Please help with PART C, D and E BR Review I ConstantsThe direction of the magnetic force is perpendicular to both the direction of the current flow and the direction of the magnetic field. Here is a "right-hand rule" to help you determine thedirection of the magnetic force. (Eigure 2)Learning Goal:To understand the magnetic force on a straight current-carrying wvire in auniform magnetic field.1. Straighten the fingers of your nght hand and point them in the direction of the current.2. Rotate your arm until you can bend your fingers to point in the direction of the magnetic field.3. Your thumb now points in the direction of the magnetic force acting on the wire.Magnetic fields exert forces on moving charged particles, whether thosecharges are moving independently or are confined to a current-carryingwire. The magnetic force F on an individual moving charged particledepends on its velocity v and charge q. In the case of a current-carryingwire, many charged particles are simultaneously in motion, so the magneticforce depends on the total current I and the length of the wire L.Part CWhat is the direction of the magnetic force acting on the wire in Part B due to the applied magnetic field?The size of the magnetic force on a straight wire of length L carrying currentI in a uniform magnetic field with strength B isdue northF = ILB sin(0).due southHere 0 is the angle between the direction of the current (along the wire) andthe direction of the magnetic field. Hence B sin(0) refers to the componentof the magnetic field that is perpendicular to the wire, B . Thus thisequation can also be written as F = ILB.The direction of the magnetic force on the wire can be described using a"right-hand rule. " This will be discussed after Part B.due eastdue weststraight upstraight downSubmitRequest AnswerFigure2 of 2Part DNow let us assume that we are able to reorientate the wire in (Figure 1) and therefore the direction of current flow.Which of the following situations would result in a magnetic force on the wire that points due north?BCheck all that apply.Current in the wire flows straight down; the magnetic field points due westCurrent in the wire flows straight up. the magnetic field points due east.Current in the wire flows due east; the magnetic field points straight down.Current in the wire flows due west and slightly up. the magnetic field points due eastCurrent in the wire flows due west and slightly down; the magnetic field points straight down.PearsonCopyright © 2022 Pearson Education Inc. All rights reserved. Terms of Use I Privacy Policy I Permissions I Contact Us I O OA Keview I ConstantSPart DLearning Goal:To understand the magnetic force on a straight current-carrying wire in auniform magnetic field.Now let us assume that we are able to reorientate the vire in (Figure 1) and therefore the direction of current flow.Magnetic fields exert forces on moving charged particles, whether thosecharges are moving independently or are confined to a current-carryingwire. The magnetic force F on an individual moving charged particledepends on its velocity v and charge q. In the case of a current-carryingwire, many charged particles are simultane ously in motion, so the magneticforce depends on the total current I and the length of the wire L.Which of the following situations would result in a magnetic force on the wire that points due north?Check all that apply.Current in the wire flows straight dowe the magnetic field points due west.The size of the magnetic force on a straight wire of length L carrying currentI in a uniform magnetic field with strength B isCurrent in the wire flows straight up. the magnetic field points due east.Qurrent in the wire flows due east the magnetic field points straight down.F = ILB sin(0).Current in the wire flows due west and slightly up. the magneticNield points due east.Here 0 is the angle between the direction of the current (along the wire) andthe direction of the magnetic field. Hence B sin(6) refers to the componentof the magnetic field that is perpendicular to the wire, Bị. Thus thisequation can also be written as F = ILB.The direction of the magnetic force on the wire can be described using a"right-hand rule." This will be discussed after Part B.Current in the wire flows due west and slightly down; the magnetic fieldQoints straight down.SubmitRequest AnswerPart EAssume that the applied magnetic field of size 0.55 T in (Figure 1) is rotated so that it points horizontaily due south What is the size of the magnetic force on the wire due to theapplied magnetic field now?Figure1 of 2Express your answer in newtons to two significant figures.• View Available Hint(s)ΑΣφ?UpNW +SubmitSDownProvide FeedbackNext P Pearson

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Please help with PART C, D and E BR Review I ConstantsThe direction of the magnetic force is perpendicular to both the direction of the current flow and the direction of the magnetic field. Here is a &#34;right-hand rule&#34; to help you determine thedirection of the magnetic force. (Eigure 2)Learning Goal:To understand the magnetic force on a straight current-carrying wvire in auniform magnetic field.1. Straighten the fingers of your nght hand and point them in the direction of the current.2. Rotate your arm until you can bend your fingers to point in the direction of the magnetic field.3. Your thumb now points in the direction of the magnetic force acting on the wire.Magnetic fields exert forces on moving charged particles, whether thosecharges are moving independently or are confined to a current-carryingwire. The magnetic force F on an individual moving charged particledepends on its velocity v and charge q. In the case of a current-carryingwire, many charged particles are simultaneously in motion, so the magneticforce depends on the total current I and the length of the wire L.Part CWhat is the direction of the magnetic force acting on the wire in Part B due to the applied magnetic field?The size of the magnetic force on a straight wire of length L carrying currentI in a uniform magnetic field with strength B isdue northF = ILB sin(0).due southHere 0 is the angle between the direction of the current (along the wire) andthe direction of the magnetic field. Hence B sin(0) refers to the componentof the magnetic field that is perpendicular to the wire, B . Thus thisequation can also be written as F = ILB.The direction of the magnetic force on the wire can be described using a&#34;right-hand rule. &#34; This will be discussed after Part B.due eastdue weststraight upstraight downSubmitRequest AnswerFigure2 of 2Part DNow let us assume that we are able to reorientate the wire in (Figure 1) and therefore the direction of current flow.Which of the following situations would result in a magnetic force on the wire that points due north?BCheck all that apply.Current in the wire flows straight down; the magnetic field points due westCurrent in the wire flows straight up. the magnetic field points due east.Current in the wire flows due east; the magnetic field points straight down.Current in the wire flows due west and slightly up. the magnetic field points due eastCurrent in the wire flows due west and slightly down; the magnetic field points straight down.PearsonCopyright © 2022 Pearson Education Inc. All rights reserved. Terms of Use I Privacy Policy I Permissions I Contact Us I O OA Keview I ConstantSPart DLearning Goal:To understand the magnetic force on a straight current-carrying wire in auniform magnetic field.Now let us assume that we are able to reorientate the vire in (Figure 1) and therefore the direction of current flow.Magnetic fields exert forces on moving charged particles, whether thosecharges are moving independently or are confined to a current-carryingwire. The magnetic force F on an individual moving charged particledepends on its velocity v and charge q. In the case of a current-carryingwire, many charged particles are simultane ously in motion, so the magneticforce depends on the total current I and the length of the wire L.Which of the following situations would result in a magnetic force on the wire that points due north?Check all that apply.Current in the wire flows straight dowe the magnetic field points due west.The size of the magnetic force on a straight wire of length L carrying currentI in a uniform magnetic field with strength B isCurrent in the wire flows straight up. the magnetic field points due east.Qurrent in the wire flows due east the magnetic field points straight down.F = ILB sin(0).Current in the wire flows due west and slightly up. the magneticNield points due east.Here 0 is the angle between the direction of the current (along the wire) andthe direction of the magnetic field. Hence B sin(6) refers to the componentof the magnetic field that is perpendicular to the wire, Bị. Thus thisequation can also be written as F = ILB.The direction of the magnetic force on the wire can be described using a&#34;right-hand rule.&#34; This will be discussed after Part B.Current in the wire flows due west and slightly down; the magnetic fieldQoints straight down.SubmitRequest AnswerPart EAssume that the applied magnetic field of size 0.55 T in (Figure 1) is rotated so that it points horizontaily due south What is the size of the magnetic force on the wire due to theapplied magnetic field now?Figure1 of 2>Express your answer in newtons to two significant figures.• View Available Hint(s)ΑΣφ?UpNW +SubmitSDownProvide FeedbackNext >P Pearson

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In this question we have to find the direction of magnetic force.
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