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Allushta [10]
2 years ago
5

What is the magnetic field strength at point 1 in the figure? (figure 1)

Physics
1 answer:
Lana71 [14]2 years ago
6 0

The magnetic field strength at point 1 in the figure will be 6.67 ×10⁻⁵ T.

<h3>What is magnetic field strength?</h3>

The number of magnetic flux lines on a unit area passing perpendicular to the given line direction is known as induced magnetic field strength .it is denoted by B.

The magnetic field strength is found as;

B = \frac{\mu_0I}{2r}  \\\\ \mu_0 = 4 \PI \times 10^{-7}

In the formula,I denote current, and r denotes the distance between the point and the current carrying wire and magnetic field due to current in the bottom wire.

At point 1, the net magnetic field is found as the sum of magnetic field due to current in the top wire.

\rm B_{net} = B1_+(-B_2)

B = \frac{ 4 \PI \times 10^{-7}I}{2r}  \\\\  \rm B_{net} = B_1_+(-B_2)\\\\ \rm B_{net} = \frac{4 \times \pi \times 10^{-7} \times 10}{2 \times \pi \times 0.02} -\frac{4 \times \pi \times 10^{-7} \times 10 }{2 \times \pi \times 0.06} \\\\\  \rm B_{net} = 6.67 \times 10^{-5} T

Hence, the magnetic field strength at point 1 in the figure will be 6.67 ×10⁻⁵ T.

To learn more about the strength of induced magnetic field, refer:

brainly.com/question/2248956

#SPJ4

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Why do you think the speed of sound is higher<br> in solids, than liquids and gases?
mr Goodwill [35]

Answer:

Explanation:

Particles of a solid are closely packed, and  when a solid is hit to produce a sound the kinetic energy present in the solid will cause the particles to vibrate vigorously in the intermolecular spaces in the solid, whereas the paticles of liquids and gases are rather further apart therefore having large intermolecular spaces between particles.

8 0
3 years ago
What is the frequency of the fundamental mode of vibration of a steel piano wire stretched to a tension of 440 N? The wire is 0.
Rasek [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

    The tension is  T  =  440 \  N

     The length of the wire is l =  0.600 \  m

     The mass is  m  =  5.60 \ g = 0.0056 \  kg

Generally the frequency is mathematically represented as

            f  =  \frac{1}{2l}  *  \sqrt{\frac{T}{\frac{m}{l} } }

=>        f  =  \frac{1}{2 * 0.600 }  *  \sqrt{\frac{440}{\frac{0.0056}{0.600} } }

=>        f = 181 \ Hz

4 0
3 years ago
The period of sound wave coming from an instrument is o.oo2. What is the frequency of the sound
Maslowich

The frequency of the sound is 500

5 0
3 years ago
Two resistors, A and B, are connected in series to a 6.0 V battery. A voltmeter connected across resistor A measures a potential
mestny [16]

Answer:

Resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

Explanation:

When the two resistors are in series, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B.

Given that V₁ + V₂ = 6.0 V and V₁ = 4.0 V,

V₂ = 6.0 V - V₁ = 6.0 V - 4.0 V = 2.0 V

Also, let the current in series be I.

So, V₁ = IR₁ and V₂ = IR₂

I = V₁/R₁ and I = V₂/R₂

equating both expressions, we have

V₁/R₁ = V₂/R₂

4.0 V/R₁ = 2.0 V/R₂

dividing through by 2.0 V, we have

2/R₁ = 1/R₂

taking the reciprocal, we have

R₂ = R₁/2

R₁ = 2R₂

From the parallel connection, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B. Since it is parallel, V₁ = V₂ = V = 6.0 V

Also, V₂ = I₂R₂ where I₂ = current in resistor B = 2.0 A and R₂ = resistance of resistor B

So, R₂ = V₂/I₂

= 6.0 V/2.0 A

= 3.0 Ω

R₁ = 2R₂

= 2(3.0 Ω)

= 6.0 Ω

So, resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

6 0
3 years ago
A pendulum of length L=36.1 cm and mass m=168 g is released from rest when the cord makes an angle of 65.4 degrees with the vert
pychu [463]

(a) -0.211 m

At the beginning the mass is displaced such that the length of the pendulum is L = 36.1 cm and the angle with the vertical is

\theta=65.4^{\circ}

The projection of the length of the pendulum along the vertical direction is

L_y = L cos \theta = (36.1 cm)(cos 65.4^{\circ})=15.0 cm

the full length of the pendulum when the mass is at the lowest position is

L = 36.1 cm

So the y-displacement of the mass is

\Delta y = 15.0 cm - 36.1 cm = -21.1 cm = -0.211 m

(b) 0.347 J

The work done by gravity is equal to the decrease in gravitational potential energy of the mass, which is equal to

\Delta U = mg \Delta y

where we have

m = 168 g = 0.168 kg is the mass of the pendulum

g = 9.8 m/s^2 is the acceleration due to gravity

\Delta y = 0.211 m is the vertical displacement of the pendulum

So, the work done by gravity is

W=(0.168 kg)(9.8 m/s^2)(0.211 m)=0.347 J

And the sign is positive, since the force of gravity (downward) is in the same direction as the vertical displacement of the mass.

(c) Zero

The work done by a force is:

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and the displacement

In this situation, the tension in the string always points in a radial direction (towards the pivot of the pendulum), while the displacement of the mass is tangential (it follows a circular trajectory): this means that the tension and the displacement are always perpendicular to each other, so in the formula

\theta=90^{\circ}, cos \theta = 0

and so the work done is zero.

5 0
3 years ago
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