Multiple Choice Questions

Assume that loop B (in the Part A figure) has length L along k^ (the z direction). What is the loop integral in Ampère's law? Assume that the top end of the loop is very far from the solenoid (even though it may not look like it in the figure), so that the field there is assumed to be small and can be ignored.
  • Torque on the loop: IABsinϕ=I(pi*r^2)Bsinϕ
  • B⃗ E is into the page.
  • integral B*dl=BinL
  • out of the screen
Ampère's law can be used to find the magnetic field around a straight current-carrying wire.
  • True
  • False
Is the force between the wires attractive or repulsive?
  • Learning GoalTo understand the origin of the torque on a current loop due to the magnetic forces on the current-carrying wires.This problem will show you how to calculate the torque on a magnetic dipole in a uniform magnetic field. We start with a rectangular current loop, the shape of which allows us to calculate the electromagnetic forces explicitly. Then we generalize our result. Even if you already know the general formula to solve this problem, you might find it instructive to discover where it comes from.
  • The net current through the loopThe positive direction of the line integral and the positive direction for the current are related by the right-hand rule:Wrap your right-hand fingers around the closed path, then the direction of your fingers is the positive direction for dl⃗ and the direction of your thumb is the positive direction for the net current.Note also that the angle the current-carrying wire makes with the surface enclosed by the loop doesn't matter. (If the wire is at an angle, the normal component of the current is decreased, but the area of intersection of the wire and the surface is correspondingly increased.)
  • B=mu(knot)I/2(pi)dF=ILB--->F/L=ID=I(mu(knot)I/2(pi)d)--->mu(knot)I^2/2(pi)dFirst some relevant formulas:FB=qvBsin(θ) = qv→×B→ on a long straight wire,F=ILBsin(θ) = I→L×B→ where L is the length of the wireSo the magnetic field at wire 2 from the current in wire 1 will be B=μ0I1 ⋅ 1/[2πd]The force on a length ΔL of wire 2 will be F=ΔLI2×BThe force per unit length in terms of the currents will be,F/ΔL = μ0I1I2 ⋅ 1/[2πd]since I1 = I2,F/L = μ0I2 / [2πd]
  • Attractive: current flows in the same directionThe second right hand rule assists us in solving this problem. Since we have two parallel wires with equal current going the same direction, that vector represents each of the wires velocity vectors. By pointing our right thumb in this direction and using our pointer finger to represent the magnetic field (pointing back toward ourselves), at any point, the Force of the wire will be represented by our middle finger. Thinking of this for both wires, the force vectors are pointing towards each other, resulting in an attractive relationship.
Ampère's law can be used to find the magnetic field inside a toroid. (A toroid is a doughnut shape wound uniformly with many turns of wire.)
  • True
  • False
Force between Moving Charges
  • (Figure 1) Two point charges, with charges q1 and q2, are each moving with speed v toward the origin. At the instant shown q1 is at position (0, d) and q2 is at ( d, 0). (Note that the signs of the charges are not given because they are not needed to determine the magnitude of the forces between the charges.)
  • A loop of wire is in the shape of two concentric semicircles as shown. (Figure 1)The inner circle has radius a; the outer circle has radius b. A current I flows clockwise through the outer wire and counterclockwise through the inner wire.
  • Consider an infinite sheet of parallel wires. The sheet lies in the xy plane. A current I runs in the -y direction through each wire. There are N/a wires per unit length in the x direction. (Figure 1)
  • You are given two infinite, parallel wires each carrying current I. The wires are separated by a distance d, and the current in the two wires is flowing in the same direction. This problem concerns the force per unit length between the wires.
Write an expression for B(d), the magnetic field a distance d above the xy plane of the sheet.Use μ0 for the permeability of free space.
  • A toroid is a solenoid bent into the shape of a doughnut. It looks similar to a toy Slinky® with ends joined to make a circle. Consider a toroid consisting of N turns of a single wire with current I flowing through it. (Figure 1)Consider the toroid to be lying in the r θ plane of a cylindrical coordinate system, with the z axis along the axis of the toroid (pointing out of the screen). Let θ represent the angular position around the toroid, and let r be the distance from the axis of the toroid.For now, treat the toroid as ideal; that is, ignore the component of the current in the θ^ direction.
  • Consider only locations where the distance from the ends is many times D.
  • path must pass through r and must have symmetryB.
  • Current through one turn=ITotal current through n turns I' = I(n/a) LUsing Ampere's lawB 2L =μo I (n/a) LSo...B_vec(d) = -[(- μoIn/a)/2] i
What is the direction of the magnetic field BB at Point B?
  • B⃗ E is into the page.
  • BA is out of the page.
  • BD is out of the page.
  • BB is into the page.
Now find the magnetic field B⃗ in(r⃗ ) inside the wire (i.e., when the distance r is less than a). (Figure 2)
  • Find the magnetic field of q1 on q2.Magnitude of the magnetic field.The Biot-Savart law, which gives the magnetic field produced by a moving charge, can be written,where is the permeability of free space and is the vector from the charge to the point where the magnetic field is produced. Note we have in the numerator, not , necessitating an extra powerof in the denominator.Determine the cross productFind the direction of the magnetic field.Computing the force.F = q1q2v^2μ0/4π2√2d^2
  • along any closed path that you choose.
  • Current enclosed of wire of radius a:pi*a^2-->II'=(Ir^2)/a^2Circuital Law: Bin(2pi*r)=mu(knot)I'after some algebra:Bin=(mu(knot)Jr/2)theta hat
  • The magnetic field outside current carrying wires:Ampere's circuital law gives:integral of Bdl=mu(knot)I(enclosed)I=current enclosedmu=permiability B=strength of magnetic field.The current density: J=I/A---> I/(pi*a^2)I=J((pi/)a^2)Bout2(pi)r=mu(knot)IBout=mu(knot)I/(2(pi)r)Insert IBout=(mu(knot)Ja^2/2r)theta hat
What is the direction of the magnetic field B⃗ E at Point E?
  • BB is into the page.
  • B⃗ E is into the page.
  • BD is out of the page.
  • BA is out of the page.
What is the direction of the magnetic field at the center of the semicircles?
  • BD is out of the page.
  • BB is into the page.
  • out of the screen
  • B⃗ E is into the page.
What is the magnitude of the electric force between the two charges?
  • Learning GoalTo understand Ampère's law and its application.Ampère's law is often written ∮B (r )⋅dl =μ0Iencl.
  • A toroid is a solenoid bent into the shape of a doughnut. It looks similar to a toy Slinky® with ends joined to make a circle. Consider a toroid consisting of N turns of a single wire with current I flowing through it. (Figure 1)Consider the toroid to be lying in the r θ plane of a cylindrical coordinate system, with the z axis along the axis of the toroid (pointing out of the screen). Let θ represent the angular position around the toroid, and let r be the distance from the axis of the toroid.For now, treat the toroid as ideal; that is, ignore the component of the current in the θ^ direction.
  • Apply coulombs lawfind the value R^2F = q1q2/4πϵ02d2
  • Learning Goal:To apply Ampère's law to find the magnetic field inside an infinite solenoid.In this problem we will apply Ampère's law, written∮B⃗ (r⃗ )⋅dl⃗ =μ0Iencl,to calculate the magnetic field inside a very long solenoid (only a relatively short segment of the solenoid is shown in the pictures). The solenoid has length L, diameter D, and n turns per unit length with each carrying current I. (Figure 1) It is usual to assume that the component of the current along the z axis is negligible. (This may be assured by winding two layers of closely spaced wires that spiral in opposite directions.)From symmetry considerations it is possible to show that far from the ends of the solenoid, the magnetic field is axial.
The magnetic field inside the toroid varies as a function of which parameters?
  • Find the direction of the magnetic field at each of the indicated points.
  • along any closed path that you choose.
  • Evaluate the line integral and find IenclB(r) = μ0IN2πr
  • Consider only locations where the distance from the ends is many times D.
Current Sheet
  • You are given two infinite, parallel wires each carrying current I. The wires are separated by a distance d, and the current in the two wires is flowing in the same direction. This problem concerns the force per unit length between the wires.
  • The magnetic field due to the length of the wire: F=BIwSo torque=bFcos(θ) Which is: τ = BIwbcos(θ)
  • Current enclosed of wire of radius a:pi*a^2-->II'=(Ir^2)/a^2Circuital Law: Bin(2pi*r)=mu(knot)I'after some algebra:Bin=(mu(knot)Jr/2)theta hat
  • Consider an infinite sheet of parallel wires. The sheet lies in the xy plane. A current I runs in the -y direction through each wire. There are N/a wires per unit length in the x direction. (Figure 1)
First find the magnetic field, B out(r ), outside the wire (i.e., when the distance r is greater than a). (Figure 1)
  • The magnetic field outside current carrying wires:Ampere's circuital law gives:integral of Bdl=mu(knot)I(enclosed)I=current enclosedmu=permiability B=strength of magnetic field.The current density: J=I/A---> I/(pi*a^2)I=J((pi/)a^2)Bout2(pi)r=mu(knot)IBout=mu(knot)I/(2(pi)r)Insert IBout=(mu(knot)Ja^2/2r)theta hat
  • Determine which force has a greater magnitude by finding the ratio of the electric force to the magnetic force and then applying the approximation. The magnitude of the electric force is greater than the magnitude of the magnetic force.This result holds quite generally: Magnetic forces between moving charges are much smaller than electric forces as long as the speeds of the charges are nonrelativistic.
  • B=mu(knot)I/2(pi)dF=ILB--->F/L=ID=I(mu(knot)I/2(pi)d)--->mu(knot)I^2/2(pi)dFirst some relevant formulas:FB=qvBsin(θ) = qv→×B→ on a long straight wire,F=ILBsin(θ) = I→L×B→ where L is the length of the wireSo the magnetic field at wire 2 from the current in wire 1 will be B=μ0I1 ⋅ 1/[2πd]The force on a length ΔL of wire 2 will be F=ΔLI2×BThe force per unit length in terms of the currents will be,F/ΔL = μ0I1I2 ⋅ 1/[2πd]since I1 = I2,F/L = μ0I2 / [2πd]
  • The magnetic field due to the length of the wire: F=BIwSo torque=bFcos(θ) Which is: τ = BIwbcos(θ)
Which of the following conditions must hold to allow you to use Ampère's law to find a good approximation?
  • Find the direction of the magnetic field at each of the indicated points.
  • Consider only locations where the distance from the ends is many times D.
  • a and b
  • Evaluate the line integral and find IenclB(r) = μ0IN2πr
What physical property does the symbol Iencl represent?
  • Find the magnetic field of q1 on q2.Magnitude of the magnetic field.The Biot-Savart law, which gives the magnetic field produced by a moving charge, can be written,where is the permeability of free space and is the vector from the charge to the point where the magnetic field is produced. Note we have in the numerator, not , necessitating an extra powerof in the denominator.Determine the cross productFind the direction of the magnetic field.Computing the force.F = q1q2v^2μ0/4π2√2d^2
  • The net current through the Ampèrian pathThe positive direction of the line integral and the positive direction for the current are related by the right-hand rule:Wrap your right-hand fingers around the closed path, then the direction of your fingers is the positive direction for dl⃗ and the direction of your thumb is the positive direction for the net current.Note also that the angle the current-carrying wire makes with the surface enclosed by the loop doesn't matter. (If the wire is at an angle, the normal component of the current is decreased, but the area of intersection of the wire and the surface is correspondingly increased.)
  • B=mu(knot)I/2(pi)dF=ILB--->F/L=ID=I(mu(knot)I/2(pi)d)--->mu(knot)I^2/2(pi)dFirst some relevant formulas:FB=qvBsin(θ) = qv→×B→ on a long straight wire,F=ILBsin(θ) = I→L×B→ where L is the length of the wireSo the magnetic field at wire 2 from the current in wire 1 will be B=μ0I1 ⋅ 1/[2πd]The force on a length ΔL of wire 2 will be F=ΔLI2×BThe force per unit length in terms of the currents will be,F/ΔL = μ0I1I2 ⋅ 1/[2πd]since I1 = I2,F/L = μ0I2 / [2πd]
  • Attractive: current flows in the same directionThe second right hand rule assists us in solving this problem. Since we have two parallel wires with equal current going the same direction, that vector represents each of the wires velocity vectors. By pointing our right thumb in this direction and using our pointer finger to represent the magnetic field (pointing back toward ourselves), at any point, the Force of the wire will be represented by our middle finger. Thinking of this for both wires, the force vectors are pointing towards each other, resulting in an attractive relationship.
A current I flows around a plane circular loop of radius r, giving the loop a magnetic dipole moment of magnitude μ. The loop is placed in a uniform magnetic field B⃗ , with an angle ϕ between the direction of the field lines and the magnetic dipole moment as shown in the figure. (Figure 3) Find an expression for the magnitude of the torque τ on the current loop.
  • The magnetic field due to the length of the wire: F=BIwSo torque=bFcos(θ) Which is: τ = BIwbcos(θ)
  • Torque on the loop: IABsinϕ=I(pi*r^2)Bsinϕ
  • Evaluate the line integral and find IenclB(r) = μ0IN2πr
  • a and b
Magnetic Field of a Current-Carrying Wire
  • You are given two infinite, parallel wires each carrying current I. The wires are separated by a distance d, and the current in the two wires is flowing in the same direction. This problem concerns the force per unit length between the wires.
  • Calculate magnetic field using ampere's law:Iencl = InL
  • Current enclosed of wire of radius a:pi*a^2-->II'=(Ir^2)/a^2Circuital Law: Bin(2pi*r)=mu(knot)I'after some algebra:Bin=(mu(knot)Jr/2)theta hat
  • Find the magnetic field a distance r from the center of a long wire that has radius a and carries a uniform current per unit area j in the positive z direction.
Magnetic Field inside a Very Long Solenoid
  • Apply coulombs lawfind the value R^2F = q1q2/4πϵ02d2
  • Learning Goal:To apply Ampère's law to find the magnetic field inside an infinite solenoid.In this problem we will apply Ampère's law, written∮B⃗ (r⃗ )⋅dl⃗ =μ0Iencl,to calculate the magnetic field inside a very long solenoid (only a relatively short segment of the solenoid is shown in the pictures). The solenoid has length L, diameter D, and n turns per unit length with each carrying current I. (Figure 1) It is usual to assume that the component of the current along the z axis is negligible. (This may be assured by winding two layers of closely spaced wires that spiral in opposite directions.)From symmetry considerations it is possible to show that far from the ends of the solenoid, the magnetic field is axial.
  • A toroid is a solenoid bent into the shape of a doughnut. It looks similar to a toy Slinky® with ends joined to make a circle. Consider a toroid consisting of N turns of a single wire with current I flowing through it. (Figure 1)Consider the toroid to be lying in the r θ plane of a cylindrical coordinate system, with the z axis along the axis of the toroid (pointing out of the screen). Let θ represent the angular position around the toroid, and let r be the distance from the axis of the toroid.For now, treat the toroid as ideal; that is, ignore the component of the current in the θ^ direction.
  • Learning GoalTo understand Ampère's law and its application.Ampère's law is often written ∮B (r )⋅dl =μ0Iencl.
What is Bout, the z component of the magnetic field outside the solenoid?
  • BB is into the page.
  • For thesemicirle,Themagnetic field at the point P due to the inner wire of radius 'a'is Ba = (1/2)(μ0I / 2a)Themagnetic field at the point P due to the inner wire of radius 'b'is Bb = (1/2)(μ0I / 2b)Now thedirections of these two magnetic field's are in the oppositedirections, as a result the resulatnat magnetic field at the pointP is B = Ba - Bb =(1/2)(μ0I / 2a) - (1/2)(μ0I /2b)
  • 0
  • B⃗ E is into the page.
The key point is that to be able to use Ampère's law, the path along which you take the line integral of B⃗ must have sufficient symmetry to allow you to pull the magnitude of B outside the integral. Whether the current distribution has symmetry is incidental.
  • True
  • False
Torque on a Current Loop in a Magnetic Field
  • Attractive: current flows in the same directionThe second right hand rule assists us in solving this problem. Since we have two parallel wires with equal current going the same direction, that vector represents each of the wires velocity vectors. By pointing our right thumb in this direction and using our pointer finger to represent the magnetic field (pointing back toward ourselves), at any point, the Force of the wire will be represented by our middle finger. Thinking of this for both wires, the force vectors are pointing towards each other, resulting in an attractive relationship.
  • You are given two infinite, parallel wires each carrying current I. The wires are separated by a distance d, and the current in the two wires is flowing in the same direction. This problem concerns the force per unit length between the wires.
  • Learning GoalTo understand the origin of the torque on a current loop due to the magnetic forces on the current-carrying wires.This problem will show you how to calculate the torque on a magnetic dipole in a uniform magnetic field. We start with a rectangular current loop, the shape of which allows us to calculate the electromagnetic forces explicitly. Then we generalize our result. Even if you already know the general formula to solve this problem, you might find it instructive to discover where it comes from.
  • B=mu(knot)I/2(pi)dF=ILB--->F/L=ID=I(mu(knot)I/2(pi)d)--->mu(knot)I^2/2(pi)dFirst some relevant formulas:FB=qvBsin(θ) = qv→×B→ on a long straight wire,F=ILBsin(θ) = I→L×B→ where L is the length of the wireSo the magnetic field at wire 2 from the current in wire 1 will be B=μ0I1 ⋅ 1/[2πd]The force on a length ΔL of wire 2 will be F=ΔLI2×BThe force per unit length in terms of the currents will be,F/ΔL = μ0I1I2 ⋅ 1/[2πd]since I1 = I2,F/L = μ0I2 / [2πd]
A current I flows in a plane rectangular current loop with height w and horizontal sides b. The loop is placed into a uniform magnetic field B⃗ in such a way that the sides of length w are perpendicular to B⃗ (Figure 1) , and there is an angle θ between the sides of length b and B⃗ (Figure 2) .Calculate τ, the magnitude of the torque about the vertical axis of the current loop due to the interaction of the current through the loop with the magnetic field.
  • the line integral along the (magnetic field along a) closed loop
  • Torque on the loop: IABsinϕ=I(pi*r^2)Bsinϕ
  • Find the magnetic field a distance r from the center of a long wire that has radius a and carries a uniform current per unit area j in the positive z direction.
  • The magnetic field due to the length of the wire: F=BIwSo torque=bFcos(θ) Which is: τ = BIwbcos(θ)
What is the magnitude of the magnetic force on q2 due to the magnetic field caused by q1?
  • Find the magnetic field of q1 on q2.Magnitude of the magnetic field.The Biot-Savart law, which gives the magnetic field produced by a moving charge, can be written,where is the permeability of free space and is the vector from the charge to the point where the magnetic field is produced. Note we have in the numerator, not , necessitating an extra powerof in the denominator.Determine the cross productFind the direction of the magnetic field.Computing the force.F = q1q2v^2μ0/4π2√2d^2
  • The net current through the Ampèrian pathThe positive direction of the line integral and the positive direction for the current are related by the right-hand rule:Wrap your right-hand fingers around the closed path, then the direction of your fingers is the positive direction for dl⃗ and the direction of your thumb is the positive direction for the net current.Note also that the angle the current-carrying wire makes with the surface enclosed by the loop doesn't matter. (If the wire is at an angle, the normal component of the current is decreased, but the area of intersection of the wire and the surface is correspondingly increased.)
  • B=mu(knot)I/2(pi)dF=ILB--->F/L=ID=I(mu(knot)I/2(pi)d)--->mu(knot)I^2/2(pi)dFirst some relevant formulas:FB=qvBsin(θ) = qv→×B→ on a long straight wire,F=ILBsin(θ) = I→L×B→ where L is the length of the wireSo the magnetic field at wire 2 from the current in wire 1 will be B=μ0I1 ⋅ 1/[2πd]The force on a length ΔL of wire 2 will be F=ΔLI2×BThe force per unit length in terms of the currents will be,F/ΔL = μ0I1I2 ⋅ 1/[2πd]since I1 = I2,F/L = μ0I2 / [2πd]
  • out of the screen
In the SI system, the unit of current, the ampere, is defined by this relationship using an apparatus called an Ampère balance. What would be the force per unit length of two infinitely long wires, separated by a distance 1m, if 1A of current were flowing through each of them?
  • If I = 1 A, & d = 1 m, plugging these values into the formula from the previous section,F/L = μ0(1 A)2 / [2π(1m)]= μ0/ [2π] ⋅ A2/mμ0 = 4π×10-7 ⋅ T⋅m/A→ 4π×10-7 / [2π] ⋅A2/m ⋅T⋅m/A → 2×10-7 ⋅ T⋅A1 T = 1 N / [A⋅m] →2×10e-7 ⋅ N/m
  • True
  • A toroid is a solenoid bent into the shape of a doughnut. It looks similar to a toy Slinky® with ends joined to make a circle. Consider a toroid consisting of N turns of a single wire with current I flowing through it. (Figure 1)Consider the toroid to be lying in the r θ plane of a cylindrical coordinate system, with the z axis along the axis of the toroid (pointing out of the screen). Let θ represent the angular position around the toroid, and let r be the distance from the axis of the toroid.For now, treat the toroid as ideal; that is, ignore the component of the current in the θ^ direction.
  • Current through one turn=ITotal current through n turns I' = I(n/a) LUsing Ampere's lawB 2L =μo I (n/a) LSo...B_vec(d) = -[(- μoIn/a)/2] i
What is the direction of the magnetic field BA at Point A?
  • BB is into the page.
  • BD is out of the page.
  • B⃗ E is into the page.
  • BA is out of the page.
Note that the bottom wire carries a current of magnitude 2I.What is the direction of the magnetic field B⃗ C at Point C?
  • BA is out of the page.
  • BC is out of the page.
  • BD is out of the page.
  • Bin = μ0nI
What is the force per unit length F/L between the two wires?
  • BA is out of the page.
  • Find the magnetic field of q1 on q2.Magnitude of the magnetic field.The Biot-Savart law, which gives the magnetic field produced by a moving charge, can be written,where is the permeability of free space and is the vector from the charge to the point where the magnetic field is produced. Note we have in the numerator, not , necessitating an extra powerof in the denominator.Determine the cross productFind the direction of the magnetic field.Computing the force.F = q1q2v^2μ0/4π2√2d^2
  • BB is into the page.
  • B=mu(knot)I/2(pi)dF=ILB--->F/L=ID=I(mu(knot)I/2(pi)d)--->mu(knot)I^2/2(pi)dFirst some relevant formulas:FB=qvBsin(θ) = qv→×B→ on a long straight wire,F=ILBsin(θ) = I→L×B→ where L is the length of the wireSo the magnetic field at wire 2 from the current in wire 1 will be B=μ0I1 ⋅ 1/[2πd]The force on a length ΔL of wire 2 will be F=ΔLI2×BThe force per unit length in terms of the currents will be,F/ΔL = μ0I1I2 ⋅ 1/[2πd]since I1 = I2,F/L = μ0I2 / [2πd]
Ampère's Law Explained
  • If I = 1 A, & d = 1 m, plugging these values into the formula from the previous section,F/L = μ0(1 A)2 / [2π(1m)]= μ0/ [2π] ⋅ A2/mμ0 = 4π×10-7 ⋅ T⋅m/A→ 4π×10-7 / [2π] ⋅A2/m ⋅T⋅m/A → 2×10-7 ⋅ T⋅A1 T = 1 N / [A⋅m] →2×10e-7 ⋅ N/m
  • Apply coulombs lawfind the value R^2F = q1q2/4πϵ02d2
  • Learning GoalTo understand Ampère's law and its application.Ampère's law is often written ∮B (r )⋅dl =μ0Iencl.
  • Learning Goal:To apply Ampère's law to find the magnetic field inside an infinite solenoid.In this problem we will apply Ampère's law, written∮B⃗ (r⃗ )⋅dl⃗ =μ0Iencl,to calculate the magnetic field inside a very long solenoid (only a relatively short segment of the solenoid is shown in the pictures). The solenoid has length L, diameter D, and n turns per unit length with each carrying current I. (Figure 1) It is usual to assume that the component of the current along the z axis is negligible. (This may be assured by winding two layers of closely spaced wires that spiral in opposite directions.)From symmetry considerations it is possible to show that far from the ends of the solenoid, the magnetic field is axial.
Magnetic Field due to Semicircular Wires
  • Consider an infinite sheet of parallel wires. The sheet lies in the xy plane. A current I runs in the -y direction through each wire. There are N/a wires per unit length in the x direction. (Figure 1)
  • You are given two infinite, parallel wires each carrying current I. The wires are separated by a distance d, and the current in the two wires is flowing in the same direction. This problem concerns the force per unit length between the wires.
  • Find the magnetic field a distance r from the center of a long wire that has radius a and carries a uniform current per unit area j in the positive z direction.
  • A loop of wire is in the shape of two concentric semicircles as shown. (Figure 1)The inner circle has radius a; the outer circle has radius b. A current I flows clockwise through the outer wire and counterclockwise through the inner wire.
The integral on the left is
  • Find the direction of the magnetic field at each of the indicated points.
  • Calculate magnetic field using ampere's law:Iencl = InL
  • Evaluate the line integral and find IenclB(r) = μ0IN2πr
  • the line integral along the (magnetic field along a) closed loop
The circle on the integral means that B (r ) must be integrated
  • BD is out of the page.
  • Calculate magnetic field using ampere's law:Iencl = InL
  • Evaluate the line integral and find IenclB(r) = μ0IN2πr
  • along any closed path that you choose.
Assuming that the charges are moving nonrelativistically (v≪c), what can you say about the relationship between the magnitudes of the magnetic and electrostatic forces?
  • Determine which force has a greater magnitude by finding the ratio of the electric force to the magnetic force and then applying the approximation. The magnitude of the electric force is greater than the magnitude of the magnetic force.This result holds quite generally: Magnetic forces between moving charges are much smaller than electric forces as long as the speeds of the charges are nonrelativistic.
  • For thesemicirle,Themagnetic field at the point P due to the inner wire of radius 'a'is Ba = (1/2)(μ0I / 2a)Themagnetic field at the point P due to the inner wire of radius 'b'is Bb = (1/2)(μ0I / 2b)Now thedirections of these two magnetic field's are in the oppositedirections, as a result the resulatnat magnetic field at the pointP is B = Ba - Bb =(1/2)(μ0I / 2a) - (1/2)(μ0I /2b)
  • simplify the problemB =μ0I2RThis is the same expression that you would derive for the magnetic field at the center of a circular loop of current-carrying wire. To see why this makes sense, imagine that the local diameter d of the coils gets so small that it is negligible in comparison to the radius of the toroid. The wire makes one complete turn around the axis of the toroid. So, to a point in the center, the toroid looks like a simple current loop.
  • Find the magnetic field of q1 on q2.Magnitude of the magnetic field.The Biot-Savart law, which gives the magnetic field produced by a moving charge, can be written,where is the permeability of free space and is the vector from the charge to the point where the magnetic field is produced. Note we have in the numerator, not , necessitating an extra powerof in the denominator.Determine the cross productFind the direction of the magnetic field.Computing the force.F = q1q2v^2μ0/4π2√2d^2
0 h : 0 m : 1 s

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