AP Physics C- Answer to Free-Response Question Involving Magnetic Field

A free response question involving magnetic field was posted on 7^th January 2008 for your practice. This was the question:

A rectangular, plane, single turn coil ABCD of length 2ℓ and breadth ℓ is arranged as shown, with the plane of the coil in the XY plane. O is the origin of coordinates and the side AB of the coil is along the X-axis OX such that the X-coordinate of the side DA is 2ℓ. A Magnetic field acing in the negative Z-direction and of magnitude B = K/x where K is a constant and x is the x-coordinate, acts in the entire region of the coil.

Now, answer the following questions:

(a) Calculate the magnetic flux linked with the coil in terms of K, x and ℓ.

(b) Calculate the change of magnetic flux through the coil when the coil is rotated through 180º about the side AB.

(c) The coil is moved in the positive X-direction, maintaining its plane in the XY plane. What is the direction of the current induced in the coil? (Put a tick mark against the correct direction)

Clockwise------

Anticlockwise------

Give reason for your answer.

(d) If the coil is moved in the positive Y-direction, maintaining its orientation in the XY plane, what is the emf induced in the coil? Justify your answer.

(e) How will you produce a magnetic field (in the region of the coil) of the type given in this question? (You must consider the magnitude as well as the direction of the magnetic field).

As promised, I give the answer below:

(a) Consider a strip of the area (of the coil) parallel to the side DA at distance x from the Y-axis (fig.).If the width dx of this strip is small, the magnetic field over the entire strip can be assumed constant (equal to K/x). The magnetic flux through the strip is (K/x) ℓdx, since the area of the strip is ℓdx and the magnetic field is directed perpendicular to the plane of the strip.

The total magnetic flux (Ф) through the entire coil is obtained by integrating the above flux between the limits 2ℓ and 4ℓ.

Ф = _2ℓ ∫^4ℓ (K/x) ℓdx = Kℓ [_2ℓ∫^4ℓ (dx/x)] = Kℓ[ln(4ℓ) – ln(2ℓ)]

Or, Ф = Kℓ[ln(4ℓ/2ℓ)] = Kℓ ln2

(b) When the coil is rotated through 180º, the magnetic flux linked with the coil changes from Ф to – Ф so that the change of flux is 2Kℓln2.

(c) When the coil is moved in the positive X-direction, maintaining its plane in the XY plane, the flux through the coil decreases. Lenz’s law demands that this decrease of flux is to be opposed. Therefore, a current is induced in the coil so as to produce a magnetic field in the negative Z-direction, thereby trying to increase the flux. Applying the right hand palm rule, we find that the induced current must be clockwise.

(d) If the coil is moved in the positive Y-direction, maintaining its orientation in the XY plane, the flux through the coil is the same in all positions. The change of flux and hence the induced emf is therefore zero.

(e) The magnetic field given in this question is of the form B = μ₀I/2πr, which is the field produced at distance ‘r’ by a long straight conductor carrying a current I. The straight conductor must be placed along the Y-axis and the current in it must flow in the positive Y- direction.

You will find some useful posts (MCQ) in this section here

AP Physics C- Free Response Question Involving Magnetic Field

The following free response practice question is meant for those preparing for AP Physics C exam:

A rectangular, plane, single turn coil ABCD of length 2ℓ and breadth ℓ is arranged as shown, with the plane of the coil in the XY-plane. O is the origin of coordinates and the side AB of the coil is along the X-axis OX such that the X-coordinate of the side DA is 2ℓ. A Magnetic field acing in the negative Z-direction and of magnitude B = K/x where K is a constant and x is the x-coordinate, acts in the entire region of the coil.

Now, answer the following questions:

(a) Calculate the magnetic flux linked with the coil in terms of K, x and ℓ.

(b) Calculate the change of magnetic flux through the coil when the coil is rotated through 180º about the side AB.

(c) The coil is moved in the positive X-direction, maintaining its plane in the XY plane. What is the direction of the current induced in the coil? (Put a tick mark against the correct direction)

Clockwise------

Anticlockwise------

Give reason for your answer.

(d) If the coil is moved in the positive Y-direction, maintaining its orientation in the XY plane, what is the emf induced in the coil? Justify your answer.

(e) How will you produce a magnetic field (in the region of the coil) of the type given in this question? (You must consider the magnitude as well as the direction of the magnetic field).

The above question carries 15 points and you will get 15 minutes to answer it. The distribution of points may be as 5+2+3+2+3 for parts a,b,c,d and e respectively.

Take it as an exercise and try to answer it within the stipulated time. I’ll be back with the answer shortly.

AP Physics B & C – Multiple Choice Questions involving Magnetic fields

The following questions will be of common interest to students preparing for AP Physics B as well as AP Physics C examinations:

(1) The magnetic field produced at a point P by a long, thick, straight cylindrical copper wire of radius R carrying a steady current I is plotted against the distance of the point P from the centre of the wire. Which graph among the five options shown in the figure represents the correct plot?

Normally, you will consider the magnetic field due to thin straight conductors and you know that the field outside at distance r is given by

B =µ₀I/2πr

If you remember that the field is directly proportional to the current I and inversely proportional to the distance r, you can solve the above problem.

Since the field is inversely proportional to the distance, if the current is constant, the graph must be a hyperbola, as is the case for points outside the wire. In the case of points within the wire ( where r is less than the radius R)

i = I(πr²/ πR²) = I r²/R²

[We have assumed that the current is uniformly distributed across the section of the wire. In the case of any point, we need consider the contribution due to the current passing through a circular section of radius r only, remembering that the magnetic field within a current carrying pipe is zero everywhere. Try and prove it!]

In the case of points within the wire, the magnetic field is obtained by replacing the constant current I with the variable current i in the equation, B =µ₀I/2πr.

Thus, B = µ₀ (I r²/R²) /2πr = µ₀I r/ 2πR²

The field within the wire is thus directly proportional to the distance r.

The correct plot is shown in figure (e).

(2) A free rectangular current carrying loop ABCD is placed as shown, near a long straight conductor PQ carrying a current I. The plane of the loop is the same as the plane containing the straight conductor and two sides of the loop are parallel to the straight conductor. The loop will

(a) move towards the straight conductor

(b) move away from the straight conductor

(c) rotate clockwise

(d) rotate anticlockwise

(e) have translational as well as rotational motion

Since PQ and DA carry like currents (currents in the same direction), they attract each other. You can easily show this by Fleming’s left hand rule (motor rule).. Since PQ and CB carry unlike currents (currents in opposite directions), they repel each other. The attractive force is greater than the repulsive force in this case since the separation between PQ and DA is greater than that between PQ and CB. [Remember that the force is directly proportional to the product of currents and inversely proportional to the separation]. The loop therefore will move towards the straight conductor.

At the loop, the magnetic field produced by PQ is directed perpendicularly into the plane of the figure. Therefore, the magnetic force on side AB is upwards and that on side CD is downwards. Since the magnitude is the same, the resultant of these two forces is zero The lines of action of these two forces are not separated . So, there is no torque. The lines of action of the forces on BC and AD also are not separated and again there is no torque.

The correct option therefore is (a)

(3) Two negative ions of the same charge but different masses are projected vertically upwards with the same velocity from point O (fig.) in a region of space where a uniform magnetic field directed perpendicularly in to the plane of the figure exists. If the gravitational pull is ignored, the trajectory of the heavier ion is

(a) OA

(b) OB

(c) OC

(d) OD

(e) OE

This is a question in which Flemings left hand rule (motor rule) will help you. It is better that you treat the ions to be positive (for the time being) so that you can hold the middle finger of your left hand along the direction of projection (upwards). The forefinger should indicate the direction of the magnetic field and hence must point into the plane of the figure. Then the direction of the thumb which is held perpendicular to the fore finger as well as the middle finger gives the direction of the magnetic force. The direction is leftwards for the positive ion. Since the ions in the question are negative, The deflection is towards right. Since the ions are of the same charge and the same velocity, the heavier ion is deflected less (because of greater inertia). Therefore its path is OD.

[If you remember the expression for radius r, which is

r = mv/qB, you will immediately note that r is greater for the larger mass m.

(4) In a mass spectrograph, positive ions moving with different velocities in the positive X-direction are admitted into a velocity filter in which an electric field of magnitude E is applied in the negative Y-direction. What should be the direction of the magnetic field so that ions of the same velocity pass through the velocity filter without deflection?

(a) Positive Z-direction

(b) Positive Y-direction

(c) Positive X-direction

(d) Negative Y-direction

(e) Negative Z-direction

As you might be knowing, the velocity filter is an arrangement in which electric and magnetic fields are applied at right angles so that the deflection produced by the electric field is cancelled by the deflection produced by the magnetic field. In other words, the electric force and the magnetic force on the ion are equal and opposite.

Since the electric field is in the negative Y-direction, the electric force (on the positive ion) is in the negative Y-direction. The magnetic force must therefore be in the positive Y-direction.

Now, to apply Fleming’s left hand rule, hold the thumb, fore finger and middle finger of your left hand in mutually perpendicular directions such that the middle finger is in the positive X-direction (to indicate the velocity of positive charge) and the thumb is in the positive Y-direction (to indicate the required direction of the magnetic force). The fore finger will then point along the required direction of the magnetic field. You will get it along the negative Z-direction

(5) In the above question, if the electric field is E, what should be the magnitude of the magnetic field so that ions of velocity v and charge q proceed undeflected through the velocity filter?

(a) q/E

(b) Eq/v

(c) E/v

(d) v/E

(e) vq/E

The electric force and the magnetic force on the ion must be equal and opposite. Equating the magnitudes, we have

qE = qvB

Therefore, B = E/v

You can expect more questions in this section in due course.

Meanwhile, find some useful posts in this section at http://www.physicsplus.in

AP Physics B & C - Magnetic Field - Equations to be Remembered

The section ‘Magnetic Fields’ in the AP Physics syllabus contains the following sub sections:

(1) Forces on moving charges in magnetic fields

(2) Forces on current-carrying wires in magnetic fields

(3) Fields of long current-carrying wires

(4) Biot-Savart’s law and Ampere's law (For AP Physics C only)

AP Physics B carries 4% of the total points in this section while AP Physics C carries 10%.

Here are the equations to be remembered in this section:

(1) The magnetic force ‘F’ acting on a particle of charge ‘q’ moving with velocity ‘v’ making an angle ‘θ’ with a magnetic field ‘B’ (fig.) is given by

F = qvB sinθ

The force F is perpendicular to both v and B. In vector form, the above equation is

F = qv×B

Note that bold face characters are used to represent vectors.

When electric and magnetic fields act simultaneously on a charge, the total force on the charge is given by Lorentz force equation,

F = q(v×B + E) where E is the electric field.

(2) The path of a charged particle of mass ‘m’ projected with a velocity ‘v’ perpendicular to a magnetic field B is a circle of radius ‘r’ given by

qvB = mv²/r, where ‘q’ is the charge

[Note that we have equated the magnetic force to the centripetal force]

Therefore, r = mv/qB

If the particle is projected into the magnetic field at an angle other than zero or 90º, the path is a helix of radius ‘r’ given by

r = mv sinθ/qB

(3) The period of circular motion as well as helical motion of a charged particle in a magnetic field is

T = 2πm/qB

(4) The frequency of revolution along the circular path (or helical path) is

f = qB/2πm

This is called the cyclotron frequency

(5) The magnetic force ‘dF’ acting on an elemental length dℓ of a conductor carrying a current ‘I’ placed in a magnetic field B, making an angle ‘θ’ with the magnetic field is given

dF = IdℓB sinθ

The force dF is perpendicular to both dℓ and B. In vector form, the above equation is

dF = I dℓ×B, treating the elemental length dℓ as a vector.

In the case of a straight conductor of length ℓ, the magnetic force is

F= IℓB sinθ which can be written in vector form as

F = I ℓ×B

(6) Force per unit length between two infinitely long parallel current carrying conductors is given by

F = µ₀I₁I₂/2πd, where µ₀ is the permeability of free space, ‘d’ is the separation between the conductors and I₁ and I₂ are the currents in the conductors.

(7) Torque on a plane current carrying coil (current loop) placed in a magnetic field B is

τ = nIAB sinθ, where ‘n’ is the number of turns in the coil, A is the area of the coil, I is the current in the coil and θ is the angle between the area vector and the magnetic field vector. Remember that the area is a vector which has magnitude equal to the area and direction perpendicular to the area.

(8) Magnetic field (dB) due to a current element of length dℓ at a point P distant ‘r’ from the current element is given by Biot-Savart law:

dB = (µ₀/4π) Idℓ sinθ/r²

This magnetic field is perpendicular to the plane containing the current element and the point P.

In vector notation the above equation is

dB = (µ₀/4π) Idℓ ×r/r³

Here r is a vector of length r directed from the current element to the

point P.

The length of the current element also is treated as a vector dℓ of length dℓ with its direction same as that of the current.

(9) The magnetic field due to a straight infinitely long current carrying conductor
at a point P at a perpendicular distance ‘r’ from the conductor is

B =µ₀I/2πr

(10) The magnetic field due to a plane circular current carrying

coil of ‘n’ turns and radius R at a point P on the axis at a distance

‘x’ from the centre of the coil is

B = µ₀nR²I /2(R² + x²)^3/2

The magnetic field at the centre of the coil is

B= µ₀nI/2R

(11) The magnetic field on the axis of an infinitely long straight solenoid at a point P well within the solenoid is

B= µ₀nI, where ‘n’ is the number of turns per metre of the solenoid.

(12) The magnetic field inside a toroid of ‘n’ turns per metre is

B= µ₀nI

(13) Ampere’s circuital law states that the line integral of magnetic flux density over any closed curve is equal to µ₀ times the total current passing through the surface enclosed by the closed curve. This is stated mathematically as

∫B. dℓ = µ₀I (The integration is over the closed path)

Amperes circuital law as modified by Maxwell to accommodate the displacement current flowing through even free space is

∫B. dℓ = µ₀ [I+ ε₀ (dφ_E/dt)], where ε₀ (dφ_E/dt) is the displacement current resulting from the rate of change of electric flux φ_E. ε₀ is the permittivity of free space.

In the next post we will discuss questions in this section.

AP Physics C– Multiple Choice Questions on Circular Motion and Rotation

In the post dated 24^th January 2008, we had discussed some multiple choice questions (on circular motion and rotation) of common interest to AP Physics B as well as AP Physics C aspirants.

As promised in that post, I give below some typical questions in this section, for the benefit of those preparing for the AP Physics C examination:

(1) A wheel starts from rest and rotates through 1000 radians in 10 s under the action of a constant torque. If the moment of inertia of the wheel is 4 kg m², the torque acting on the wheel is

(a) 10 Nm

(b) 20 Nm

(c) 40 Nm

(d) 80 Nm

(e) 100 Nm

The angular displacement (θ) at the instant t is given by

θ = θ₀ + ω₀t + (½) αt² where θ₀ is the initial displacement (at t =0), ω₀ is the initial angular velocity and α is the angular acceleration.

Substituting the known values in the above equation, we have

1000 = 0 + 0 + (½) α × 10², so that α = 20 radian per second².

Torque, τ = I α = 4×20 = 80 Nm

(2) A small solid cylinder rolls up along a curved surface (fig.) with an initial velocity v. It will ascend up to a height ‘h’ equal to

(a) 3v²/2g

(b) 3v²/4g

(c) v²/2g

(d) v²/4g

(e)3v²/g

Since the cylinder is rolling, it has translational and rotational kinetic energies. The total kinetic energy is K = (½) Mv² + (½) Iω² where M is the mass, I is the moment of inertia and ω is the angular velocity of the cylinder. At the highest point, the kinetic energy will be zero since the entire energy will be converted into gravitational potential energy Mgh where h is the height. So. we have

(½) Mv² + (½) Iω² = Mgh

The moment of inertia of the cylinder about its own axis is I = (½) MR² where R is the radius of the cylinder. The angular velocity ω = v/R. Substituting these in the above equation,

(½) Mv² + (½)×(½) MR² ×v²/R² = Mgh

Or, (¾)Mv² = Mgh, from which h = 3v²/4g

Now, suppose we change the above question as follows:

A small cylinder rolling with a velocity v along a horizontal surface encounters a smooth inclined surface. The height ‘h’ up to which the cylinder will ascend is

(a) 3v²/2g

(b) 3v²/4g

(c) v²/2g

(d) v²/4g

(e)3v²/g

You should remember that a body can roll along a surface only if the surface is rough (so that there is frictional force). On a smooth surface the body can slide; but it cannot have a linear displacement by rolling. [You might have seen how a car tyre rotates in mud without producing any movement of the car].

In the present problem, the body will roll up to the foot of the inclined smooth surface. It will continue to spin with the angular speed it has acquired, and will slide up to a certain height, maintaining its spin motion throughout the smooth surface. Its translational kinetic energy alone is responsible for its upward motion along the smooth incline so that the height up to which it will rise is given by

(½) Mv² = Mgh

Therefore, h = v²/2g.

(3) Two inclined planes have same height but different lengths (and therefore different angles). If a solid sphere is allowed to roll down from the top of these inclined planes

(a) the time of descent will be same, but the speed at the bottom of the plane will be greater for the longer plane

(b) the time of descent will be same, but the speed at the bottom of the plane will be smaller for the longer plane

(c) the time of descent and the speed at the bottom will be same in both cases

(d) time of descent will be different, but the speed at the bottom will be same in both cases

(e) speeds and the times of descent will be different

A body rolling down an inclined plane has greater acceleration if the angle (θ) of the plane is greater.

[The expression for the acceleration is a = gsinθ / [1 + (k²/R²)], as given in he post dated 20^th January 2008, but you can solve this problem even without remembering this equation]

The length of the steeper inclined plane is smaller since the planes have the same height. Because of the larger value of θ and the smaller distance to be traveled, the time taken in the case of the shorter plane is smaller.

The velocity of the sphere at the bottom of the plane is determined by the height of the plane since the kinetic energy at the bottom is equal to the gravitational potential energy at the top, as we have seen in the previous question.

[In the case of a sphere, the equation will be (½) Mv² + (½)×(²/₅)) MR² ×v²/R² = Mgh, since the moment of inertia is (²/₅)) MR²].

Since the height is the same, the speed at the bottom will be the same. Therefore, the correct option is (d).

(4) A metre stick AB hinged (without friction) at the end A as shown in the figure is kept horizontal by means of a string tied to the end B, the other end of the string being tied to a hook. The string is carefully cut (or burnt), and the scale executes angular oscillations about an axis passing through the end A. What is the speed of the end B when the metre stick assumes vertical position immediately after the string is burnt? (Acceleration due to gravity = 10 ms^–2)

(a) 2.5 ms^–1

(b) 4.2 ms^–1

(c) 5.4 ms^–1

(d) 6.8 ms^–1

(e) 9.8ms^–1

The centre of gravity of the metre stick is at its middle. When the string is burnt, the height of the centre of gravity is reduced by 0.5 m, thereby reducing the gravitational potential energy of the metre stick by Mgh = M×10×0.5 = 5M joule, where M is the mass of the metre stick.

The decrease in the potential energy is equal to the increase in kinetic energy. Therefore in the vertical position of the metre stick, its kinetic energy

(½)Iω² = 5M

Since the moment of inertia of a uniform bar of length L about a normal axis through ite mid point is ML²/12, its moment of inertia about a parallel axis through its end is ML²/3. [You will get this by applying the parallel axis theorem (see the post dated 2o^th January 2008)]

Therefore, we have (½)(ML²/3)ω² = 5M

Since ω = v/L and L = 1 metre, the above equation becomes

v²/6 = 5, from which v = √30 = 5.4 ms^–1 (approximately).

You will find some useful posts in this section at http://www.physicsplus.in