1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
vodka [1.7K]
3 years ago
6

Two wires are perpendicular to each other and form a coordinate axis. The current in the vertical wire is going up (in the posit

ive y direction) and the current in the horizontal wire is going to the right(in the positive x direction). Where is the net magnetic field equal to zero?
Physics
1 answer:
sergij07 [2.7K]3 years ago
4 0

Answer:

Magnetic field shall be zero at exactly in between the wires.

Explanation:

We can find the magnetic field by biot Savart law as follows

\overrightarrow{dB}=\frac{\mu _{0}I}{4\pi }\int \frac{\overrightarrow{dl}\times \widehat{r}}{r^{2}}

For current carrying wire in positive y direction we have

\overrightarrow{dB_{1}}=\frac{\mu _{0}Idl}{4\pi }\int \frac{\widehat{j}\times \widehat{r_{1}}}{r_{1}^{2}}

Similarly for wire carrying current in -y direction we have \overrightarrow{dB_{2}}=\frac{-\mu _{0}Idl}{4\pi }\int \frac{\widehat{j}\times \widehat{r_{2}}}{r_{2}^{2}}

Thus the net magnetic field at any point in space is given by

\overrightarrow{dB_{1}}+\overrightarrow{dB_{2}}

\frac{\mu _{0}Idl}{4\pi }\int \frac{\widehat{j}\times \widehat{r_{1}}}{r_{1}^{2}}+\frac{-\mu _{0}Idl}{4\pi }\int \frac{\widehat{j}\times \widehat{r_{2}}}{r_{2}^{2}}=0\\\\\Rightarrow \overrightarrow{r_{1}}=\overrightarrow{r_{2}}

For points with same position vectors from the 2 wires we have a net zero magnetic field. These points are exactly midway between the 2 wires

You might be interested in
A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c
saw5 [17]

Answer:

The final velocity of the thrower is \bf{3.88~m/s} and the final velocity of the catcher is \bf{0.029~m/s}.

Explanation:

Given:

The mass of the thrower, m_{t} = 70~Kg.

The mass of the catcher, m_{c} = 55~Kg.

The mass of the ball, m_{b} = 0.0480~Kg.

Initial velocity of the thrower, v_{it} = 3.90~m/s

Final velocity of the ball, v_{fb} = 33.5~m/s

Initial velocity of the catcher, v_{ic} = 0~m/s

Consider that the final velocity of the thrower is v_{ft}. From the conservation of momentum,

&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s

Consider that the final velocity of the catcher is v_{fc}. From the conservation of momentum,

&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s

Thus, the final velocity of thrower is 3.88~m/s and that for the catcher is 0.029~m/s.

8 0
3 years ago
What material would you use to build countertops
babunello [35]

Answer:   Natural stone

Explanation:

6 0
2 years ago
Read 2 more answers
The speed at which a light aircraft can take off is 120 km/h. (A) What is the minimum constant acceleration required for the pla
kondor19780726 [428]

Answer:

A) a = 2.31[m/s^2]; B) t = 14.4 [s]

Explanation:

We can solve this problem using the kinematic equations, but firts we must identify the data:

Vf= final velocity = take off velocity = 120[km/h]

Vi= initial velocity = 0, because the plane starts to move from the rest.

dx= distance to run = 240 [m]

v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\  a=2.31[m/s^2]\\

To find the time we must use another kinematic equation.

v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]

7 0
3 years ago
If the man has initial velocity of 0.4 m/s, acceleration is 0.343 m/s^2, a time of 5.8 seconds, and experimental final velocity
agasfer [191]

Answer:

The answer is going to be C.

Explanation:

Trust me. Im an expert in physics

6 0
3 years ago
A paper clip that has a mass of 1.5 grams is thrown into the air and initially has a kinetic energy
Vsevolod [243]

Answer:

v = 4.2 \ m/s

Explanation:

Given data:

Mass of the paper clip, m = 1.5 \ g = 0.0015 \ kg

Kinetic energy, K = 0.013 \ \rm J

Let the velocity of the paper clip when it is thrown be <em>v</em>.

Thus,

K = \frac{1}{2}mv^{2}

0.013 = 0.5 \times 0.0015 \times v^{2}

\Rightarrow \ v = 4.16 \ m/s

v = 4.2 \ m/s.  (rounding to nearest tenth)

3 0
3 years ago
Other questions:
  • Explain how earths lithosphere and asthenosphere work together
    6·1 answer
  • Calculate the acceleration of a galloping horse going from 2 m/s to 12 m/s in 2 seconds.
    5·1 answer
  • Drivers should train themselves to scan ahead two seconds looking for ________ hazards and from 10-12 seconds down the road for
    11·1 answer
  • you are doing an experiment outside on a sunny day you find the temperature of some sand is 28 degress Celsius you also find the
    14·1 answer
  • Why gravitational force at the center og earth is zero
    11·2 answers
  • Two boats are operating in the same general area. who is responsible for avoiding a collision?
    13·2 answers
  • What type of electromagnetic waves do heat lamps give off?
    14·2 answers
  • How does communication relate to measuring? cite examples​
    7·1 answer
  • A 10kg box putting pushed with a force of 24 Newtons and a fraction of 12N.
    14·2 answers
  • What is the speed of light in a vacuum?
    9·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!