Electromagnetism

Sample questions

Showing 20 of 70 questions from this chapter. The full bank is in the chapter test below.

1. 1

   In Fleming's left-hand rule, the first (fore) finger is pointed in the direction of the

   • A.✓ magnetic field (correct answer)
   • B.current in the conductor
   • C.force on the conductor
   • D.motion of the conductor

   Why: The forefinger gives the direction of the magnetic field, the second finger the direction of the current, and the thumb the direction of the force on the conductor.
2. 2

   A current of $5.0\ \mathrm{A}$ flows through a straight wire $0.20\ \mathrm{m}$ long that lies at right angles to a uniform magnetic field of $0.40\ \mathrm{T}$. The force acting on the wire is

   • A.$4.0\ \mathrm{N}$
   • B.✓ $0.40\ \mathrm{N}$ (correct answer)
   • C.$0.04\ \mathrm{N}$
   • D.$2.0\ \mathrm{N}$

   Why: $F = BIL = 0.40 \times 5.0 \times 0.20 = 0.40\ \mathrm{N}$. The value $4.0\ \mathrm{N}$ is obtained if the length is wrongly taken as $2.0\ \mathrm{m}$.
3. 3

   The magnetic flux density $B$ is defined as the force acting per unit

   • A.length of a wire placed parallel to the magnetic field
   • B.area of a coil placed parallel to the magnetic field
   • C.✓ current per unit length on a wire placed at right angles to the magnetic field (correct answer)
   • D.charge moving with unit velocity along the magnetic field

   Why: Rearranging $F = BIL$ gives $B = F/(IL)$, so $B$ is the force per unit current per unit length when the wire is at right angles to the field. The flux per unit area is equal to $B$ only when that area is perpendicular to the field.
4. 4

   The SI unit of magnetic flux density is

   • A.$\mathrm{N\,m\,A^{-1}}$
   • B.$\mathrm{N\,A\,m^{-1}}$
   • C.$\mathrm{N\,A^{-1}\,m^{-2}}$
   • D.✓ $\mathrm{N\,A^{-1}\,m^{-1}}$ (correct answer)

   Why: Flux density is force per unit current per unit length, so its unit is newton per ampere per metre, which is named the tesla. The combination $\mathrm{N\,m\,A^{-1}}$ is the weber, the unit of magnetic flux.
5. 5

   The SI unit of magnetic flux is the

   • A.tesla
   • B.newton per ampere metre
   • C.✓ weber (correct answer)
   • D.volt second per ampere

   Why: The unit of flux is the weber, which equals one newton metre per ampere. The tesla is the unit of flux density, and the volt second per ampere is the henry, the unit of inductance.
6. 6

   The magnetic flux through a flat surface of area $0.020\ \mathrm{m^{2}}$ placed perpendicular to a uniform magnetic field is $0.010\ \mathrm{Wb}$. The magnetic flux density is

   • A.$2.0 \times 10^{-4}\ \mathrm{T}$
   • B.✓ $0.50\ \mathrm{T}$ (correct answer)
   • C.$5.0\ \mathrm{T}$
   • D.$2.0\ \mathrm{T}$

   Why: For a surface perpendicular to the field $\Phi = BA$, so $B = \Phi/A = 0.010/0.020 = 0.50\ \mathrm{T}$. The value $2.0 \times 10^{-4}\ \mathrm{T}$ comes from multiplying the flux by the area instead of dividing.
7. 7

   A uniform magnetic field of $0.20\ \mathrm{T}$ makes an angle of $60^\circ$ with the normal to a flat surface of area $0.50\ \mathrm{m^{2}}$. The magnetic flux through the surface is

   • A.$0.10\ \mathrm{Wb}$
   • B.$0.087\ \mathrm{Wb}$
   • C.$0.025\ \mathrm{Wb}$
   • D.✓ $0.050\ \mathrm{Wb}$ (correct answer)

   Why: $\Phi = BA\cos 60^\circ = 0.20 \times 0.50 \times 0.50 = 0.050\ \mathrm{Wb}$. Using $\sin 60^\circ$ in place of $\cos 60^\circ$ gives the incorrect value $0.087\ \mathrm{Wb}$.
8. 8

   The magnetic flux through a flat coil placed in a uniform magnetic field is zero when

   • A.✓ the plane of the coil is parallel to the magnetic field (correct answer)
   • B.the plane of the coil is perpendicular to the magnetic field
   • C.the normal to the coil is parallel to the magnetic field
   • D.the coil is placed in a very strong magnetic field

   Why: Flux is $\Phi = BA\cos\theta$, where $\theta$ is the angle between the field and the normal to the coil. When the plane of the coil lies parallel to the field the normal makes $90^\circ$ with the field, so $\cos 90^\circ = 0$ and the flux vanishes.
9. 9

   A coil of $200$ turns has a magnetic flux of $5.0 \times 10^{-4}\ \mathrm{Wb}$ through each turn. The magnetic flux linkage of the coil is

   • A.$0.010\ \mathrm{Wb}$
   • B.$5.0 \times 10^{-4}\ \mathrm{Wb}$
   • C.✓ $0.10\ \mathrm{Wb}$ (correct answer)
   • D.$2.5 \times 10^{-6}\ \mathrm{Wb}$

   Why: Flux linkage is $N\Phi = 200 \times 5.0 \times 10^{-4} = 0.10\ \mathrm{Wb}$. Dividing the flux by the number of turns instead of multiplying gives $2.5 \times 10^{-6}\ \mathrm{Wb}$.
10. 10

    The magnetic flux linkage of a coil of $N$ turns, each linked with a flux $\Phi$, is

    • A.$\Phi/N$
    • B.✓ $N\Phi$ (correct answer)
    • C.$BA$
    • D.$BIL$

    Why: Flux linkage is the flux through one turn multiplied by the number of turns, that is $N\Phi$. The quantity $BA$ is the flux through a single turn and takes no account of the number of turns.
11. 11

    A charged particle moving with velocity $v$ in a uniform magnetic field $B$ experiences the maximum force when the angle between $v$ and $B$ is

    • A.$0^\circ$
    • B.$45^\circ$
    • C.$180^\circ$
    • D.✓ $90^\circ$ (correct answer)

    Why: $F = qvB\sin\theta$ is greatest when $\sin\theta = 1$, that is when the velocity is at right angles to the field. The force is zero at $0^\circ$ and at $180^\circ$, where the particle moves along the field.
12. 12

    An electron ($q = 1.6 \times 10^{-19}\ \mathrm{C}$) moves with a speed of $2.0 \times 10^{6}\ \mathrm{m\,s^{-1}}$ perpendicular to a uniform magnetic field of $0.50\ \mathrm{T}$. The force on the electron is

    • A.✓ $1.6 \times 10^{-13}\ \mathrm{N}$ (correct answer)
    • B.$3.2 \times 10^{-13}\ \mathrm{N}$
    • C.$8.0 \times 10^{-14}\ \mathrm{N}$
    • D.$1.6 \times 10^{-19}\ \mathrm{N}$

    Why: $F = qvB\sin 90^\circ = 1.6 \times 10^{-19} \times 2.0 \times 10^{6} \times 0.50 = 1.6 \times 10^{-13}\ \mathrm{N}$. The option $1.6 \times 10^{-19}\ \mathrm{N}$ is simply the charge itself and is not a force.
13. 13

    An electron enters a uniform magnetic field of $2.5\ \mathrm{T}$ at right angles to it with a speed of $1.0 \times 10^{7}\ \mathrm{m\,s^{-1}}$. The radius of the circular path it follows is about ($m_e = 9.11 \times 10^{-31}\ \mathrm{kg}$, $e = 1.6 \times 10^{-19}\ \mathrm{C}$)

    • A.$2.3 \times 10^{-6}\ \mathrm{m}$
    • B.$4.6 \times 10^{-5}\ \mathrm{m}$
    • C.✓ $2.3 \times 10^{-5}\ \mathrm{m}$ (correct answer)
    • D.$2.3 \times 10^{-4}\ \mathrm{m}$

    Why: The magnetic force supplies the centripetal force, so $r = \frac{mv}{eB} = \frac{9.11 \times 10^{-31} \times 1.0 \times 10^{7}}{1.6 \times 10^{-19} \times 2.5} = 2.3 \times 10^{-5}\ \mathrm{m}$.
14. 14

    The radius of the circular path of a charged particle moving at right angles to a uniform magnetic field is directly proportional to the particle's

    • A.charge
    • B.✓ momentum (correct answer)
    • C.kinetic energy
    • D.magnetic flux density

    Why: Since $Bqv = mv^{2}/r$, the radius is $r = mv/(qB)$, which is proportional to the momentum $mv$ and inversely proportional to the charge and the flux density.
15. 15

    A proton and an electron enter the same uniform magnetic field with the same speed, each moving at right angles to the field. Which is deflected more, and why?

    • A.The proton, because it has the greater mass
    • B.Neither, because both have the same charge magnitude
    • C.Neither, because a magnetic field does no work on either particle
    • D.✓ The electron, because its much smaller mass gives it a much smaller radius of curvature (correct answer)

    Why: Both particles have the same charge magnitude and speed, so they feel the same force, but $r = mv/(qB)$ is far smaller for the lighter electron, so the electron curves much more sharply.
16. 16

    Two particles of the same mass and the same charge enter a uniform magnetic field at right angles to it. The particle that is deflected more is the one moving

    • A.✓ more slowly (correct answer)
    • B.more quickly
    • C.The deflection is the same at both speeds
    • D.The deflection depends only on the charge and not on the speed

    Why: The radius is $r = mv/(qB)$, so the slower particle has the smaller radius and bends through a greater angle over the same distance.
17. 17

    A charged particle travelling in a circle inside a uniform magnetic field keeps a constant speed because

    • A.the magnetic force acts along the velocity of the particle
    • B.the magnetic field does positive work on the particle
    • C.✓ the magnetic force is always perpendicular to the velocity and so does no work (correct answer)
    • D.the charge of the particle stays constant

    Why: The force $q(\mathbf{v}\times\mathbf{B})$ is at right angles to $\mathbf{v}$, so it changes the direction of motion but not the speed, and the kinetic energy stays the same.
18. 18

    An electron enters a uniform magnetic field at right angles to it. Inside the field its path is

    • A.a straight line
    • B.a parabola
    • C.a helix
    • D.✓ a circle (correct answer)

    Why: The force is always perpendicular to the velocity and has constant magnitude, so it acts as a centripetal force and the electron moves in a circle of radius $mv/(eB)$.
19. 19

    In a velocity selector the electric and magnetic forces on a charged particle cancel each other only for one speed. That speed is

    • A.$EB$
    • B.✓ $E/B$ (correct answer)
    • C.$B/E$
    • D.$1/(EB)$

    Why: Undeviated motion requires $qE = Bqv$, which gives $v = E/B$. Taking the inverse ratio $B/E$ gives a quantity with the wrong units for a speed.
20. 20

    Alpha particles of various speeds enter a velocity selector in which $E = 300\ \mathrm{V\,m^{-1}}$ and $B = 0.20\ \mathrm{T}$. The particles that pass through undeviated have a speed of

    • A.✓ $1500\ \mathrm{m\,s^{-1}}$ (correct answer)
    • B.$60\ \mathrm{m\,s^{-1}}$
    • C.$150\ \mathrm{m\,s^{-1}}$
    • D.$600\ \mathrm{m\,s^{-1}}$

    Why: The selected speed is $v = E/B = 300/0.20 = 1500\ \mathrm{m\,s^{-1}}$; particles of any other speed are deflected aside and do not leave the tube at the far end.