One long wire carries current 30.0 A to the left along the x...

One long wire carries current 30.0 A to the left along the $x$ axis. A second long wire carries current 50.0 A to the right along the line $(y=0.280 \mathrm{m}, z=0) .$ (a) Where in the plane of the two wires is the total magnetic field equal to zero? (b) A particle with a charge of $-2.00 \mu \mathrm{C}$ is moving with a velocity of 150$\hat{\mathrm{i}} \mathrm{Mm} / \mathrm{s}$ along the line $(y=0.100 \mathrm{m},$ warticle. (c) What If? A uniform electric field is applied to allow this particle to pass through this region undeflected. Calculate the required vector electric field.

Key Concepts

Magnetic Field from Straight Current-Carrying Wires

This concept involves understanding that a current-carrying wire produces a magnetic field in the surrounding space. The magnitude of the field at a distance from a long straight wire is given by a relationship derived using the Biot–Savart Law or Ampere’s Law, typically expressed as B = ??I/(2?r). The field circulates around the wire, and its direction can be determined using the right?hand rule.

Superposition Principle for Magnetic Fields

The superposition principle states that if more than one source of magnetic fields exists, the total magnetic field at any point is the vector sum of the individual fields produced by each current-carrying wire. This principle is key to solving problems where fields from multiple wires interact, as one must consider both their magnitudes and directions to find points of cancellation or reinforcement.

Right-Hand Rule

The right-hand rule is a mnemonic used to determine the direction of the magnetic field generated by a current-carrying conductor. By aligning the thumb of the right hand with the current direction, the curl of the fingers indicates the circular direction of the magnetic field lines around the wire. This concept is essential for correctly analyzing the directions of fields produced by different current flows.

Lorentz Force

The Lorentz force is the force experienced by a charged particle moving in electric and magnetic fields, expressed as F = q(E + v × B). In the context of magnetic fields generated by currents, the v × B component describes how the particle is deflected by the magnetic field. Understanding this concept helps in analyzing the motion of charged particles, particularly how their trajectories change under the influence of magnetic forces.

Equilibrium of Electric and Magnetic Forces

This concept pertains to the condition where a charged particle moves without deflection in regions with both electric and magnetic fields. It involves setting the electric force equal and opposite to the magnetic force, ensuring that the net force is zero. This is crucial in problems where an additional electric field is applied to counteract the magnetic force, allowing the particle to maintain a straight-line motion.

Transcript

00:09 Hello everyone in this problem.

00:12 There is a current carrying conductor along x -axis in which current is 30 ampere along negative x -axis.

00:19 There is another long conductor at the position 0 .28 meter on y -axis, z is 0 and x is 0 and it carries the current 50 ampere along positive x -axis.

00:34 There are three parts in the question.

00:36 In the first part we have to find the position in the plane of the wire position in the plane of the wire at which to net magnetic field your total magnetic field is to be zero in second part we have to calculate the force on the charge of two microculum moving with velocity 156 into 10 to the power 6 meter per second along positive x -axis at a position 0 .1 and 0.

01:38 And in the third part, we have to calculate electric field is to be applied so that charge remains undeflected from its paths.

02:06 Now b is start solving it one by one.

02:09 First b will solve a part.

02:15 Diagram again this is the first wire in which 50 ampere current is flowing this is the second wire this is origin in which 30 ampere current is flowing this is your y -axis this is origin this is your x -axis suppose we have taken any point p x distance below x -axis and we are considering a magnetic field at p point is to be now we will calculate it.

03:00 Magnetic field at p due to first by having the magnitude mu not i1 upon 2 pi x and direction if you find it will be outward that is k -camp.

03:25 Similarly magnetic field due to second by mu not i2 upon 2 pi 0 .28 plus x and it will be mild k cap and this value is given to 0 for us now simplify substitute the value so from this equation by calculation you will get x to be 0 .4 2 meter or you may write its coordinate to be minus answer you may write because it is 4 to meter at which magnetic field is to be 0 now we will solve the second part again we have to draw the diagram to understand.

04:44 This is the origin, this is the second wire having 50 mpere current and its coordinate is 0 .28 meter comma 0.

05:02 This is your y -axis.

05:05 This is first wire having the current 30 ampere and this is your xxis.

05:14 Here it is a charge q at this position.

05:19 It is moving along this direction and this distance is given 0 .1 meter so at the position of charge we require the magnetic field so here we will write the magnetic field it will be mu not due to this wire it will be mu not i upon 2 pi into 0 and direction you can find by right and thumbroon so it is outward so it will be plus k cap similarly mu not i2 2 pi 0 .18 this distance will be 0 .18 meter and direction will be k -cap now substitute the value you will get magnetic field at the position of the earth moving will be 4 pi into 10 to the power of minus 7 30 upon 2 pi into 0 .1 k cap 4 pi into 10 to the power of minus 7 2 pi 0 .18 50 k cap.

06:49 So on solving this will be 1 .16 into 10 to the power minus 4 k cap and unit will be tesla...

Please give Ace some feedback