Definitions [14]
Define ampere.
Current passed through each of the two infinitely long parallel straight conductors kept at a distance of one meter apart in vacuum causes each conductor to experience a force of 2 × 10-7 newton per meter length of the conductor.
A long solenoid is a coil whose length is much greater than its radius, producing a uniform magnetic field inside and nearly zero field outside.
Define the term ‘current sensitivity’ of a moving coil galvanometer.
The current sensitivity of a galvanometer is defined as the deflection produced in the galvanometer when a unit current flows through it.
Mathematically, it can be given by:
IS = `(NBA)/k`
Where k is the couple per unit twist.
Current sensitivity is defined as the deflection e per unit current.
1 ampere is the current which when flowing in each of two infinitely-long parallel conductors 1 metre apart in vacuum, produces between them a force of exactly 2 × 10-7 newton per metre of length.
If a charge of 1 coulomb moving with a velocity of 1 metre per second perpendicular to a uniform magnetic field experiences a force of 1 newton, then the magnitude of the field is 1 tesla.
The region near a magnet, where a magnetic needle experiences a torque and rests in a definite direction, is called 'magnetic field'.
If we stretch our right-hand palm such that the thumb points in the direction of the current (I) and the stretched fingers in the direction of the magnetic field \[\vec B\], then the force \[\vec F\] on the conductor will be perpendicular to the palm in the direction of pushing by the palm.
A charged particle when moving in a region experiences a deflecting force then a magnetic field is said to exist in that region. This field is denoted by \[\vec B\] and is also called ‘magnetic induction’.
If the forefinger, the middle finger, and the thumb of the left hand are stretched at right angles to one another, such that the forefinger points in the direction of the magnetic field \[\vec B\] and the middle finger in the direction of the current I, then the thumb will point in the direction of the force \[\vec F\] on the conductor.
The direction in the magnetic field along which a current-carrying conductor does not experience any force is called the direction of the magnetic field.
The current sensitivity of a galvanometer is defined as the deflection produced in the galvanometer when a unit current flows through it.
“The magnetic moment of a coil is equal to the maximum torque acting on a coil when placed in a uniform magnetic field of unit strength".
Units: NIA as A-m²
The voltage sensitivity of a galvanometer is defined as the deflection produced in the galvanometer when a unit voltage is applied across its coil.
A moving-coil galvanometer is a sensitive instrument used to detect and measure electric current, based on the principle that a current-carrying coil placed in a magnetic field experiences a deflecting torque proportional to the current.
Formulae [12]
\[\vec{E}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\hat{r}\]
\[\vec{F}=I\vec{l}\times\vec{B}\]
\[\vec{B}=\frac{\mu_0IR^2}{2(x^2+R^2)^{3/2}}\hat{i}\]
Where:
- I = current
- R = radius of loop
- x = distance from centre along axis
- μ0 = permeability of free space
B = μ0nI
Where:
- μ0 = permeability of free space
- n = number of turns per unit length
- I = current
\[\vec{m}=I\vec{A}\]
For N turns:
\[\vec{m}=NI\vec{A}\]
\[B=\frac{\mu_0}{4\pi}\frac{2m}{x^3}\]
Magnetic Field on the Equatorial Line (for x ≫ R)
\[B=\frac{\mu_0}{4\pi}\frac{m}{x^3}\]
\[\vec F\] = q(\[\vec E\] + \[\vec v\] × \[\vec B\])
\[\frac{F}{L}=\frac{\mu_0}{2\pi}\frac{I_1I_2}{r}\]
OR
\[F=\frac{\mu_0I_aI_bL}{2\pi d}\]
F = Bq v sin θ
or,
B = \[\frac{F}{qv\sin\theta}\]
Units: 1 T = 1 NA-1 m-1 = 1 Wb m-2
Dimensions: [M T-2A-1].
\[\frac{\phi}{V}=\frac{NAB}{cR}\]
OR
\[\frac{\phi}{V}=\frac{NAB}{kR}\]
τ = Ι Α B sin θ
Vector form:
\[\vec τ\] = \[\vec m\times\vec B\]
\[\frac{\phi}{I}=\frac{NAB}{c}\]
OR
\[\frac{\phi}{I}=\frac{NAB}{k}\]
Theorems and Laws [2]
Statement
The magnetic field produced at a point due to a small current element is directly proportional to the current through the element, the length of the element, and the sine of the angle between the element and the line joining it to the point, and inversely proportional to the square of the distance of the point from the element.
Proof
Consider a conductor of arbitrary shape carrying a current I. Let dl be a small current element of the conductor and r the distance of this element from a point P.

According to experimental observations:
-
The magnetic field dB at point P is proportional to the current I:
dB ∝ I
-
It is proportional to the length of the current element dl:
dB ∝ dl -
It is proportional to sin θ, where θ is the angle between dl and the line joining the element to point P:
dB ∝ sinθ
-
It is inversely proportional to the square of the distance r:
dB ∝ \[\frac {1}{r^2}\]
Combining all these relations: dB ∝ \[\frac{Idl\sin\theta}{r^2}\]
For air or vacuum, this proportionality is written as:
dB = \[\frac{\mu_{0}}{4\pi}\frac{Idl\sin\theta}{r^{2}}\]
The direction of dB is perpendicular to the plane containing dl and the position vector \[\vec r\]. Hence, in vector form:
d\[\vec B\] = \[\frac{\mu_{0}}{4\pi}\frac{I(d\vec l\times \vec r)}{r^{3}}\]
The magnetic field due to the entire conductor is obtained by integrating over its length:
\[\vec B\] = ∫ d\[\vec B\] = \[\frac{\mu_0I}{4\pi}\int\frac{d\vec l\times\mathbf{\vec r}}{r^3}\]
Conclusion
Biot–Savart’s law gives the magnitude and direction of the magnetic field produced by a current-carrying conductor. It shows that the magnetic field depends on the current element, its orientation, and its distance from the point, and that it forms the basis for calculating magnetic fields due to finite-sized conductors.
Biot-Savart Law in Terms of Current Density j
\[j=\frac{I}{A}=\frac{Idl}{Adl}=\frac{Idl}{dV},\]
where dV is the volume of current element.
\[\therefore\] Idl = j dV
d\[\vec B\] = \[\frac{\mu_0}{4\pi}\frac{\vec j\times\vec r}{r^3}\]
Units: kg m s2 A-2
Dimensions: [M L T2A-2].
Statement:
The line integral of the magnetic field B around any closed path in free space is equal to μ0 times the net steady current enclosed by the path.
\[\oint\vec{B}\cdot d\vec{l}=\mu_0I\]
Explanation/Proof (for a long straight conductor):
Consider a long, straight conductor carrying a steady current I.
The magnetic field at a distance r from the conductor is
B = \[\frac{\mu_0I}{2\pi r}\]
The field lines are circular, and B is tangential and constant in magnitude along a circular path of radius r.
Hence,
\[\oint\vec{B}\cdot d\vec{l}=B\oint dl=B(2\pi r)\]
Substituting the value of B,
\[\oint\vec{B}\cdot d\vec{l}=\mu_0I\]
Thus, Ampere’s circuital law is verified.
Conclusion:
- Ampere’s circuital law is valid for steady currents and time-independent magnetic fields.
- It is applicable to any closed path.
- It is especially useful for conductors with high symmetry, where calculating the magnetic field is easier.
- It is analogous to Gauss’s law in electrostatics.
Key Points
- A current-carrying conductor placed in a magnetic field experiences a force when the direction of current is not parallel to the magnetic field.
- The direction of force reverses when the direction of current or the direction of magnetic field is reversed, and no force acts when current flows parallel to the magnetic field.
- Existence of Force: A current-carrying conductor placed in a uniform magnetic field experiences a force perpendicular to both the direction of current and the magnetic field.
- Direction of Force: The direction of force can be determined by Fleming’s Left-Hand Rule or Right-Hand Palm Rule.
- Expression for Force
The magnitude of force acting on a straight conductor of length l is
F = B I l sin θ
where θ is the angle between the conductor and the magnetic field. - Special Cases:
When θ = 0∘, the force is zero.
When θ = 90∘, the force is maximum, Fmax = B I l. - Vector Form: The magnetic force on a current-carrying conductor can be written as
\[\vec F\] = I \[\vec l\] × \[\vec B\]
- For a long straight current-carrying wire, the magnetic field at a distance r is
B = \[\frac{\mu_0I}{2\pi r}\]The field lines are concentric circles and B ∝ \[\frac {1}{r}\] - Inside a long, straight solenoid, the magnetic field is uniform and given by
B = μ0nI
where n is the number of turns per unit length; the field outside the solenoid is nearly zero. - The magnetic field at the end of a long solenoid is half of that at the centre:
Bend = \[\frac {1}{2}\]μ0nI - For a solenoid with a magnetic core of relative permeability μr, the field becomes
B = μ0μrnI - In a toroidal (endless) solenoid, the magnetic field exists only inside the core and is
B = \[\frac{\mu_0NI}{2\pi a}\] = μ0nI
The field is zero outside the toroid and varies with the radial distance a.
- Magnetic force on a moving charge: A charged particle of charge q moving with velocity v in a uniform magnetic field B experiences a force
F = qv B sin θ,
where θ is the angle between \[\vec V\] and \[\vec B\]. - Case θ = 0∘(velocity parallel to field): When the particle moves parallel to the magnetic field, the magnetic force is zero, and the particle moves in a straight line without deviation.
- Case θ = 90∘ (velocity perpendicular to field): When the velocity is perpendicular to the magnetic field, the particle experiences a constant force perpendicular to its velocity and moves in a circular path at a constant speed.
- Radius of circular path: In perpendicular entry, the magnetic force provides the centripetal force, giving the radius:
r = \[\frac {mv}{qB}\]Hence, the radius is proportional to the momentum of the particle. - Time period and frequency: The time period of the revolution is
T = \[\frac {2πm}{qB}\]and is independent of the speed of the particle. The angular frequency is
ω = \[\frac {qB}{m}\]. - Case 0∘ < θ < 90∘ (oblique entry): When the particle enters at an angle, it follows a helical path formed by the combination of circular motion (due to perpendicular component) and linear motion (due to parallel component).
- Speed and energy remain constant: For all angles between \[\vec v\] and \[\vec B\], the speed, kinetic energy, and time period remain unchanged, but the direction of momentum changes when θ ≠ 0∘.
- Principle: A cyclotron accelerates charged particles using an alternating electric field and a magnetic field to bring them repeatedly to the accelerating gap.
- Construction: It has two D-shaped hollow electrodes (dees) placed in a uniform magnetic field perpendicular to their plane.
- Motion of Particle: Inside the dees, the particle moves in a circular path with radius
r = \[\frac {mv}{qB}\]and gains energy only in the gap. - Resonance Condition: For continuous acceleration, the frequency of the applied voltage equals the particle’s cyclotron frequency:
ν = \[\frac {qB}{2πm}\] - Limitations: A cyclotron cannot accelerate neutral particles or electrons and is ineffective at very high (relativistic) speeds.
- A charged particle q moving with velocity v in a uniform magnetic field B experiences a force
F = Bqv sinθ or \[\vec F\] = q(\[\vec v\] × \[\vec B\])
where θ is the angle between \[\vec v\] and \[\vec B\]. - The magnetic force acts perpendicular to both the velocity \[\vec v\] and the magnetic field \[\vec B\]; its direction is given by the right-hand screw rule for positive charges and is opposite for negative charges.
- Fleming’s left-hand rule can also be used: forefinger → magnetic field, middle finger → current (direction of positive charge motion), and thumb → direction of magnetic force.
- Current produces magnetism: Oersted showed that a current-carrying conductor produces a magnetic field around it, as indicated by the deflection of a magnetic needle.
- Direction and strength: Reversing the direction of the current reverses the needle’s deflection, and increasing the current or reducing the distance increases the deflection.
- Moving charges and force: Since current is the flow of moving charges, the experiment shows that moving charges create magnetic fields and exert forces.
- Right-hand palm rule: Thumb shows current direction; perpendicular from the palm gives magnetic field direction.
- Right-hand thumb rule: Thumb points along current; curled fingers show magnetic field direction.
- Maxwell’s right-hand screw rule: Direction of screw motion gives current; direction of rotation gives magnetic field.
- The magnetic field at a point P at a distance r from a finite straight current-carrying conductor is
B = \[\frac{\mu_0I}{4\pi r}(\sin\phi_1+\sin\phi_2)\]
where ϕ1 and ϕ2 are the angles subtended by the conductor at P. - For special cases of a straight conductor:
Infinite length: B = \[\frac{\mu_0I}{2\pi r}\]
Point near one end: B = \[\frac{\mu_0I}{4\pi r}\]
The magnetic field is directly proportional to current I and inversely proportional to distance r. - The magnetic field at a point on the axis of a circular loop of radius a, carrying current I, at a distance x from the centre is
B = \[\frac{\mu_0Ia^2}{2(a^2+x^2)^{3/2}}\]For a coil of N turns, the field is multiplied by N. - At the centre of a circular current-carrying loop, the magnetic field is
B = \[\frac{\mu_0I}{2a}\]For a coil of N turns,
B = \[\frac{\mu_0NI}{2a}\] - The direction of the magnetic field is given by the right-hand rule. At the centre of a circular coil, magnetic field lines are nearly straight, parallel, and perpendicular to the plane of the coil, indicating a nearly uniform magnetic field.
- Coulomb’s law describes the electric field due to charges, while Biot–Savart law describes the magnetic field due to a current.
- In both cases, the field strength decreases with the square of the distance.
- Biot–Savart law is the magnetic counterpart of Coulomb’s law.
- The electric field depends only on distance, but the magnetic field also depends on the angle of the current element.
- The electric field acts along the line joining the source and the point, while the magnetic field acts perpendicular to it.
- A voltmeter is used to measure potential difference and is always connected in parallel across the points of measurement.
- An ideal voltmeter has infinite resistance so that it draws no current from the circuit.
- A galvanometer is converted into a voltmeter by connecting a high resistance in series with it.
- The resistance of a voltmeter is much greater than the resistance of the galvanometer.
- The higher the voltmeter's range, the greater its resistance; a lower-range voltmeter has lower resistance.
- An ammeter is used to measure electric current and is always connected in series in a circuit.
- An ideal ammeter has zero resistance, but a galvanometer has appreciable resistance and cannot be used directly as an ammeter.
- To convert a galvanometer into an ammeter, a low resistance shunt is connected in parallel with the galvanometer.
- The shunt allows most of the current to bypass the galvanometer, so only a small, safe current flows through the coil.
- The shunt's value depends on the ammeter's range and ensures that full-scale deflection corresponds to the desired maximum current.
- A current-carrying loop placed in a uniform magnetic field experiences a turning effect (torque).
- Equal and opposite forces act on opposite sides of the loop, forming a couple that tends to rotate the loop.
- The forces on the remaining sides cancel each other, so the net force on the loop is zero.
- The torque depends on the loop's orientation in the magnetic field and is zero at a particular position.
- This effect is the basic principle of galvanometers and electric motors.
- A moving-coil galvanometer is used to detect and measure small electric currents.
- It works on the principle that a current-carrying coil in a magnetic field experiences a torque.
- There are two types: suspended-coil (more sensitive) and pivoted-coil (Weston; more convenient).
- A radial magnetic field is used so that deflection is directly proportional to the current.
- The coil comes to rest when the deflecting torque equals the restoring torque.
- A shunt is connected in parallel to protect the galvanometer from high currents and to enable null-point measurements.
Important Questions [123]
- Find the Condition Under Which the Charged Particles Moving with Different Speeds in the Presence of Electric and Magnetic Field Vectors Can Be Used to Select Charged Particles of a Particular Speed.
- Two Identical Circular Wires P and Q Each of Radius R and Carrying Current ‘I’ Are Kept in Perpendicular Planes Such that They Have a Common Centre as Shown in the Figure.
- Two Long Straight Parallel Conductors Carrying Steady Currents I1 And I2 Are Separated by a Distance 'D'. Explain Briefly, with the Help of a Suitable Diagram, How the Magnetic Field Due to One
- Depict the Behaviour of Magnetic Field Lines in the Presence of a Diamagnetic Material?
- A Point Charge Q Moving with Speed V Enters a Uniform Magnetic Field B that is Acting into the Plane of the Paper as Shown. What is the Path Followed by the Charge Q and in Which Plane Does It Move?
- Two Identical Coils P and Q Each of Radius R Are Lying in Perpendicular Planes Such that They Have a Common Centre.
- If an Electric Field → E is Also Applied Such that the Particle Continues Moving Along the Original Straight Line Path, What Should Be the Magnitude and Direction of the Electric Field → E ?
- Sketch a Schematic Diagram Depicting Oscillating Electric and Magnetic Fields of an Em Wave Propagating Along + Z-direction ?
- Show with the Help of a Diagram How the Force Between the Two Conductors Would Change When the Currents in Them Flow in the Opposite Directions?
- The Motion of Copper Plate is Damped When It is Allowed to Oscillate Between the Two Poles of a Magnet. What is the Cause of this Damping?
- Two horizontal thin long parallel wires, separated by a distance r carry current I each in the opposite directions. The net magnetic field at a point midway between them will be ______.
- Using Biot-savart Law, Deduce the Expression for the Magnetic Field at a Point (X) on the Axis of a Circular Current Carrying Loop of Radius R.
- Use Biot-savart Law to Derive the Expression for the Magnetic Field on the Axis of a Current Carrying Circular Loop of Radius R. Draw the Magnetic Field Lines Due to a Circular Wire Carrying Current I.
- Two Identical Circular Coils, P and Q Each of Radius R, Carrying Currents 1 a and √3a Respectively, Are Placed Concentrically and Perpendicular to Each Other Lying in the Xy and Yz Planes.
- At a place, the horizontal component of earth's magnetic field is B and angle of dip is 60°. What is the value of horizontal component of the earth's magnetic field at equator?
- Derive the Expression for the Torque on a Rectangular Current Carrying Loop Suspended in a Uniform Magnetic Field.
- Use Biot-savart'S Law to Find the Expression for the Magnetic Field Due to a Circular Loop of Radius 'R' Carrying Current 'I', at Its Centre ?
- Read the following paragraph Consider the experimental set-up shown in the figure. Explain the reason for the jumping of the ring when the switch is closed in the circuit.
- A Long Straight Wire of a Circular Cross-section of Radius ‘A’ Carries a Steady Current ‘I’. the Current is Uniformly Distributed Across the Cross-section. Apply Ampere’S Circuital Law to Calculate
- Briefly explain various ways to increase the strength of the magnetic field produced by a given solenoid.
- A long straight wire of radius 'a' carries a steady current 'I'. The current is uniformly distributed across its area of cross-section. The ratio of the magnitude of magnetic field
- State Ampere’s circuital law.
- Write Maxwell'S Generalization of Ampere'S Circuital Law.
- Electron Drift Speed is Estimated to Be of the Order of mm s^−1. Yet Large Current of the Order of Few Amperes Can Be Set up in the Wire. Explain Briefly.
- Write Maxwell'S Generalization of Ampere'S Circuital Law
- Using Ampere’S Circuital Law, Obtain the Expression for the Magnetic Field Due to a Long Solenoid at a Point Inside the Solenoid on Its Axis ?
- Obtain the Expression for Mutual Inductance of a Pair of Long Coaxial Solenoids
- A Wire Ab is Carrying a Steady Current of 12 a and is Lying on the Table. Another Wire Cd Carrying 5 a is Held Directly Above Ab at a Height of 1 Mm.
- How is the Magnetic Field Inside a Given Solenoid Made Strong?
- Define Self-inductance of a Coil.
- Obtain the Expression for the Magnetic Energy Stored in an Inductor of Self-inductance L to Build up a Current I Through It.
- Derive the Expression for the Magnetic Field Due to a Solenoid
- Draw and Compare the Pattern of the Magnetic Field Lines in the Two Cases ?
- Use this Law to Obtain the Expression for the Magnetic Field Inside an Air Cored Toroid of Average Radius 'r', Having 'n' Turns per Unit Length and Carrying a Steady Current I.
- In What Respect is a Toroid Different from a Solenoid?
- An Observer to the Left of a Solenoid of N Turns Each of Cross Section Area 'A' Observes that a Steady Current I in It Flows in the Clockwise Direction. Depict the Magnetic Field Lines Due to the Solenoid Specifying Its Polarity and Show that It Acts as a Bar Magnet of Magnetic Moment M = NIA
- Derive an Expression for the Mutual Inductance of Two Long Co-axial Solenoids of Same Length Wound One Over the Other
- Obtain an Expression for the Energy Stored in a Solenoid of Self-inductance ‘L’ When the Current Through It Grows from Zero to ‘I’.
- Two Long Coaxial Insulated Solenoids, S1 and S2 of Equal Lengths Are Wound One Over the Other as Shown in the Figure. a Steady Current "I" Flow Thought the Inner Solenoid S1 to the Other End B, Which is Connected to the Outer Solenoid S2 Through Which the Same Current "I" Flows in the Opposite Direction So as to Come Out at End A.
- Define the Term Self-inductance of a Solenoid
- Define Mutual Inductance Between Two Long Coaxial Solenoids. Find Out the Expression for the Mutual Inductance of Inner Solenoid of Length L Having the Radius R1
- A Wire Ab is Carrying a Steady Current of 6 a and is Lying on the Table. Another Wire Cd Carrying 4 a is Held Directly Above Ab at a Height of 1 Mm.
- A Wire Ab is Carrying a Steady Current of 10 a and is Lying on the Table. Another Wire Cd Carrying 6 a is Held Directly Above Ab at a Height of 2 Mm. Find the Mass per Unit Length of the Wire Cd
- How Does One Understand this Motional Emf by Invoking the Lorentz Force Acting on the Free Charge Carriers of the Conductor?
- Two Infinitely Long Straight Parallel Wires, '1' and '2', Carrying Steady Currents I1 and I2 in the Same Direction Are Separated by a Distance D. Obtain the Expression for the Magnetic Field `B`Due to the Wire '1' Acting on Wire '2'. Hence Find Out, with the Help of a Suitable Diagram,
- Using the Concept of Force Between Two Infinitely Long Parallel Current Carrying Conductors, Define One Ampere of Current.
- The Figure Shows Three Infinitely Long Straight Parallel Current Carrying Conductors. Find the (I) Magnitude and Direction of the Net Magnetic Field at Point a Lying on Conductor 1
- Two Long Straight Parallel Conductors 'A' and 'B', Carrying Steady Currents Ia And Ib Are Separated by a Distance D. Write the Magnitude and Direction of the Magnetic Field Produced by the Conductor
- Two Infinitely Large Plane Thin Parallel Sheets Having Surface Charge Densities σ1 And σ2 (σ1 > σ2) Are Shown in the Figure.
- Derive the Expression for Force per Unit Length Between Two Long Straight Parallel Current Carrying Conductors. Hence Define One Ampere.
- Answer the Following Question. Two Infinitely Long Straight Wire A1 And A2 Carrying Currents I and 2i Flowing in the Same Direction Are Kept' Distance Apart.
- Beams of electrons and protons move parallel to each other in the same direction. They ______.
- Two long parallel wires kept 2 m apart carry 3A current each, in the same direction. The force per unit length on one wire due to the other is ______.
- A Square Loop of Side 'A' Carrying a Current I2 is Kept at Distance X from an Infinitely Long Straight Wire Carrying a Current I1 as Shown in the Figure. Obtain the Expression for the Resultant
- Assertion (A): The deflecting torque acting on a current-carrying loop is zero when its plane is perpendicular to the direction of the magnetic field. Reason (R): The deflecting torque
- A Magnetised Needle of Magnetic Moment 4.8 × 10−2 Jt−1 is Placed at 30° with the Direction of Uniform Magnetic Field of Magnitude 3 × 10−2 T. Calculate the Torque Acting on the Needle.
- A Rectangular Loop of Wire of Size 2 Cm × 5 Cm Carries a Steady Current of 1 A. a Straight Long Wire Carrying 4 a Current is Kept Near the Loop as Shown.
- A Rectangular Loop of Wire of Size 2.5 Cm × 4 Cm Carries a Steady Current of 1 A. a Straight Wire Carrying 2 a Current is Kept Near the Loop as Shown
- A planar loop of rectangular shape is moved within the region of a uniform magnetic field acting perpendicular to its plane. What is the direction and magnitude of the current induced in it?
- A Rectangular Loop of Wire of Size 4 Cm × 10 Cm Carries a Steady Current of 2 A. a Straight Long Wire Carrying 5 a Current is Kept Near the Loop as Shown. If the Loop and the Wire Are Coplanar, Find
- A Rectangular Loop of Size L × B Carrying a Steady Current I is Placed in a Uniform Magnetic Field → B . Prove that the Torque → τ Acting on the Loop is Give by → τ = → M × → B , Where → M
- With the Help of a Neat and Labelled Diagram, Explain the Principle and Working of a Moving Coil Galvanometer ?
- Define the Current Sensitivity of a Galvanometer ?
- Write Current Sensitivity of a Galvanomete S.I. Unit.
- Figure Shows Two Circuits Each Having a Galvanometer and a Battery of 3v.When the Galvanometers in Each Arrangement Do Not Show Any Deflection, Obtain the Ratio R1/R2.
- Write the Underlying Principle of a Moving Coil Galvanometer.
- A Galvanometer of Resistance G is Converted into a Voltmeter to Measure Upto V Volts by Connecting a Resistance R1 in Series with the Coil.
- Define the term ‘current sensitivity’ of a moving coil galvanometer.
- Increasing the Current Sensitivity of a Galvanometer May Not Necessarily Increase Its Voltage Sensitivity. Explain, Giving Reason.
- Why is It Necessary to Introduce a Cylindrical Soft Iron Core Inside the Coil of a Galvanometer?
- Draw a Labelled Diagram of a Moving Coil Galvanometer. Describe Briefly Its Principle and Working.
- Obtain the Expression for Current Sensitivity of Moving Coil Galvanometer.
- Explain the Significance of a Radial Magnetic Field When a Current-carrying Coil is Kept in It.
- State the Underlying Principle of Working of a Moving Coil Galvanometer. Write Two Reasons Why a Galvanometer Can Not Be Used as Such to Measure Current in a Given Circuit.
- A galvanometer shows full-scale deflection for current Ig. A resistance R1 is required to convert it into a voltmeter of range (0 - V) and a resistance R2 to convert it
- State how a moving coil galvanometer can be converted into an ammeter.
- Why Does a Galvanometer Show a Momentary Deflection at the Time of Charging Or Discharging a Capacitor?
- Why is it necessary to introduce a radial magnetic field inside the coil of a galvanometer?
- Why Does a Galvanometer When Connected in Series with a Capacitor Show a Momentary Deflection, When It is Being Charged Or Discharged?
- Define current sensitivity of a galvanometer.
- Explain, Giving Reasons, the Basic Difference in Converting a Galvanometer into (I) a Voltmeter and (Ii) an Ammeter?
- How is current sensitivity increased?
- Draw a Labelled Diagram of a Moving Coil Galvanometer and Explain Its Working. What is the Function of Radial Magnetic Field Inside the Coil?
- In the Meter Bridge Experiment, Balance Point Was Observed at J with Aj = L.(I) The Values of R and X Were Doubled and Then Interchanged. What Would Be the New Position of Balance Point?
- State the Principle of the Working of a Moving Coil Galvanometer, Giving Its Labeled Diagram ?
- Outline the Necessary Steps to Convert a Galvanometer of Resistance Rg into an Ammeter of a Given Range ?
- Can a galvanometer as such be used for measuring the current? Explain.
- What Can Be the Causes of Helical Motion of a Charged Particle?
- Magnetic Field Lines Can Be Entirely Confined Within the Core of a Toroid, but Not Within a Straight Solenoid. Why?
- An Iron Ring of Relative Permeability µR Has Windings of Insulated Copper Wire of N Turns per Meter. When the Current in the Windings is I, Find the Expression for the Magnetic Field in the Ring.
- A Bar Magnet of Magnetic Moment 6 J/T is Aligned at 60° with a Uniform External Magnetic Field of 0·44 T (A) the Work Done in Turning the Magnet to Align Its Magnetic Moment
- Sketch the Change in Flux, Emf and Force
- A Small Compass Needle of Magnetic Moment ‘M’ is Free to Turn About an Axis Perpendicular to the Direction of Uniform Magnetic Field ‘B’.
- In the Circuit Shown in the Figure, Find the Value of the Current Shown in the Ammeter A.
- A Short Bar Magnet of Magnetic Moment 0.9 J/T is Placed with Its Axis at 30° to a Uniform Magnetic Field. It Experiences a Torque of 0.063 J.(I) Calculate the Magnitude of the Magnetic Field.
- Show that the kinetic energy of the particle moving in a magnetic field remains constant.
- An Electron Moving Horizontally with a Velocity of 4 ✕ 104 M/S Enters a Region of Uniform Magnetic Field of 10−5 T Acting Vertically Upward as Shown in the Figure. Draw Its
- A Straight Wire of Mass 200 G and Length 1.5 M Carries a Current of 2 A. It is Suspended in Mid Air by a Uniform Magnetic Field B. What is the Magnitude of the Magnetic Field?
- Write the expression for the force F acting on a particle of mass m and charge q moving with velocity V in a magnetic field B , Under what conditions will it move in (i) a circular path and (ii) a helical path?
- A Flexible Wire of Irregular Shape, Abcd, as Shown in the Figure, Turns into a Circular Shape When Placed in a Region of Magnetic Field Which is Directed Normal to the Plane of the Loop Away
- A Proton and a Deuteron Having Equal Momenta Enter in a Region of a Uniform Magnetic Field at Right Angle to the Direction of a the Field. Depict Their Trajectories in the Field.
- Write the Expression for the Force, → F Acting on a Charged Particle of Charge ‘Q’, Moving with a Velocity → V in the Presence of Both Electric Field → F and Magnetic Field → B .
- Write the Expression for Lorentz Magnetic Force on a Particle of Charge ‘Q’ Moving with Velocity → V in a Magnetic Field → B . Show that No Work is Done by this Force on the Charged Particle.
- Write the Expression, in a Vector Form, for the Lorentz Magnetic Force → F Due to a Charge Moving with Velocity → V in a Magnetic Field → B . What is the Direction of the Magnetic Force?
- A Proton and an α-particle Move Perpendicular to a Magnetic Field. Find the Ratio of Radii of Circular Paths Described by Them When Both Have (I) Equal Velocities, and (Ii) Equal Kinetic Energy.
- A Neutron, an Electron and an Alpha Particle, Moving with Equal Velocities, Enter a Uniform Magnetic Field Going into the Plane of the Paper, as Shown. Trace Their Paths in the Field and Justify Your Answer.
- A potential difference V is applied across a conductor of length L and diameter D. How is the drift velocity, vd, of charge carriers in the conductor affected when (i) V is halved, (ii) L is doubled
- Velocity of Light in Glass is 2 × 108 M/S and in Air is 3 × 108 M/S. If the Ray of Light Passes from Glass to Air, Calculate the Value of Critical Angle.
- How Does the Path Followed by the Charge Get Affected If Its Velocity Has a Component Parallel to → B .
- What is the Ratio of the Velocity of the Wave in the Two Media of Refractive Indices μ1 and μ2 ?
- Write the Relation Between the Velocity of Propagation and the Magnitudes of Electric and Magnetic Fields.
- An α-particle and a proton are released from the centre of the cyclotron and made to accelerate.(i) Can both be accelerated at the same cyclotron frequency?Give reason to justify your answer.
- Draw a Schematic Sketch of a Cyclotron. Explain Clearly the Role of Crossed Electric and Magnetic Field in Accelerating the Charge.
- State the Principle of a Cyclotron.
- Obtain the Expression for the Cyclotron Frequency.
- A deuteron and a proton are accelerated by the cyclotron. Can both be accelerated with the same oscillator frequency? Give reason to justify your answer.
- A Proton and an Electron Travelling Along Parallel Paths Enter a Region of Uniform Magnetic Field, Acting Perpendicular to Their Paths. Which of Them Will Move in a Circular Path with Higher Frequency?
- Explain the Principle and Working of a Cyclotron with the Help of a Schematic Diagram. Write the Expression for Cyclotron Frequency.
- Why is this Property Necessary for the Operation of a Cyclotron?
- Answer the Following Question. State the Underlying Principle of a Cyclotron. Explain Its Working with the Help of a Schematic Diagram. Obtain the Expression for Cyclotron Frequency.
- State the underlying principle of a cyclotron. Write briefly how this machine is used to accelerate charged particles to high energies
- Deduce an Expression for the Frequency of Revolution of a Charged Particle in a Magnetic Field
- A Cyclotron'S Oscillator Frequency is 10 Mhz. What Should Be the Operating Magnetic Field for Accelerating Protons? If the Radius of Its 'Dees' is 60 Cm, Calculate the Kinetic Energy
Concepts [25]
- Magnetic force
- Sources and Fields of Magnetic Force
- Magnetic Field, Lorentz Force
- Force on a Current Carrying Conductor in a Magnetic Field
- Motion in a Magnetic Field
- Biot-Savart Law
- Magnetic Field on the Axis of a Circular Current Loop
- Ampere’s Circuital Law
- Solenoid and the Toroid - the Solenoid
- Force Between Two Parallel Currents, the Ampere
- Circular Current Loop as a Magnetic Dipole
- Torque on a Rectangular Current Loop in a Uniform Magnetic Field
- Moving Coil Galvanometer
- Oersted's Experiment
- Solenoid and the Toroid - the Toroid
- Magnetic Diapole
- Torque on a Current-Loop in a Uniform Magnetic Field
- Force on a Current - Carrying Conductor in a Uniform Magnetic Field
- Force on a Moving Charge in Uniform Magnetic and Electric Fields
- Straight and Toroidal Solenoids (Only Qualitative Treatment)
- The Magnetic Dipole Moment of a Revolving Electron
- Velocity Selector
- Cyclotron
- Overview: Moving Charges and Magnetic Field
- Overview: Torque on a Current-Loop : Moving-Coil Galvanometer
