What is the magnetic field at the position of the dot in FIGURE E...

Question

What is the magnetic field at the position of the dot in FIGURE EX29.6? Give your answer as a vector.

Accepted Answer

Hi, everyone. In this problem statement, we are being asked to determine what is the factor representation of the magnetic field at the location of the dot depicted in the given diagram, we were given a figure or a diagram where we have a velocity factor equals five times 10 to the power of 6 m per second. And a proton located four centimeters to the right and four centimeters to the bottom of a dot The options given for the magnetic field are a 3.54 times 10 to the power of negative 16 J Tesla B 1.77 times 10 to the power of negative 16 K tesla, C 1.41 times 10 to the power of negative 17 I tesla and D 7.07 times 10 to the power of negative 18 J tesla. So in this practice problem, we can use the Bio Saar Law where the formula for the magnetic field due to a moving charge is going to be given by B for the magnetic field equals to mu not divided by four pi multiplied by Q multiplied by V multiplied by sine of theta, all of that divided by R square just like. So in this case, B is the magnetic field factor mu knot is the permeability of free space where mu knot is just four pi multiplied by 10 to the power of negative seven sla meter divided by M. And then Q is the charge of the particle which in this case, it is given to be a proton where the charge of a proton is 1.6 times 10 to the power of negative 19. And then V is the velocity factor of the particle sine theta is the angle between the velocity and the factor to the point in question which in this problem statement, sine theta is going to be equal to sine of 45 degrees which is obtained by because we have the um dot and the proton separated by a 45 degree angle just like. So, so sine theta equals to sine 45 degrees. And R is going to be the position factor from the charge to the point of interest, which is going to be this uh distance right here just like so awesome. So now that we have determined each single variable here, let's actually substitute the values that we have. So first, we will have B to then be equal to um you not divided by four pi which is four pi times 10 to the power of negative seven tesla meter divided by a divided by four pi close parentheses multiplied by Q is 1.6 times 10 to the power of negative 19 coon multiplied that with the velocity factor which is five times 10 to the power of 6 m per second multiplied that again with sine of data which is sine of 45 degrees just like. So all of that is going to be divided by R squared, which in this case, R squared is going to be obtained from the Pythagorean theorem, which is um going to be the square or square root of four centimeter squared plus four centimeter squared squared just like. So, so the square and the square is going to cancel each other. So it's essentially just four centimeter squared plus four centimeter squared. So the next thing that we need to do is to convert the four centimeters, which in this case, we have to convert that into meter in order for us to obtain or in order for us to be able to calculate this. So then the B will then be equal to four pi can be canceled out. So then to the power of negative seven Tesla meter divided by M multiplied by 1.6 times 10 to the power of negative 19 Coolum multiplied by five times 10 to the power of m per second multiplied by a sign of 45 degrees. All of that is going to be divided by. So the square and the square root is going to cancel each other up. So now it will just be 0.04 m squared plus 0.04 m squared which is converting from centimeter to meter. So RB or the magnetic field is then going to be equal to 1.77 times 10 to the power of negative tesla. So next we want to apply your right hand rule to the positive charge at X equals to four centimeter. So that we get the magnetic field factor B to point out of the page. Therefore, B is going to be equal to 1.77 times 10 to the power of negative 17 K because it is pointing out of the page Tesla. So that the correct answer answer is going to be equal to option B which is going to be the answer to this particular practice problem. So that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our other lesson videos on similar topics available on our website. But other than that, that will be it for this one. Thank you.

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What is the magnetic field at the position of the dot in FIGURE EX29.6? Give your answer as a vector.

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