex]

Hence

[tex]B=(\frac{4*\p *10^{-7}}{4\pi}*{\frac{(30)(2*10^{-3}sin45)}{\sqrt{(3*10^{-2})^2+((3*10^{-2})^2)}/2})[/tex]

B=1.9*10^{-5}T

In conclusion, To determine the magnetic field direction, use the right-hand -rule on the page magnetic field is going.

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Answer:

The magnitude of the magnetic field at the point P is [tex]1.57*10^{-5}T[/tex] and the field is pointing into the page.

Explanation:

The general form of a similar question to this is:

[tex]\vec{B} = \frac{\mu _{0} }{4\pi } * \oint \frac{Id\vec{l} \times \hat{r}}{r^{2} }[/tex]

where [tex]\vec{B}[/tex] is the vector of the Magnetic Field, [tex]\mu _{0}[/tex] is the Free Space Permeability Constant (equal to [tex]4\pi * 10^{-7} \frac{N}{A^2}[/tex]), [tex]I[/tex] is the current, and [tex]r[/tex] is the distance from the segment to the point P. (I will get to the [tex]d\vec{l} \times \hat{r}[/tex] term in a bit)

This equation is fairly complicated. Luckily, it can be simplified by looking at the magnitude and direction separately.

The first thing to simplify is the cross product. Due to the fact that a cross product can be simplified from [tex]\vec{x} \times \vec{y}[/tex] to [tex]xy\sin(\theta)[/tex], where [tex]\theta[/tex] is the angle between the 2 vectors, and [tex]\hat{r}[/tex] is the unit vector of [tex]r[/tex] (i.e. [tex]\hat{r} = \vec{r}/r[/tex]) we can simplify [tex]d\vec{l} \times \hat{r}[/tex] to just [tex]dl \sin(\theta)[/tex].

Next, we will look at the integral. In this scenario, everything will function as a constant, so we can essentially ignore it.

Finally, [tex]\frac{\mu_{0}}{4\pi}[/tex] simplifies down to [tex]10^{-7}[/tex].

This gives us our new equation for the Magnetic Field produced by a single segment at a point:

[tex]B = \frac{Il\sin\theta}{r^{2}}*10^{-7}[/tex]

Now we need to find values for [tex]r[/tex] and [tex]\theta[/tex]. Luckily, we are dealing with a 45-45-90 triangle with sides of [tex]1.5 cm[/tex]. This means the distance [tex]r[/tex] is [tex](1.5\sqrt2)cm[/tex]! Similarly, because it is a 45-45-90 triangle, our [tex]\theta[/tex] is [tex]45\textdegree[/tex]!

Now we can start plugging things in:

[tex]B = \frac{(25A)(2*10^{-3}m)\sin(45\textdegree)}{(1.5\sqrt2*10^{-2}m)^2}*10^{-7}\frac{N}{A^2}[/tex]

[tex]B = 7.86^{-6} \frac{N}{A}[/tex] or [tex]T[/tex]

This is the magnitude due to only one single segment. In order to find the total field, we need to know the direction of the field due to each segment.

Finding the direction is really easy. Just use the right hand rule. Point your thumb in the direction of the current and curl the rest of your fingers around an imaginary pole. The direction your fingers point is the direction of the field. In this case, the field lines due to the segments point into the page in the 4th quadrant (the origin is the bend). This means that at point P, both segments induce the same field in the same direction. Therefore, we can take our value from before and double it, giving us our final answer:

[tex]B = 1.57*10^{-5} T[/tex]; into the page.