1. Oersted's observation is best represented by which statement?
ⓐ. A stationary charge near a compass needle always deflects the needle
ⓑ. A current-carrying conductor produces a magnetic field around it
ⓒ. A magnetic field can exist only inside a permanent magnet
ⓓ. A compass needle deflects only when the wire is made of magnetic material
Correct Answer: A current-carrying conductor produces a magnetic field around it
Explanation: Oersted observed that a compass needle placed near a current-carrying wire gets deflected. This showed that electric current has a magnetic effect. The important point is not that the wire itself must be a permanent magnet, but that moving charges inside the wire produce a magnetic field around it. When no current flows, the magnetic effect associated with the current disappears. This observation connected electricity and magnetism and became the starting point of electromagnetism. The common mistake is to think that only permanent magnets produce magnetic fields, while a current-carrying conductor also produces a magnetic field.
2. A straight wire is placed parallel to a small compass needle. When current is passed through the wire, the compass needle deflects. If the current direction in the wire is reversed, what is expected?
ⓐ. The compass deflection remains exactly the same in direction
ⓑ. The compass needle stops responding because current is reversed
ⓒ. The compass deflects in the opposite direction
ⓓ. The compass deflects only if the wire is heated
Correct Answer: The compass deflects in the opposite direction
Explanation: A compass needle aligns according to the magnetic field at its position. A current-carrying wire produces magnetic field lines around the wire. The direction of this magnetic field depends on the direction of current. When the current is reversed, the direction of the magnetic field around the wire also reverses. Since the compass responds to the direction of \(\vec{B}\), its deflection is reversed. The direction change is the key observation; current reversal does not destroy the magnetic effect, it reverses it.
3. Consider the following statements about the magnetic field produced by a current-carrying conductor.
Statement I: The magnetic field around the conductor has direction as well as magnitude.
Statement II: Reversing the current reverses the direction of the magnetic field around the conductor.
Statement III: The magnetic field near the conductor is independent of whether current is flowing or not.
Which statements are correct?
ⓐ. II and III only
ⓑ. I and III only
ⓒ. I and II only
ⓓ. I, II and III
Correct Answer: I and II only
Explanation: Magnetic field \(\vec{B}\) is a vector quantity, so it has both magnitude and direction. Around a current-carrying conductor, the direction of \(\vec{B}\) is linked with the direction of current. If the current direction is reversed, the magnetic field direction around the wire also reverses. Statement III is not correct because the magnetic field due to the wire is produced by current in the conductor. Without current, this current-produced magnetic field is absent. The important distinction is that the conductor is not behaving like a permanent magnet by itself; the current is responsible for the magnetic field.
4. The blank in the sentence should be filled correctly: The SI unit of magnetic field \(\vec{B}\) is ______.
ⓐ. \(\text{coulomb}\)
ⓑ. \(\text{tesla}\)
ⓒ. \(\text{ampere}\)
ⓓ. \(\text{newton}\)
Correct Answer: \(\text{tesla}\)
Explanation: Magnetic field is denoted by \(\vec{B}\). Its SI unit is \(\text{tesla}\), written as \(\text{T}\). The unit \(\text{coulomb}\) is used for electric charge, not magnetic field. The unit \(\text{ampere}\) is used for electric current, which can produce a magnetic field but is not the unit of \(\vec{B}\). The unit \(\text{newton}\) is the unit of force. A common unit-related mistake is to confuse the cause of the magnetic field, current in \(\text{A}\), with the magnetic field itself, measured in \(\text{T}\).
5. Match the physical ideas with the correct descriptions.
| Column I | Column II |
|---|
| P. Magnetic field \(\vec{B}\) | 1. SI unit \(\text{T}\) |
| Q. Electric current \(I\) | 2. Moving charges in a conductor |
| R. Oersted's observation | 3. Compass deflection near a current-carrying wire |
| S. Reversing current direction | 4. Reversal of magnetic field direction around the wire |
ⓐ. P-2, Q-1, R-4, S-3
ⓑ. P-1, Q-3, R-2, S-4
ⓒ. P-4, Q-2, R-3, S-1
ⓓ. P-1, Q-2, R-3, S-4
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: Magnetic field \(\vec{B}\) is measured in \(\text{tesla}\), whose symbol is \(\text{T}\). Electric current \(I\) in a conductor is due to moving charges. Oersted's observation was that a compass needle deflects near a current-carrying wire. Reversing the current reverses the magnetic field direction around the conductor, so the compass deflection also reverses. These four links connect the basic physics language here: current is the cause, \(\vec{B}\) is the magnetic effect, and the compass shows the direction-sensitive response. A common mistake is to mix up the unit of current, \(\text{A}\), with the unit of magnetic field, \(\text{T}\).
6. Assertion: A magnetic field is treated as a vector quantity.
Reason: A compass placed in a magnetic field shows a definite direction at that point.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: A vector quantity must have both magnitude and direction. Magnetic field \(\vec{B}\) has a definite direction at every point where the field is defined. A compass needle helps reveal this direction because it tends to align along the magnetic field direction. Therefore, the Reason correctly supports why \(\vec{B}\) is treated as a vector quantity. The compass does not measure only the strength of the field; it also indicates the direction of the field. The sign or symbol of \(\vec{B}\) should not be treated as merely decorative because its direction is physically meaningful.
7. Study the table and identify the row that contains an incorrect statement.
| Row | Situation | Statement |
|---|
| P | Wire carrying current \(I\) | It produces a magnetic field around it |
| Q | Current direction reversed | Direction of magnetic field around the wire reverses |
| R | Magnetic field \(\vec{B}\) | It has SI unit \(\text{T}\) |
| S | Wire without current | It produces the same current-related magnetic field as before |
ⓐ. Row S
ⓑ. Row P
ⓒ. Row Q
ⓓ. Row R
Correct Answer: Row S
Explanation: A wire carrying current \(I\) produces a magnetic field around it, so row P is correct. When the current direction is reversed, the direction of the magnetic field around the wire also reverses, so row Q is correct. Magnetic field \(\vec{B}\) is measured in \(\text{T}\), so row R is correct. Row S is incorrect because the magnetic field being discussed here is due to current in the conductor. If the current is absent, the current-related magnetic effect of that wire is absent. The common misconception is to treat an ordinary current-carrying wire like a permanent magnet even after the current is switched off.
8. Use the arrangement described below and answer the question.
A straight vertical wire passes near a compass needle. Initially, no current flows and the compass remains along Earth's magnetic field. Then a steady current is switched on in the wire.
Which conclusion is most directly supported by the observed deflection of the compass needle?
ⓐ. The compass needle becomes electrically charged
ⓑ. The wire must have become a permanent magnet
ⓒ. Earth's magnetic field becomes zero near the wire
ⓓ. Current produces a magnetic field around the wire
Correct Answer: Current produces a magnetic field around the wire
Explanation: The compass needle is a magnetic needle, so its deflection indicates the presence of a magnetic field direction different from its original alignment. When current is switched on, moving charges in the wire produce a magnetic field around the wire. The compass needle responds to the resultant magnetic field at its location. The observation does not require the wire to become a permanent magnet. It also does not mean that Earth's magnetic field disappears; instead, the field due to the current changes the net magnetic field experienced by the compass. A common mistake is to confuse compass deflection with electric charging of the compass needle.
9. Which option best distinguishes the magnetic effect of current from the electric effect of stationary charges?
ⓐ. A stationary charge produces the same circular magnetic field around it as a steady current
ⓑ. A current-carrying conductor deflects a compass by its magnetic field
ⓒ. Magnetic field \(\vec{B}\) is scalar, while electric field \(\vec{E}\) is vector
ⓓ. Current can produce only heating effect and cannot produce a magnetic effect
Correct Answer: A current-carrying conductor deflects a compass by its magnetic field
Explanation: The magnetic effect of current is linked with charges in motion. In a current-carrying conductor, charges move in an organized way, and the conductor produces a magnetic field around it. A compass needle is deflected because it responds to this magnetic field. A stationary charge is associated with an electric field, but the introductory magnetic effect here is connected with moving charges or current. Both \(\vec{E}\) and \(\vec{B}\) are vector fields, so calling \(\vec{B}\) scalar is wrong. The key contrast is that electric effects may be discussed for charges, while magnetic effects in this context arise from moving charges.
10. Use the graph description below.
A graph is imagined between the qualitative compass deflection near a straight wire and the current in the wire. The direction of current is first taken as positive. Then the same magnitude of current is passed in the opposite direction.
Which graph behaviour is physically most reasonable for the current-produced magnetic effect near the wire?
ⓐ. The deflection keeps the same sign for both current directions
ⓑ. The deflection becomes permanently zero after current reversal
ⓒ. The deflection changes sign when the current direction is reversed
ⓓ. The deflection becomes unrelated to the current direction
Correct Answer: The deflection changes sign when the current direction is reversed
Explanation: \( \textbf{Known idea:} \) A current-carrying wire produces a magnetic field around it, and a compass responds to the direction of the local magnetic field.
\( \textbf{Direction condition:} \) The direction of the magnetic field around the wire depends on the direction of current \(I\).
\( \textbf{Current reversal:} \) If \(I\) is changed to the opposite direction, the magnetic field direction at the compass position also reverses.
\( \textbf{Graph interpretation:} \) A sign change in deflection represents reversal of direction, not disappearance of the magnetic effect.
\( \textbf{Physical meaning:} \) Equal magnitudes of opposite currents produce equal-strength magnetic effects in opposite directions at the same point.
\( \textbf{Common mistake:} \) Reversal of current should not be interpreted as making the magnetic field directionless; it reverses the field direction.
\( \textbf{Final answer:} \) The deflection changes sign when the current direction is reversed.
11. A charged particle of charge \(q\) moves with velocity \(\vec{v}\) in a uniform magnetic field \(\vec{B}\). Which expression correctly gives the magnetic force on the particle?
ⓐ. \(\vec{F}=q(\vec{B}\times\vec{v})\)
ⓑ. \(\vec{F}=q(\vec{v}\times\vec{B})\)
ⓒ. \(\vec{F}=qvB\)
ⓓ. \(\vec{F}=q(\vec{v}\cdot\vec{B})\)
Correct Answer: \(\vec{F}=q(\vec{v}\times\vec{B})\)
Explanation: Magnetic force on a moving charge is a vector force, so its complete expression must include direction information. The cross product \(\vec{v}\times\vec{B}\) shows that the force is perpendicular to the plane containing \(\vec{v}\) and \(\vec{B}\). The charge \(q\) also matters because the force direction reverses for a negative charge. The expression \(qvB\) gives only the magnitude in the special case when \(\vec{v}\perp\vec{B}\), so it is not the full vector law. The dot product form would give a scalar-like projection effect, not the magnetic force direction. The order of the cross product is important because \(\vec{B}\times\vec{v}=-(\vec{v}\times\vec{B})\), which would reverse the force direction.
12. The magnitude of magnetic force on a charge moving in a magnetic field is \(F=|q|vB\sin\theta\). Which statement is correct?
ⓐ. \(F\) is maximum when \(\theta=0^\circ\)
ⓑ. \(F\) is zero when \(\theta=90^\circ\)
ⓒ. \(F\) depends on the component of \(\vec{v}\) perpendicular to \(\vec{B}\)
ⓓ. \(F\) depends on the component of \(\vec{v}\) parallel to \(\vec{B}\) only
Correct Answer: \(F\) depends on the component of \(\vec{v}\) perpendicular to \(\vec{B}\)
Explanation: The magnetic force magnitude is \(F=|q|vB\sin\theta\), where \(\theta\) is the angle between \(\vec{v}\) and \(\vec{B}\). The factor \(v\sin\theta\) is the component of velocity perpendicular to the magnetic field. A velocity component parallel to \(\vec{B}\) does not contribute to magnetic force because it gives \(\sin0^\circ=0\) or \(\sin180^\circ=0\). The force is maximum when \(\theta=90^\circ\), since \(\sin90^\circ=1\). The common misconception is to use the full speed \(v\) in every case without checking the angle between motion and field.
13. A particle with charge magnitude \(3.2\times10^{-19}\,\text{C}\) moves with speed \(2.0\times10^6\,\text{m s}^{-1}\) in a uniform magnetic field of \(0.50\,\text{T}\). The angle between \(\vec{v}\) and \(\vec{B}\) is \(30^\circ\). What is the magnitude of the magnetic force on the particle?
ⓐ. \(3.2\times10^{-13}\,\text{N}\)
ⓑ. \(4.8\times10^{-13}\,\text{N}\)
ⓒ. \(6.4\times10^{-13}\,\text{N}\)
ⓓ. \(1.6\times10^{-13}\,\text{N}\)
Correct Answer: \(1.6\times10^{-13}\,\text{N}\)
Explanation: \( \textbf{Given:} \) \(|q|=3.2\times10^{-19}\,\text{C}\), \(v=2.0\times10^6\,\text{m s}^{-1}\), \(B=0.50\,\text{T}\), and \(\theta=30^\circ\).
\( \textbf{Required:} \) Magnitude of magnetic force \(F\).
\( \textbf{Useful relation:} \)
\[
F=|q|vB\sin\theta
\]
\( \textbf{Why this applies:} \) The force is magnetic force on a moving charge, so only the component of velocity perpendicular to \(\vec{B}\) contributes.
\( \textbf{Angle value:} \) \(\sin30^\circ=\frac{1}{2}\).
\( \textbf{Substitution:} \)
\[
F=(3.2\times10^{-19})(2.0\times10^6)(0.50)\left(\frac{1}{2}\right)
\]
\( \textbf{First simplification:} \) \((2.0\times10^6)(0.50)=1.0\times10^6\).
\( \textbf{Second simplification:} \) \((1.0\times10^6)\left(\frac{1}{2}\right)=5.0\times10^5\).
\( \textbf{Force value:} \)
\[
F=(3.2\times10^{-19})(5.0\times10^5)=16.0\times10^{-14}\,\text{N}
\]
\( \textbf{Scientific notation:} \)
\[
F=1.6\times10^{-13}\,\text{N}
\]
\( \textbf{Unit check:} \) The magnetic force must be in \(\text{N}\), not in \(\text{T}\) or \(\text{C}\).
\( \textbf{Final answer:} \) The force magnitude is \(1.6\times10^{-13}\,\text{N}\).
14. The magnetic force on a moving charged particle becomes zero when the angle between \(\vec{v}\) and \(\vec{B}\) is ______.
ⓐ. \(30^\circ\) or \(60^\circ\)
ⓑ. \(45^\circ\) or \(90^\circ\)
ⓒ. \(0^\circ\) or \(180^\circ\)
ⓓ. \(90^\circ\) only
Correct Answer: \(0^\circ\) or \(180^\circ\)
Explanation: The magnitude of magnetic force is \(F=|q|vB\sin\theta\). The force becomes zero when \(\sin\theta=0\). This happens at \(\theta=0^\circ\) and \(\theta=180^\circ\), meaning the velocity is parallel or antiparallel to the magnetic field. In both cases, there is no velocity component perpendicular to \(\vec{B}\). At \(\theta=90^\circ\), the perpendicular component is maximum, so the force is maximum, not zero. A useful check is to remember that magnetic force depends on perpendicular motion, not on motion along the field.
15. Assertion: A magnetic field alone cannot change the kinetic energy of a charged particle.
Reason: The magnetic force on the charged particle is always perpendicular to its instantaneous velocity.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Magnetic force on a moving charge is \(\vec{F}=q(\vec{v}\times\vec{B})\), so \(\vec{F}\) is perpendicular to \(\vec{v}\). Work done by a force depends on the component of force along displacement. Since the instantaneous displacement of the particle is along \(\vec{v}\), the magnetic force has no component along the displacement. Hence the magnetic force does no work on the particle. If no work is done, kinetic energy and speed cannot be changed by the magnetic field alone. The magnetic field can change the direction of motion, so confusing change in direction with change in speed is the common mistake.
16. A positive charge and a negative charge have equal charge magnitudes and move with the same velocity \(\vec{v}\) through the same magnetic field \(\vec{B}\). Which comparison is correct if \(\vec{v}\) is not parallel to \(\vec{B}\)?
ⓐ. Their force magnitudes are unequal, but their force directions are the same
ⓑ. The positive charge experiences force, but the negative charge experiences no force
ⓒ. The negative charge experiences force only if its speed is greater
ⓓ. Their force magnitudes are equal, but their force directions are opposite
Correct Answer: Their force magnitudes are equal, but their force directions are opposite
Explanation: The magnitude of magnetic force is \(F=|q|vB\sin\theta\). If the charge magnitudes, speeds, magnetic fields, and angles are the same, the force magnitudes are equal. However, the vector formula is \(\vec{F}=q(\vec{v}\times\vec{B})\), so the sign of \(q\) affects the direction of force. A negative charge experiences force opposite to the direction obtained for a positive charge with the same \(\vec{v}\) and \(\vec{B}\). The negative charge is not force-free merely because its charge has a negative sign. The sign of charge changes direction, while \(|q|\) controls the magnitude.
17. Study the table and identify the row that gives the correct force condition.
| Row | Angle between \(\vec{v}\) and \(\vec{B}\) | Magnetic force magnitude |
|---|
| P | \(\theta=0^\circ\) | Maximum |
| Q | \(\theta=90^\circ\) | Maximum |
| R | \(\theta=180^\circ\) | Same as at \(\theta=90^\circ\) |
| S | \(\theta=30^\circ\) | Zero |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row Q
Explanation: The magnetic force magnitude is \(F=|q|vB\sin\theta\). At \(\theta=0^\circ\), \(\sin0^\circ=0\), so row P is incorrect. At \(\theta=90^\circ\), \(\sin90^\circ=1\), so the force is maximum and row Q is correct. At \(\theta=180^\circ\), \(\sin180^\circ=0\), so row R is incorrect. At \(\theta=30^\circ\), \(\sin30^\circ=\frac{1}{2}\), so the force is not zero and row S is incorrect. The deciding quantity is \(\sin\theta\), not the numerical size of the angle alone.
18. A charged particle enters a region of uniform magnetic field with speed \(v\). The graph of magnetic force magnitude \(F\) versus \(\sin\theta\) is plotted for fixed \(|q|\), \(v\), and \(B\), where \(\theta\) is the angle between \(\vec{v}\) and \(\vec{B}\). What does the slope of this graph represent?
ⓐ. \(|q|vB\)
ⓑ. \(\frac{|q|B}{v}\)
ⓒ. \(\frac{vB}{|q|}\)
ⓓ. \(|q|v\sin\theta\)
Correct Answer: \(|q|vB\)
Explanation: \( \textbf{Graph relation:} \) The magnetic force magnitude is
\[
F=|q|vB\sin\theta
\]
\( \textbf{Fixed quantities:} \) In this graph, \(|q|\), \(v\), and \(B\) are constant.
\( \textbf{Variable:} \) The horizontal variable is \(\sin\theta\).
\( \textbf{Linear form:} \) The equation can be written as
\[
F=(|q|vB)(\sin\theta)
\]
\( \textbf{Slope comparison:} \) This matches \(y=mx\), where \(F\) is like \(y\), \(\sin\theta\) is like \(x\), and \(|q|vB\) is the slope.
\( \textbf{Unit check:} \) Since \(\sin\theta\) is dimensionless, the slope must have the same unit as force, \(\text{N}\).
\( \textbf{Common mistake:} \) The slope is not \(|q|v\sin\theta\), because \(\sin\theta\) is the plotted variable and should not remain inside the constant slope.
\( \textbf{Final answer:} \) The slope of the \(F\) versus \(\sin\theta\) graph is \(|q|vB\).
19. Two particles \(P\) and \(Q\) move in the same uniform magnetic field. Particle \(P\) has charge magnitude \(q\), speed \(2v\), and angle \(30^\circ\) with \(\vec{B}\). Particle \(Q\) has charge magnitude \(2q\), speed \(v\), and angle \(90^\circ\) with \(\vec{B}\). What is the ratio \(\frac{F_P}{F_Q}\)?
ⓐ. \(\frac{1}{4}\)
ⓑ. \(\frac{1}{2}\)
ⓒ. \(1\)
ⓓ. \(2\)
Correct Answer: \(\frac{1}{2}\)
Explanation: \( \textbf{Given for particle \(P\):} \) \(|q_P|=q\), \(v_P=2v\), and \(\theta_P=30^\circ\).
\( \textbf{Given for particle \(Q\):} \) \(|q_Q|=2q\), \(v_Q=v\), and \(\theta_Q=90^\circ\).
\( \textbf{Same condition:} \) Both particles move in the same magnetic field, so \(B\) is common.
\( \textbf{Force formula:} \)
\[
F=|q|vB\sin\theta
\]
\( \textbf{Force on \(P\):} \)
\[
F_P=(q)(2v)B\sin30^\circ
\]
\( \textbf{Use \(\sin30^\circ=\frac{1}{2}\):} \)
\[
F_P=(q)(2v)B\left(\frac{1}{2}\right)=qvB
\]
\( \textbf{Force on \(Q\):} \)
\[
F_Q=(2q)(v)B\sin90^\circ
\]
\( \textbf{Use \(\sin90^\circ=1\):} \)
\[
F_Q=2qvB
\]
\( \textbf{Ratio:} \)
\[
\frac{F_P}{F_Q}=\frac{qvB}{2qvB}=\frac{1}{2}
\]
\( \textbf{Final answer:} \) The ratio \(\frac{F_P}{F_Q}\) is \(\frac{1}{2}\).
20. Read the passage and answer the question.
An electron enters a region where only a uniform magnetic field is present. Its velocity is perpendicular to the magnetic field at entry. During its motion, the magnetic field keeps exerting a force on it.
Which statement about the electron's motion is correct?
ⓐ. Its speed increases continuously because force is present
ⓑ. Its kinetic energy decreases because the magnetic field opposes motion
ⓒ. Its speed remains constant, but its direction of motion changes
ⓓ. Its velocity remains completely unchanged because magnetic force does no work
Correct Answer: Its speed remains constant, but its direction of motion changes
Explanation: A force can change velocity either by changing speed, direction, or both. Magnetic force on a moving charge is perpendicular to its instantaneous velocity. Because of this perpendicularity, the magnetic force does no work on the electron. With no work done, the kinetic energy and speed remain constant. However, the perpendicular force can still change the direction of velocity, producing curved motion when the velocity has a perpendicular component to \(\vec{B}\). The subtle point is that zero work does not mean zero force; it means the force has no component along the instantaneous displacement.
