CHAPTER 4: MOVING CHARGES AND MAGNETISM
SAMPLE QUESTION PAPER (ASSIGNMENT-1)
- There are 33 questions in all. All questions are compulsory.
- Section A: 16 questions (12 MCQs and 4 Assertion-Reasoning) of 1 mark each.
- Section B: 5 short answer questions of 2 marks each.
- Section C: 7 short answer questions of 3 marks each.
- Section D: 2 case study-based questions of 4 marks each.
- Section E: 3 long answer questions of 5 marks each.
SECTION A (16 Marks)
Q.1. The direction of the magnetic field around a straight current-carrying conductor is determined by:
Explanation: Point the thumb of the right hand in the direction of current; the curled fingers give the direction of magnetic field lines around the straight wire.
Q.2. The magnetic force acting on a charge $\displaystyle {q}$ moving with velocity $\displaystyle {v}$ parallel to a uniform magnetic field $\displaystyle {B}$ is:
Explanation: Magnetic force is given by $\displaystyle {F = q v B \sin\theta}$. When particle moves parallel to field, $\displaystyle {\theta = 0^\circ \implies \sin 0^\circ = 0 \implies F = 0}$.
Q.3. The radius $\displaystyle {r}$ of the circular path described by a charged particle moving perpendicular to a uniform magnetic field depends on:
Explanation: Centripetal force is provided by Lorentz magnetic force: $\displaystyle {{\frac{{m v^2}}{{r}}} = q v B \implies r = {\frac{{m v}}{{q B}}}}$. Thus $\displaystyle {r}$ depends on mass $m$, velocity $v$, and magnetic field $B$.
Q.4. Two long straight parallel conductors carrying currents in the same direction:
Explanation: Currents in the same direction produce attractive magnetic forces between parallel wires ($\displaystyle {f = {\frac{{\mu_0 I_1 I_2}}{{2\pi d}}}}$).
Q.5. The magnetic field $\displaystyle {B}$ at the centre of a circular current-carrying loop of radius $\displaystyle {R}$ is directly proportional to:
Explanation: $\displaystyle {B = {\frac{{\mu_0 N I}}{{2 R}}} \implies B \propto I}$.
Q.6. A moving coil galvanometer works on the fundamental principle of:
Explanation: Deflecting torque $\displaystyle {\tau = N I A B}$ balances restoring torque $\displaystyle {C \theta}$, giving angular deflection $\displaystyle {\theta \propto I}$.
Q.7. The magnetic field inside an ideal long current-carrying solenoid is:
Explanation: Magnetic field inside an ideal long solenoid is uniform ($\displaystyle {B = \mu_0 n I}$) directed parallel to its central axis.
Q.8. The torque $\displaystyle {\tau}$ acting on a rectangular current loop placed in a uniform magnetic field is maximum when angle $\displaystyle {\theta}$ between area vector $\displaystyle {\vec{A}}$ and field $\displaystyle {\vec{B}}$ is:
Explanation: Torque is given by $\displaystyle {\tau = M B \sin\theta = N I A B \sin\theta}$. Torque is maximum when $\displaystyle {\sin\theta = 1 \implies \theta = 90^\circ}$ (plane of coil is parallel to magnetic field).
Q.9. A galvanometer can be converted into an ammeter by connecting a:
Explanation: Connecting a low resistance shunt $\displaystyle {S = {\frac{{I_g R_g}}{{I - I_g}}}}$ in parallel allows most current to bypass the delicate galvanometer coil.
Q.10. The trajectory of a charged particle entering a uniform magnetic field at an angle $\displaystyle {\theta}$ other than $\displaystyle {0^\circ, 90^\circ, 180^\circ}$ is a:
Explanation: Parallel velocity component $\displaystyle {v\cos\theta}$ moves the particle along the field lines, while perpendicular component $\displaystyle {v\sin\theta}$ rotates it circularly, forming a helical path.
Q.11. A proton and an alpha particle ($\displaystyle {\text{He}^{2+}}$) enter a uniform magnetic field $\displaystyle {B}$ with the same speed $\displaystyle {v}$ perpendicular to $\displaystyle {B}$. Which particle describes a path of larger radius?
Explanation: $\displaystyle {r = {\frac{{m v}}{{q B}}} \implies r \propto {\frac{{m}}{{q}}}}$. For proton $\displaystyle {m_p/e}$; for alpha particle $\displaystyle {4m_p/(2e) = 2(m_p/e)}$. Thus $\displaystyle {r_\alpha = 2 r_p}$, so alpha particle has a larger radius.
Q.12. A straight conductor of length $\displaystyle {0.5\text{ m}}$ carrying current $\displaystyle {4\text{ A}}$ is placed at right angles ($\displaystyle {90^\circ}$) to a uniform magnetic field of $\displaystyle {0.3\text{ T}}$. Magnetic force on conductor is:
Explanation: $\displaystyle {F = I L B \sin 90^\circ = 4 \times 0.5 \times 0.3 \times 1 = 0.6\text{ N}}$.
Directions for Q.13 to Q.16:
(A) Both Assertion and Reason are true and Reason is correct explanation of Assertion.
(B) Both Assertion and Reason are true but Reason is NOT correct explanation.
(C) Assertion is true but Reason is false.
(D) Both Assertion and Reason are false.
Q.13. Assertion (A): In electric circuits, wires carrying currents in opposite directions are often twisted together.
Reason (R): Twisting wires cancels magnetic fields generated by opposite currents, preventing interference with adjacent circuits.
Q.14. Assertion (A): The magnetic field produced by a long current-carrying solenoid is independent of its length and cross-sectional area.
Reason (R): The magnetic field inside an ideal long solenoid depends only on turn density $\displaystyle {n}$ and current $\displaystyle {I}$ ($\displaystyle {B = \mu_0 n I}$).
Q.15. Assertion (A): A stationary electric charge produces a magnetic field around it.
Reason (R): Stationary charges produce only electric field in the surrounding space.
Explanation: A stationary charge produces ONLY an electric field; a moving charge produces BOTH electric and magnetic fields.
Q.16. Assertion (A): To convert a galvanometer into an ammeter, a low resistance (shunt) is connected in parallel with it.
Reason (R): Connecting shunt resistance in parallel increases the effective resistance of the combination.
Explanation: Connecting a shunt in parallel DECREASES the net combined resistance ($\displaystyle {R_{\text{net}} < S}$), turning it into a low-resistance ammeter.
SECTION B (10 Marks)
Q.17. State Biot-Savart Law. Express it in vector notation.
$\displaystyle {d B = {\frac{{\mu_0}}{{4\pi}}} {\frac{{I d l \sin\theta}}{{r^2}}}}$.
In vector form: $\displaystyle {d\vec{B} = {\frac{{\mu_0}}{{4\pi}}} {\frac{{I (d\vec{l} \times \hat{r})}}{{r^2}}} = {\frac{{\mu_0}}{{4\pi}}} {\frac{{I (d\vec{l} \times \vec{r})}}{{r^3}}}}$.
Q.18. An electron moving with velocity $\displaystyle {v = 1.5 \times 10^7\text{ m/s}}$ enters a uniform magnetic field $\displaystyle {B = 0.2\text{ T}}$ along a direction parallel to the field lines. What is its trajectory and work done by magnetic field?
Since velocity is parallel to field ($\displaystyle {\theta = 0^\circ}$), magnetic force $\displaystyle {F = q v B \sin 0^\circ = 0}$.
1. Trajectory is an un-deflected straight line path.
2. Work done by magnetic field is Zero ($\displaystyle {W = \vec{F} \cdot \vec{d} = 0}$).
Q.19. Define magnetic dipole moment of a current loop. State its SI unit and direction.
$\displaystyle {\vec{M} = N I \vec{A}}$.
SI Unit: $\displaystyle {\text{A}\cdot\text{m}^2}$ or $\displaystyle {\text{J/T}}$.
Direction: Perpendicular to plane of loop, given by Right-Hand Thumb Rule (curled fingers along current, thumb gives $\displaystyle {\vec{M}}$).
Q.20(I). A positive charge is moving vertically upwards. When it enters a region of uniform magnetic field directed towards North, what is the direction of magnetic force acting on the charge?
Q.20(II). Define current sensitivity and voltage sensitivity of a moving coil galvanometer.
Part I: Using Fleming's Left-Hand Rule or Right-Hand Cross Product ($\displaystyle {\vec{F} = q(\vec{v} \times \vec{B})}$): Velocity $\displaystyle {\vec{v}}$ is Upward ($+\hat{k}$), Field $\displaystyle {\vec{B}}$ is North ($+\hat{j}$). $\displaystyle {\vec{F} \propto \hat{k} \times \hat{j} = -\hat{i}}$ (directed towards West).
OR Part II:
Current Sensitivity: Deflection produced per unit current $\displaystyle {S_I = {\frac{{\theta}}{{I}}} = {\frac{{N A B}}{{C}}}}$.
Voltage Sensitivity: Deflection produced per unit applied voltage $\displaystyle {S_V = {\frac{{\theta}}{{V}}} = {\frac{{N A B}}{{C R}}}}$.
Q.21. A galvanometer coil has resistance $\displaystyle {R_g = 15\,\Omega}$ and shows full-scale deflection for current $\displaystyle {I_g = 4\text{ mA}}$. How will you convert it into an ammeter of range $\displaystyle {0-6\text{ A}}$?
$\displaystyle {S = {\frac{{I_g R_g}}{{I - I_g}}} = {\frac{{(4 \times 10^{-3}) \times 15}}{{6 - 0.004}}} = {\frac{{0.06}}{{5.996}}} \approx 0.01\,\Omega}$.
By connecting a shunt resistance of $\displaystyle {0.01\,\Omega}$ in parallel across the galvanometer.
SECTION C (21 Marks)
Q.22. A thin circular wire loop of radius $\displaystyle {r}$ carrying current $\displaystyle {I}$ has magnetic dipole moment $\displaystyle {M}$. If the same wire is reshaped into a square loop carrying same current $\displaystyle {I}$, calculate new magnetic moment $\displaystyle {M_1}$ and ratio $\displaystyle {M / M_1}$.
Initial circular area $\displaystyle {A_c = \pi r^2 = \pi {\left( {\frac{{L}}{{2\pi}}} \right)}^2 = {\frac{{L^2}}{{4\pi}}} \implies M = I A_c = {\frac{{I L^2}}{{4\pi}}}}$.
For square loop of side $a$: $\displaystyle {4a = L \implies a = L/4}$.
Square area $\displaystyle {A_s = a^2 = {\frac{{L^2}}{{16}}} \implies M_1 = I A_s = {\frac{{I L^2}}{{16}}}}$.
Ratio $\displaystyle {{\frac{{M}}{{M_1}}} = {\frac{{I L^2 / 4\pi}}{{I L^2 / 16}}} = {\frac{{16}}{{4\pi}}} = {\frac{{4}}{{\pi}}}}$.
Q.23. A circular arc of wire of radius of curvature $\displaystyle {r}$ subtends an angle of $\displaystyle {\pi/5\text{ rad}}$ at its centre. If current $\displaystyle {I}$ flows through it, find the magnetic field induction $\displaystyle {B}$ at its centre.
For an arc subtending angle $\displaystyle {\theta}$ in radians, field is fraction $\displaystyle {{\frac{{\theta}}{{2\pi}}}}$ of full loop:
$\displaystyle {B = {\left( {\frac{{\theta}}{{2\pi}}} \right)} B_0 = {\left( {\frac{{\pi/5}}{{2\pi}}} \right)} {\left( {\frac{{\mu_0 I}}{{2 r}}} \right)} = {\frac{{1}}{{10}}} {\left( {\frac{{\mu_0 I}}{{2 r}}} \right)} = {\frac{{\mu_0 I}}{{20 r}}}}$.
Q.24. State Ampere’s Circuital Law. Use it to derive an expression for the magnetic field $\displaystyle {B}$ inside a long straight solenoid having $\displaystyle {n}$ turns per unit length carrying current $\displaystyle {I}$.
Consider rectangular Amperian loop $abcd$ of length $h$. Field outside is zero; along sides $bc$ and $da$, $\displaystyle {\vec{B} \perp d\vec{l}}$.
$\displaystyle {\oint \vec{B} \cdot d\vec{l} = \int_a^b \vec{B} \cdot d\vec{l} = B h}$.
Total enclosed current $\displaystyle {I_{\text{enclosed}} = (n h) I}$.
Applying Ampere's Law: $\displaystyle {B h = \mu_0 (n h I) \implies B = \mu_0 n I}$.
Q.25. Two long straight parallel conductors carrying currents $\displaystyle {I_1}$ and $\displaystyle {I_2}$ are separated by distance $\displaystyle {d}$ in vacuum. Derive expression for force per unit length between them. Hence define '1 Ampere'.
Force on length $L$ of wire 2: $\displaystyle {F = I_2 L B_1 = I_2 L {\left( {\frac{{\mu_0 I_1}}{{2\pi d}}} \right)}}$.
Force per unit length: $\displaystyle {f = {\frac{{F}}{{L}}} = {\frac{{\mu_0 I_1 I_2}}{{2\pi d}}}}$.
Definition of 1 Ampere: 1 Ampere is that steady current which, when flowing through two infinitely long straight parallel conductors of negligible cross-section placed 1 metre apart in vacuum, produces between them a force of $\displaystyle {2 \times 10^{-7}\text{ N/m}}$ of length.
Q.26. Derive an expression for the torque $\displaystyle {\vec{\tau}}$ acting on a rectangular current loop of area $\displaystyle {A}$ carrying current $\displaystyle {I}$ placed in a uniform magnetic field $\displaystyle {\vec{B}}$. Draw diagram for radial field in MCG.
$\displaystyle {F = I a B}$. Perpendicular arm distance is $\displaystyle {b \sin\theta}$.
Torque $\displaystyle {\tau = F \times (b \sin\theta) = (I a B) (b \sin\theta) = I (a b) B \sin\theta = I A B \sin\theta}$.
For $N$ turns: $\displaystyle {\tau = N I A B \sin\theta \implies \vec{\tau} = \vec{M} \times \vec{B}}$.
Radial Field: Cylindrical pole pieces with soft iron core ensure magnetic field lines are always parallel to loop plane ($\displaystyle {\theta = 90^\circ \implies \sin\theta = 1}$), making torque maximum and linear: $\displaystyle {\tau = N I A B}$.
Q.27(I). Explain with circuit diagram how a galvanometer of resistance $\displaystyle {R_g}$ is converted into (i) an ammeter of range $\displaystyle {0-I}$, (ii) a voltmeter of range $\displaystyle {0-V}$.
Q.27(II). A solenoid of 1000 turns is wound uniformly on a glass tube $\displaystyle {4\text{ m}}$ long and $\displaystyle {0.3\text{ m}}$ in diameter. Find magnetic field $\displaystyle {B}$ and magnetic intensity $\displaystyle {H}$ at its centre for current $\displaystyle {4\text{ A}}$.
Part I: (i) Ammeter: Connect low resistance shunt $\displaystyle {S = {\frac{{I_g R_g}}{{I - I_g}}}}$ in parallel.
(ii) Voltmeter: Connect high resistance $\displaystyle {R = {\frac{{V}}{{I_g}}} - R_g}$ in series.
OR Part II: Turn density $\displaystyle {n = {\frac{{N}}{{L}}} = {\frac{{1000}}{{4}}} = 250\text{ turns/m}}$.
Magnetic intensity $\displaystyle {H = n I = 250 \times 4 = 1000\text{ A/m} = 10^3\text{ A/m}}$.
Magnetic field $\displaystyle {B = \mu_0 H = (4\pi \times 10^{-7}) \times 1000 = 4\pi \times 10^{-4}\text{ T} \approx 1.257 \times 10^{-3}\text{ T}}$.
Q.28. Torque on a 50 turn coil of area $\displaystyle {200\text{ cm}^2}$ carrying current $\displaystyle {20\text{ A}}$ held with its axis at right angles ($\displaystyle {90^\circ}$) to uniform field $\displaystyle {0.2\text{ T}}$ is:
$\displaystyle {\tau = N I A B \sin 90^\circ = 50 \times 20 \times 0.02 \times 0.2 \times 1 = 4\text{ N}\cdot\text{m}}$.
SECTION D: CASE STUDY QUESTIONS (8 Marks)
Q.29. Case Study 1: Velocity Selector & Cyclotron Motion.
When crossed electric and magnetic fields ($\displaystyle {\vec{E} \perp \vec{B}}$) act on a moving charge $\displaystyle {q}$, electric force $\displaystyle {q\vec{E}}$ balances magnetic force $\displaystyle {q(\vec{v} \times \vec{B})}$. Only particles with specific velocity $\displaystyle {v = E/B}$ pass undeflected.
(i) Condition for a charged particle to pass undeflected through crossed fields:
(a) $\displaystyle {v = E B}$ (b) $\displaystyle {v = E/B}$ (c) $\displaystyle {v = B/E}$ (d) $\displaystyle {v = \sqrt{E/B}}$
(ii) Work done by magnetic field on a particle undergoing helical motion is:
(a) Maximum (b) Zero (c) Minimum (d) Dependent on pitch
(iii) Kinetic energy of particle of mass $m$ in magnetic field $B$ with radius $r$:
(a) $\displaystyle {{\frac{{q^2 B^2 r^2}}{{2 m}}}}$ (b) $\displaystyle {{\frac{{q B r}}{{2 m}}}}$ (c) $\displaystyle {{\frac{{q^2 B r^2}}{{m}}}}$ (d) $\displaystyle {{\frac{{q B^2 r}}{{2 m}}}}$
(iv) Frequency of revolution of charged particle in magnetic field is independent of:
(a) Charge (b) Magnetic field (c) Speed and Radius (d) Mass
Q.30. Case Study 2: Moving Coil Galvanometer (MCG).
A moving coil galvanometer uses a coil suspended in a radial magnetic field created by concave magnetic poles and a soft iron core.
(i) Purpose of radial magnetic field in MCG:
(a) Make field strong (b) Keep plane of coil parallel to field ($\theta = 90^\circ$) for linear scale (c) Reduce resistance
(ii) Role of soft iron core inside MCG coil:
(a) Increases magnetic field strength and makes field radial (b) Prevents heating
(iii) To double current sensitivity without changing voltage sensitivity, one must:
(a) Double turns $N$ (b) Double turns $N$ and double resistance $R$ (c) Double area $A$
SECTION E (15 Marks)
Q.31(I). (a) Using Biot-Savart law, derive an expression for the magnetic field $\displaystyle {B}$ at a point on the axis of a circular current loop of radius $\displaystyle {R}$ at distance $\displaystyle {x}$ from centre.
(b) Hence evaluate field at centre ($\displaystyle {x = 0}$).
Q.31(II). (a) Derive expression for magnetic field inside an ideal solenoid $\displaystyle {B = \mu_0 n I}$.
(b) An electron moving with speed $\displaystyle {3 \times 10^7\text{ m/s}}$ enters perpendicular to field $\displaystyle {6 \times 10^{-4}\text{ T}}$. Calculate radius and frequency of circular orbit.
Part I: Axis field: $\displaystyle {B = {\frac{{\mu_0 N I R^2}}{{2 (R^2 + x^2)^{3/2}}}}}$. At centre ($\displaystyle {x=0}$): $\displaystyle {B = {\frac{{\mu_0 N I R^2}}{{2 R^3}}} = {\frac{{\mu_0 N I}}{{2 R}}}}$.
OR Part II (b): Radius $\displaystyle {r = {\frac{{m v}}{{q B}}} = {\frac{{(9.1 \times 10^{-31}) \times (3 \times 10^7)}}{{(1.6 \times 10^{-19}) \times (6 \times 10^{-4})}}} = 0.284\text{ m} = 28.4\text{ cm}}$.
Frequency $\displaystyle {f = {\frac{{q B}}{{2\pi m}}} = {\frac{{(1.6 \times 10^{-19}) \times (6 \times 10^{-4})}}{{2\pi \times (9.1 \times 10^{-31})}}} = 1.68 \times 10^7\text{ Hz} = 16.8\text{ MHz}}$.
Q.32(I). (a) Describe principle, construction, and working of a Moving Coil Galvanometer with diagram.
(b) Define current sensitivity and voltage sensitivity. Why does increasing current sensitivity not necessarily increase voltage sensitivity?
Q.32(II). (a) Derive expression for torque on current loop in uniform magnetic field.
(b) Two long parallel wires separated by $\displaystyle {10\text{ cm}}$ carry currents $\displaystyle {2\text{ A}}$ and $\displaystyle {5\text{ A}}$ in opposite directions. Find force per unit length on either wire.
Part I (b): $\displaystyle {S_I = {\frac{{N A B}}{{C}}}}$, $\displaystyle {S_V = {\frac{{N A B}}{{C R}}}}$. Doubling turns $N$ doubles $S_I$. But doubling $N$ also doubles wire length and resistance $R$, leaving $\displaystyle {S_V = {\frac{{(2N) A B}}{{C (2R)}}} = S_V}$ unchanged.
OR Part II (b): Currents in opposite directions repel:
$\displaystyle {f = {\frac{{\mu_0 I_1 I_2}}{{2\pi d}}} = {\frac{{(4\pi \times 10^{-7}) \times 2 \times 5}}{{2\pi \times 0.1}}} = 2 \times 10^{-5}\text{ N/m}}$ (Repulsive).
Q.33(I). (a) State Ampere's circuital law.
(b) Derive magnetic field due to an infinitely long straight current-carrying wire at distance $r$: $\displaystyle {B = {\frac{{\mu_0 I}}{{2\pi r}}}}$.
(c) Plot graph of $B$ vs $r$.
Q.33(II). Explain trajectory of charged particle entering magnetic field at angle $\displaystyle {\theta}$ with field. Derive expressions for (i) radius of circle, (ii) time period, (iii) pitch of helix.
Part I: Amperian loop of radius $r$: $\displaystyle {\oint \vec{B} \cdot d\vec{l} = B (2\pi r) = \mu_0 I \implies B = {\frac{{\mu_0 I}}{{2\pi r}}}}$. Graph of $B$ vs $r$ is a rectangular hyperbola ($\displaystyle {B \propto 1/r}$).
OR Part II: Components: $\displaystyle {v_\parallel = v \cos\theta, v_\perp = v \sin\theta}$.
(i) Radius $\displaystyle {r = {\frac{{m v_\perp}}{{q B}}} = {\frac{{m v \sin\theta}}{{q B}}}}$.
(ii) Period $\displaystyle {T = {\frac{{2\pi m}}{{q B}}}}$.
(iii) Pitch $\displaystyle {p = v_\parallel \times T = (v \cos\theta) {\left( {\frac{{2\pi m}}{{q B}}} \right)}}$.
ASSIGNMENT – 2 (MCQ PRACTICE)
1. Magnetic field at centre of an $N$-turn circular coil of radius $R$ carrying current $I$ is proportional to:
2. Current in a circular loop is doubled and its radius is halved. Magnetic field at centre becomes:
Explanation: $\displaystyle {B' = {\frac{{\mu_0 (2I)}}{{2 (R/2)}}} = 4 {\left( {\frac{{\mu_0 I}}{{2 R}}} \right)} = 4B}$.
3. Magnetic field inside an ideal long solenoid depends on:
4. Which law is fundamentally based on a current element $\displaystyle {I d\vec{l}}$?
5. Acceleration of a charge $\displaystyle {q}$ moving parallel to uniform magnetic field $\displaystyle {B}$ is:
ASSIGNMENT – 3 (ASSERTION & REASON)
1. Assertion (A): When current in a solenoid is reversed in direction while keeping magnitude constant, stored magnetic energy remains unchanged.
Reason (R): Magnetic field energy density is proportional to square of magnetic field ($\displaystyle {u = B^2 / 2\mu_0}$).
2. Assertion (A): A proton and an alpha particle having same kinetic energy move in circular paths of equal radii in uniform magnetic field.
Reason (R): Radius of circular path in magnetic field is given by $\displaystyle {r = \sqrt{2 m K} / (q B)}$.
Explanation: $\displaystyle {r \propto \sqrt{m}/q}$. For proton $\displaystyle {\sqrt{m_p}/e}$; for alpha particle $\displaystyle {\sqrt{4m_p}/(2e) = 2\sqrt{m_p}/(2e) = \sqrt{m_p}/e}$. Radii are equal!
ASSIGNMENT – 4 (CASE STUDY QUESTIONS)
Case Study: Force Between Parallel Currents and Definition of Ampere
Parallel current-carrying wires exert magnetic forces on each other due to the interaction of their magnetic fields.
Q1. Force between antiparallel currents is: (a) Attractive (b) Repulsive (c) Zero
Q2. Force per unit length between $1\text{ A}$ currents $1\text{ m}$ apart in air: (a) $\displaystyle {2 \times 10^{-7}\text{ N/m}}$ (b) $\displaystyle {10^{-7}\text{ N/m}}$
Q3. If distance between parallel wires is doubled, force per unit length: (a) Doubled (b) Halved
ASSIGNMENT – 5 (CONCEPTUAL & DERIVATIONS)
SHORT ANSWER QUESTIONS
1. An electron does not suffer any deflection while passing through a region of magnetic field. What can you conclude about direction of magnetic field?
2. Why is a soft iron core placed inside the coil of a moving coil galvanometer?
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