# 16 good questions on electromagnetism

(See attached file for full problem description with proper symbols and diagrams)

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3.1 A long wire carries a current of 2 A along the +z axis. Calculate B in free space at (3,4,9).

3.2 Inside a long conductor of radius a, the magnetic field is H = 5 rho;/(2pi a2) a(phi) A/m. (a) Determine the current density J. (b) Determine the total current carried by the conductor.

3.3 A three-conductor transmission line is configured as shown below. Determine the magnitude and the direction of the magnetic field H at P.

3.5 A transmission line is composed of two parallel (infinitely) long conductors spaced 2-m apart in air. (a) If the magnetic field H at the midpoint between the lines has a magnitude of 100 A/m, determine the current in each conductor of the transmission line. (b) If the direction of the current in each conductor were reversed because of incorrect wiring, how would the magnitude and the direction of the magnetic field at the midpoint change?

3.7 A square loop (0.02 m x 0.02 m) has 100 turns and it lies in the x-y plane. It carries a current of 5 mA. If the loop is exposed to a magnetic flux density B = 2 x 10-3 a Wb/m2 , determine: (a) the magnetic force exerted on the loop; (b) the maximum torque experienced by the loop; (c) If the loop is free to rotate, sketch its equilibrium position.

3.8 A transmission line is composed of two parallel (infinitely) long conductors spaced 2-m apart in air. The conductors carry equal and opposite currents of 100 A each. (a) Determine the magnetic field H (magnitude and direction) at the midpoint between the lines. (b) If both conductors were carrying currents in the same direction, how would the magnitude and the direction of the magnetic field at the midpoint change?

3.10 Two circular loops are centered at the origin in the x-y plane in air. The inner loop has a radius a = 5 cm and it carries a counterclockwise current I1 = 2.5 A. The outer loop has a radius b = 8 cm and it carries a current I2. (a) If the magnetic field at the origin is measured to be zero, determine the magnitude and the direction of I2. (b) If the magnitude of the current in each loop were doubled, how would the magnitude and the direction of the magnetic field at the origin change? Explain.

3.11 A toroid has the following parameters: radius R = 0.1 m; coil radius r = 0.01 m; number of turns N = 1,000; core material = air. (a) If the toroid carries a current of 10 mA, determine the stored magnetic energy. (b) What should be the relative permeability of the core, if we want to increase the stored energy by a factor of 1,000?

3.12 Two long straight wires carry currents of 2 A each in the + a¬z direction. The wires cross the x-y plane at (0, 0, 0) and (4, 0, 0). Determine the magnetic field vector H at: (a) (2, 0, 0); (b) (8, 0, 0).

3.14 Two long straight wires carry currents of 12 A each in the + ax direction. The wires cross the y-z plane at (0, 0, -4) and (0, 0, 4). Determine the magnetic field vector H at: (a) (2, 0, 0) and (b) (0, 0, 8).

3.15 The magnetic field H is given as: H = J0 /3 a A/m for  a and H = J0 a/(3  a A/m for  a. (a) Determine the current density J everywhere. (b) Determine the total current I passing through the annular region a b, located in the x-y plane.

4.1 A square loop (0.1 m x 0.1 m) has 3 turns and it lies in the x-y plane. If the loop is exposed to a time-varying magnetic flux density B = 2 sin (120 pi t) Wb/m2 , determine the maximum value of the induced emf.

4.2 A circular metal loop (radius = 0.1 m) has 100 turns and it lies in the x-y plane. If the loop is exposed to a magnetic flux density B = 2 cos (100 pi t) Wb/m2 , determine: (a) the magnitude of the maximum emf induced in the loop; (b) the direction of current flow during 0 < t < 1/100 s.

4.3 A square loop (0.2 m x 0.2 m) lies in the x-y plane. When the loop is exposed to a magnetic flux density B = 5 x 10-3 cos (2 pi t) Wb/m2 , a sinusoidal emf with a peak value of 62.8 mV is induced in the loop. (a) Determine the frequency f. (b) If the loop had 10 turns, how would the induced emf change?

4.4 A rectangular loop (0.3 m x 0.2 m) has 10 turns and it lies in the y-z plane. It carries a counterclockwise current of 1 A. The loop is exposed to a magnetic flux density B = 5 x 10-3 Wb/m2. (a) Determine the magnitude and the direction of the torque experienced by the loop. (b) If the loop is free to rotate, sketch its equilibrium position.

4.5 A rectangular loop (0.3 m x 0.2 m) has 3 turns and it lies in the x-y plane. If the loop is exposed to a magnetic flux density B = 2 sin (1000 pi t) Wb/m2, determine: (a) the maximum value of the induced emf and (b) the direction of current flow at t = 0. Explain.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please see the attachment.

3.1 A long wire carries a current of 2 A along the +z axis. Calculate B in free space at (3,4,9). (F-00, Exam #2)

Answer:

Magnetic field due to a current carrying conductor is given by Biot-Savart rule and the derivation for field at a point P at (perpendicular) distance r from an infinitely long current carrying wire is given in your text. As the point P is (3,4,9) r is (32 +42) = 5m. The magnitude of this field is given as

and its direction is perpendicular to r given by ampere's right hand rule.

'If the thumb of right hand is showing the direction of current then the fingers shows the direction of magnetic flux (field).'

The direction is shown in the figure.

The field at P may be given as T.

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3.2 Inside a long conductor of radius a, the magnetic field is H = 5 /(2a2) a A/m. (a) Determine the current density J. (b) Determine the total current carried by the conductor. (F-00, Exam #2)

Answer:

(I think r is the distance r from the origin, you have not given meaning of notation)

Using circuital law , where I is the current within the loop we get for a circular loop

H2r = I or H = I/2r where I is the current within the loop of radius r. gives

I = 2rH.

If the current within the loop of radius is r+dr be I+dI and the field is H +dH then

H + dH = (I+dI)/2(r+dr) gives us

I+dI =2 (r+dr)(H + dH) = 2(rH+Hdr+rdH) {drdH is very small}

Subtracting we get

dI = 2 (Hdr +rdH)

Now if the current density at distance r from the center is j, then

J = dI/2 rdr (where 2 r dr is the area of the element between r and r+dr)

Or j = (H/r + dH/dr). ....................................(1)

But H = 5 r/(2a2)a A/m or H = 5 r/(2a2) gives dH/dr = 5 /(2a2), hence

J = 5/(2a2) +5/(2a2) = 10/(2a2) constant.

Total current carried by the conductor is J.a2 = 5 A/m2.

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3.3 A three-conductor transmission line is configured as shown below. Determine the magnitude and the direction of the magnetic field H at P. (F-01, Exam #2)

Answer:

The field due two the three wires at P is along y axis and hence we can find the resultant field just by the algebraic sum.

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3.5 A transmission line is composed of two parallel (infinitely) long conductors spaced 2-m apart in air. (a) If the magnetic field H at the midpoint between the lines has a magnitude of 100 A/m, determine the current in each conductor of the transmission line. (b) If the direction of the current in each conductor were reversed because of incorrect wiring, how would the magnitude and the direction of the magnetic field at the midpoint change? (S-02, Exam #2)

Answer:

a) The current in the wires of the transmission line is in opposite direction hence using right hand rule we find that the field at midpoint are in same direction and hence

b) If the direction of the current in each wire is reversed the field will be in opposite direction.

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3.7 A square loop (0.02 m x 0.02 m) has 100 turns and it lies in the x-y plane. It carries a current of 5 mA. If the loop is exposed to a magnetic flux density B = 2 x 10-3 a¬x Wb/m2 , determine: (a) the magnetic force exerted on the loop; (b) the maximum torque experienced by the loop; (c) If the loop is free to rotate, sketch its equilibrium position. (S-02, Exam #2)

Answer:

a) As the current in the opposite direction in the opposite sides of the loop, the force on the wires is in opposite direction and hence the net force on the loop is zero.

b) The torque on the loop is given by the relation

where is the magnetic moment of the loop (nIA)

hence the maximum torque is given by

max = nIAB =100*5*4*10-4*2*10-3 = 4*10-4 Nm.

c) If the loop is in equilibrium, net force is zero and torque is zero. For the torque to be zero the magnetic moment must be in the direction of the field =0. The direction of magnetic moment is given by right hand rule.

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3.8 A transmission line is composed of two parallel (infinitely) long conductors spaced 2-m apart in air. The conductors carry equal and opposite currents of 100 A each. (a) Determine the magnetic field H (magnitude and direction) at the midpoint between the lines. (b) If both conductors were carrying currents in the same direction, how would the magnitude and the direction of the magnetic field at the midpoint change? (F-02, Exam #2)

Answer:

a)The current in the wires of the transmission line is in opposite direction hence using right hand rule we find that the field at midpoint are in same direction and hence

b) If the current in the two wires is in same direction their fields will be equal and opposite and hence the resultant field will be zero.

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3.10 Two circular loops are centered at the origin in the x-y plane in air. The inner loop has a radius a = 5 cm and it carries a counterclockwise current I1 = 2.5 A. The outer loop has a radius b = 8 cm and it carries a current I2. (a) If the magnetic field at the origin is measured to be zero, determine the magnitude and the direction of I2. (b) If the magnitude of the current in each loop were doubled, how would the magnitude and the direction of the magnetic field at the origin change? Explain. (F-03, Exam #2)

Answer:

The direction of magnetic field at the center of the loop can be given by right hand rule i.e. if the fingers of right hand are showing the direction of current then the thumb shows the direction of the magnetic field.

The magnitude of the field is given by the relation B = 0I/2R.

(a) If the net field is zero then the direction of I2 must be opposite to I1 and hence clockwise

and the magnitudes of the fields must be equal and hence

0I1/2R1 =0I2/2R2 gives I2 = I1R2/R1 = 2.5*8/5 = 4 A.

(b) If the magnitude of the current in doubled the fields will be doubled but the resultant will still remains zero as both field will be still equal and opposite.

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3.11 A toroid has the following parameters: radius R = 0.1 m; coil radius r = 0.01 m; number of turns N = 1,000; core material = air. (a) If the toroid carries a current of 10 mA, determine the stored magnetic energy. (b) What should be the relative permeability of the core, if we want to increase the stored energy by a factor of 1,000? (F-03, Exam #2)

Answer:

The toroid can be considered as an infinitely long solenoid and the magnetic field inside it is given by B = 0 n I where n is the number of turns per unit length. Hence

B = 0 (N/2R) I = 2*10-7*1000*10/0.1 = 0.02T

The energy stored in a magnetic field per unit volume is B2/20r, hence the energy store is given by

U = (B2/20)(r2R) = B22r2R/0 = (4*10-4*3.14*3.14*10-4*0.1)/(4*3.14*10-7)=3.14*10-2J.

As the r appears in denominator to increase energy r = 10-3.

3.12 Two long straight wires carry currents of 2 A each in the + a¬z direction. The wires cross the x-y plane at (0, 0, 0) and (4, 0, 0). Determine the magnetic field vector H at: (a) (2, 0, 0); (b) (8, 0, 0). (S-04, Exam #2)

Answer:

(a) As the two wires carrying current in same direction the fields at mid point(2,0.0)are equal and opposite hence resultant is zero.

(b) The point (8,0,0,) is on same side of the two wires so the fields will be in the same direction(-Y) thus will be added and hence

.

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3.14 Two long straight wires carry currents of 12 A each in the + ax direction. The wires cross the y-z plane at (0, 0, -4) and (0, 0, 4). Determine the magnetic field vector H at: (a) (2, 0, 0) and (b) (0, 0, 8). (F-04, Exam #2)

Answer:

a) I think now you don't need explanation on this

In negative y direction

b)

In positive y direction.

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3.15 The magnetic field H is given as: H = J0 /3 a A/m for a and H = J0 a/(3 a A/m for a. (a) Determine the current density J everywhere. (b) Determine the total current I passing through the annular region a b, located in the x-y plane. (F-04, Exam #2)

Answer:

(I think r is the distance r from the origin, you have not given meaning of notation)

Using circuital law , where I is the current within the loop we get for a circular loop

H2r = I or H = I/2r where I is the current within the loop of radius r. gives

I = 2rH.

If the current within the loop of radius is r+dr be I+dI and the field is H +dH then

H + dH = (I+dI)/2(r+dr) gives us

I+dI =2 (r+dr)(H + dH) = 2(rH+Hdr+rdH) {drdH is very small}

Subtracting we get

dI = 2 (Hdr +rdH)

Now if the current density at distance r from the center is j, then

J = dI/2 rdr (where 2 r dr is the area of the element between r and r+dr)

Or j = (H/r + dH/dr). ....................................(1)

But for r < a: H = J0 r/3 a A/m or H = J0 r2/3 gives dH/dr = 2J0r/3, hence

For r < a current density j1 = (J0 r/3 + 2J0 r/3) = J0r.

And for r > a: H = J0 a/(3r a A/m or H = J0 a/(3 rgives dH/dr = - J0a3/3r2 hence

For r > a current density j2 = (J0a3/3r2 - J0a3/3r2) =0

b) definitely the current density in that region is zero because r > a

4.1 A square loop (0.1 m x 0.1 m) has 3 turns and it lies in the x-y plane. If the loop is exposed to a time-varying magnetic flux density B = 2 sin (120 π t) a¬z¬ Wb/m2 , determine the maximum value of the induced emf. (Exam #2, Fall-01)

Answer:

According to faraday's law of electro magnetic induction the induced emf is given by

e = - dm/dt = - d(A.B)/dt where fm is magnetic flux, A area of the loop and B is the magnetic field. Negative sign shows the direction of induced current according to Lenz's law.

B = 2 sin (120 π t) a¬z¬ gives dB/dt = 2*120 cos 120t whose maximum value is 240

hence the maximum induced emf is e = - 3*0.01*240. = 22.62 volts.

4.2 A circular metal loop (radius = 0.1 m) has 100 turns and it lies in the x-y plane. If the loop is exposed to a magnetic flux density B = 2 cos (100 π t) a¬z Wb/m2 , determine: (a) the magnitude of the maximum emf induced in the loop; (b) the direction of current flow during 0 < t < 1/100 s. (Exam #2, Fall-02)

Answer:

Again B = 2 cos (100 π t) a¬z or B = 2cos(100t) gives dB/dt = -200sin(100t) with a max of-200. Hence the maximum induced emf = 100*0.01*200. = 1974 volt.

Between 0<t<0.01s the value of cos100t and hence tat of B is decreasing and therefore the current will in such direction that it will produce field in the direction of B.

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4.3 A square loop (0.2 m x 0.2 m) lies in the x-y plane. When the loop is exposed to a magnetic flux density B = 5 x 10-3 cos (2 π f t) a¬z Wb/m2 , a sinusoidal emf with a peak value of 62.8 mV is induced in the loop. (a) Determine the frequency f. (b) If the loop had 10 turns, how would the induced emf change? (Exam #2, Fall-03)

Answer

In the very similar way dB/dt = - 5*10-3*2f*sin(2ft)

Gives the induced emf = 0.04*5*10-3*2f*sin(2ft) =0.2*10-3*2f*sin(2ft)

The peak value of this emf is 0.2*10-3*2f = 62.8*10-3 volt gives

f = 62.8/0.4. = 49.97 Hz.

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4.4 A rectangular loop (0.3 m x 0.2 m) has 10 turns and it lies in the y-z plane. It carries a counterclockwise current of 1 A. The loop is exposed to a magnetic flux density B = 5 x 10-3 a¬z Wb/m2. (a) Determine the magnitude and the direction of the torque experienced by the loop. (b) If the loop is free to rotate, sketch its equilibrium position. (Exam #2, Sp-04)

Answer

A current carrying loop is equivalent of a magnetic dipole of moment =(IA) whose direction is according to right hand rule.

The torque acting on the loop is given by torque = X Bm whose magnitude is Bsin.

Clearly the torque is zero when is zero or or the direction of the field is normal to the surface of the loop.

4.5 A rectangular loop (0.3 m x 0.2 m) has 3 turns and it lies in the x-y plane. If the loop is exposed to a magnetic flux density B = 2 sin (1000 π t) a¬z Wb/m2, determine: (a) the maximum value of the induced emf and (b) the direction of current flow at t = 0. Explain. (Exam #2, F-04)

Answer

Again B = 2 sin (1000 π t) a¬z or B = 2sin(1000t) gives dB/dt = 2000cos(1000t) with a max of 2000. Hence the maximum induced emf = 3*0.06*2000. = 3553 volt.

At t=0, dB/dt is positive hence induced emf is negative means that the direction of current is in such a way to give the field in direction opposite to B.

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