Texts science and humanities college jubail physics phre307...

Texts: SCIENCE AND HUMANITIES COLLEGE JUBAIL PHYSICS PHRE307-HW-2 A.U: 2022/2023 TERM 3 Exercise 1: A steady current I passes through a conducting wire in the form of a closed path composed of two half-circles with radius R and 2R, related by two straight segments AB and CD as shown in the following figure. Determine the magnetic field Bo at the center O as a function of I, R, and by applying the Biot-Savart law. Exercise 2: An infinite solenoid composed of N turns and containing many coils as given in the following figure. The internal radius of the solenoid is R and its external radius is R. By applying the Ampere's law and if you know that the current in each turn is I and that the external magnetic field is equal to zero: 1. Determine the magnetic field at a point M internal to the solenoid (r < R, with r representing the distance of the point M from the axis). What do you conclude? 2. Determine the magnetic field at a point M situated between the coils of the solenoid (R < r < R), with r representing the distance of the point M from the axis.

Transcript

00:01 Hello students, so in this question, according to vector form of ohm's law, the current density j will be equal to sigma into e, where sigma is the conductivity and e is the electric field intensity.

00:09 So from here, e will be equal to j divided by sigma.

00:12 So current density is equal to current flowing per unit area.

00:14 So this will be i divided by a into sigma.

00:16 So 1 by sigma, so reciprocal of conductivity is resistivity.

00:19 So this will be i into rho, that is our resistivity, divided by area, that will be pi r square.

00:23 So we can substitute the values, this will be equal to current is 1 .3 ampere into resistivity is given as 1 point, sorry it is 8 .17 into 10 raised to minus 3 divided by 3 .14 into r square.

00:40 So this will be equal to 3 .38 into 10 raised to minus 3 divided by r square.

00:46 So if you take r equal to 1 unit, then it will be, so we will get the electric field, e will be equal to 3 .38 into 10 raised to minus 3 newton per coulomb.

00:58 So next part, we have to find out the force experienced by a wire in a magnetic field.

01:03 So here, a current carrying a straight wire carrying conductor is placed on x axis.

01:08 So it is having a current i equal to 3 ampere, and the magnetic field is making an angle 30 degree with the x axis.

01:16 So this is the magnetic field, and it is making an angle 30 degree with the x axis.

01:21 So this is the y axis.

01:22 So we have to find out the force experienced by this current carrying conductor.

01:25 So that will be lorentz force, f is equal to ibl sin theta.

01:30 So we can substitute the values.

01:33 So this will be, so this will be equal to current, current is given as 3 ampere, that is 3 into magnetic field is given as 0 .02 tesla into length will be length of the wire is 3 into 10 raised to minus 3 meter into sin 30 degree.

01:48 So this will be equal to, the force experienced will be equal to 9 into 10 raised to minus 5 newton k cap.

01:56 So this is the force experienced by the current conductor.

02:00 So next part, so next part, we have to find out, so we have to find out the velocity of the proton.

02:09 So velocity v will be equal to qbr, qbr divided by m.

02:14 So we can substitute the values, the charge of a proton will be 1 .6 into 10 raised to minus 19 into magnetic field is given as 4000 tesla and the radius of the path is given as 21 divided by the mass of a proton is 1 .673 into 10 raised to minus 27.

02:33 So if we calculate, we will get the velocity of the proton in a magnetic field will be equal to 8 .03 into 10 raised to 12 meter per second.

02:44 Here, since the proton is entering perpendicular to magnetic field, it will undergo a cyclotron motion.

02:49 So this is the velocity of a particle undergoing cyclotron motion.

02:52 So that will be 8 .03 into 10 raised to 12 meter per second.

02:56 So in next part, next part, we have to find out, we have to find out the magnetic field at a point due to a current carrying circular wire.

03:05 So if you take this is a circular wire, it is carrying current i.

03:09 So we have to find out the magnetic field at a distance x outside the wire...

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