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A proton moves in the magnetic field B=0.50^Twith a speed of 1.0107m/sin the directions shown in FIGURE. For each, what is magnetic force Fon the proton? Give your answers in component form.

Short Answer

Expert verified

(a) The magnetic force on the proton=0.5610-12j^N=0.5610-12j^N

(b) The magnetic force on the proton =810-13k^N=81013k^N

Step by step solution

01

Introduction

Amp猫re's invention is essentially a long, protracted wire that transmits an electrical current. A little low compass needle prints out a sequence of concentric circular loops within the plane perpendicular to an electrical conductor wire, Amp猫re expertise in the areas.

02

Find the rate vector

The moving charge is polarized by the field, which is depending on the speed vector. The flux increases as the angle between the speed and the magnetic field increases. Equation gives the magnetic attraction.

F=qvB

Where qis the charge. The angle between the velocity and the magnetic field is 45in the xzplane

v=1107m/scos45i^+1107m/ssin45k^

The field of force has component B=0.50i^T, so we plug the values for B,qand vinto equation to induce Fwhere (i^i^)=0

F=qvB

=1.610-19C1107m/scos45i^+1107m/ssin45k^(0.50i^T)

=1.610-19C0.5107sin45(k^i^))

=0.5610-12j^N

03

find a field of force that is travelling in the opposite direction

The angle between the rate and also the field is 90but its direction is within the negative y-direction so therefore the velocity vector is

v=-1107m/sj^

The magnetic flux has component B=0.50i^T, so we plug the values for B,qand vinto equation to induceFwhere (j^i^)=-k^

F=qvB

=1.61019C1107m/sj^(0.50i^T)]

=1.610-19C-0.5107(j^i^)]

=810-13k^N

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