Bihar Board - Class 12 Physics- Chapter 4: MOVING CHARGES AND MAGNETISM Long Answer Question
Long Question Answer
1. Write any two important points of similarities and differences each between Coulomb’s law for the electrostatic field and . Biot-Savart’s law for the magnetic field.
Answer: Similarities: Both electrostatic field and magnetic field
follows the principle of superposition.
depends inversely on the square of distance from source to the point of interest.
Differences :
Electrostatic field is produced by a scalar source (q) and the magnetic
field is produced by a vector source (Idl).
Electrostatic field is along the displacement vector between source and point of interest; while magnetic field is perpendicular to the plane, containing the displacement vector and vector source.
Electrostatic field is angle independent, while magnetic field is angle
dependent between source vector and displacement vector.
2. Define magnetic field induction at a point in terms of the force experienced by a charge moving in a magnetic field and hence define one Tesla.
Answer:
Definition of B :
We know that F = Bq υ sin q
If. sin θ = 1 i.e., θ = 90°, υ = 1, q = 1, then F = B x 1 x 1 x 1
or B = F
Thus B is defined to be numerically equal to the force experienced by a unit charge moving perpendicular to the magnetic field with unit velocity.
Definition of IT :
If q = 1 C, υ = 1 ms-1θ = 90°, F = 1 N,
Then B = 1T
∴ 1T = 1 N C-1 m-1 s = 1 N-1 A-1 m-1.
Thus the magnetic field at a’ point' is said to be 1 T if 1 C charge moving perpendicular to the magnetic field with a velocity of 1 ms-1 experiences a force of 1 N at that point.
3. State some important features of Biot-Savart’s law.
Answer:
Following are some important features of the law :
It is applicable only to very small length current carrying conductors.
This law can’t be easily verified experimentally.
This law is analogous to Coulomb’s law in electrostatics.
It is valid for a symmetrical current distribution.
The direction of dB is perpendicular to both Idl (current element) and r (position vector).
dB = 0 on the axis of a thin linear current carrying conductor.
dB = 0 i.e., minimum when 0 = 0 or 180° and maximum when 0 = 90°.
4. Explain, giving reasons, the basic difference in converting a galvanometer into
(i) a voltmeter and (ii) an ammeter.
Answer:
(i) In converting a galvanometer into a voltmeter, a very high suitable resistance is connected in series to its coil. So, the galvanometer gives full scale deflection.
(ii) In converting a galvanometer into an ammeter, a very small suitable resistance is connected in parallel to its coil. The remaining pair of the current i.e., (I – Ig) flows through the resistance. Here I = Circuit current
and Ig = Current through galvanometer.
5. (a) How will you identify whether the magnetic field at any point is due to earth or due to some current carrying conductor?
(b) An electron beam is deflected in a given electric or magnetic field. How will you detect whether it has been deflected in an electric or magnetic field?
Answer:
(a) If the magnetic field is due to earth’s magnetism only then the magnetic needle will always lie along the N-S direction On the other hand if the magnetic field is due to a current carrying conductor, then its direction may differ from N-S direction and it will come back to N-S direction when the current passing through the conductor is switched off.
(b) When the electron beam is passed through an electric field or two parallel plates having a potential difference V across them, then the electrons are deflected towards the positive plate and the path of the electron beam is parabola. On the other hand, when the electron beam enters into a uniform magnetic field perpendicularly then its path is circular. If it enters at any . angle θ (≠ 0° and 180°), then the path will be helical.
6. What is a shunt? Write its uses. What are the advantages and disadvantages of shunt?
Answer: Shunt:
It is a wire of low resistance, connected in parallel to the coil of a galvanometer.
Uses:
A galvanometer is converted into an ammeter by using a shunt.
Advantages:
It protects the coil of the galvanometer from burning as well as the breaking of the pointer.
As the shunt is connected in parallel, the resultant resistance becomes less. So, when the shunted galvanometer (ammeter) is joined in series, then the value of the current does not change.
Disadvantages:
Due to shunt, the sensitivity of the galvanometer is reduced. So, it should be removed from the galvanometer when we have to obtain a null point.
7. A wire through which a current of 8A is flowing makes an angle of 30° with the direction of magnetic field of 0.15 tesla. Calculate the force acting per unit length of wire.
Solution:
Given : I = 8A, B = 0.15T, θ = 30°, F =?
formula: F = BIlsinθ
⇒ Fl = BIsinθ
= 0.15 x 8 x sin 30°
= 0.15 x 8 x 12
= 0.6N/m.
8. What is a cyclotron? Write its principle.
Answer: Principle:
It is based on the principle that when a positively charged particle is made to move again and again in a high frequency electric field and using a strong magnetic field, then it gets accelerated and acquires a sufficiently large amount of energy.
Construction:
It consists of two hollow D – shaped metallic chambers D1 and D2 called dees. These dees are separated by a small gap where a source of positively charged particles is placed. Dees are connected to a high frequency oscillator, which provides a high frequency electric field across the gap of the dees. This arrangement is placed between two poles of a strong electromagnet. The magnetic field due to this electromagnet is perpendicular to the plane of the dees.
Working:
If a positively charged particle (proton) is emitted from O, when D2 is negatively charged and the dee D1 is positively charged, it will accelerate towards D2. As soon as it enters D2 It is shielded from the electric field by a metallic chamber (enclosed space). Inside D2, it moves at right angles to the magnetic field and hence describes a semi – circle inside it. After completing the semicircle, it enters the gap between the dees at the time when the polarities of the dees have been reversed.
Now, the proton is further accelerated towards D1. Then it enters D1 and again describes the semicircle due to the magnetic field which is perpendicular to the motion of the proton. This motion continues till the proton reaches the periphery of the dee system. At this stage, the proton is deflected by the deflecting plate which then comes out through the window and hits the target.
9. A particle having 100 times the charge of an electron is revolving in a circle of radius 0-8 meter in each second. Calculate the intensity of the magnetic field.
Solution: Formula: B = 04 . 2IR
Given : q = 100 x 1.6 x 10-19
= 1.6 x 10-17 coulomb, R = 0.8 meter
Putting the value in formula, we get I = qt
= 1.6 x 10-171 = 1.6 x 10-17
B = 04 . 21.6 10-170.8
or B = 10-2 0
10. Calculate the torque on a 20 turns square coil of side 10 cm carrying a current of 12 A, when placed, making an angle of 30° with the magnetic field of 0.8 T.
Solution:
Given : A = 100cm2 100 x 10-4m2 = 10-2m2
n = 20,
I = 12A,
B = 0.8T,
ϕ = 30°
Formula: = nIABsinϕ
= 20 x 12 x 102x 0. 8sin30°
= 20 x 12 x 0.8 x 102 x 12
= 96 x 102
= 0.96 N – m.
11. An electron and a proton crossing along parallel routes enter a territory of the uniform magnetic field, working perpendicular to their ways. Which of them will go in a circular route with a higher frequency?
Answer:
The frequency of revolution is presented by
v = Bq/2πm ⇒ v ∝ 1/m.
As for me < mp, therefore, ve >vp
The proton will move in a circular path with a higher frequency. This is because the proton has more mass than the electron. The magnetic field strength (B) is the same for both particles, but the proton has a larger radius of revolution (rp) than the electron (re). This means that the proton will move in a circle with a higher frequency (v) than the electron.
12. (a) What is the potential difference between two points if a charge of 1 coulomb is placed at one point and a charge of 2 coulombs is placed at the other point?
(b) A point charge of 1 coulomb is placed at the origin of an x-y coordinate system. A second point charge of 2 coulombs is placed at the point (2, 3). What is the electric potential at the point (4, 5)?
Answer:
(a) The potential difference between the two points is 1 volt.
The potential difference between two points is equal to the work done in moving a unit of charge from one point to the other point. In this case, 1 coulomb of charge is moved from one point to the other. The work done is 1-volt x 1 coulomb = 1 volt.
(b) The electric potential at the point (4, 5) is 4 volts.
The electric potential at a point in space is determined by the amount of charge at that point and the strength of the electric field. In this case, there are two point charges: 1 coulomb at the origin and 2 coulombs at (2, 3). The electric potential at (4, 5) is the sum of the electric potentials at those two points.
13. (a) What is the electric flux through a square with sides of length L if the electric potential at one corner is V?
(b) What is the purpose of an electric field?
Answer:
(a) The electric flux through a square with sides of length L is Φ = EL.
The electric flux through a square is proportional to the electric potential at each corner. In this case, the electric potential at one corner is V. Therefore, the electric flux through the square is Φ = EL = (V)L.
(b) The purpose of an electric field is to force charged objects to move in a particular direction. It also creates an electric potential difference between the two points.
An electric field is a force that acts on charged objects. It creates an electric potential difference between the two points.
The greater the charge of the object, the greater the force of the electric field. The electric field also causes charged objects to move in a particular direction.
14. What is the Magnitude of Magnetic Force Per Unit Length on a Wire Carrying a Current of 8 and Making an Angle of 30° with the Direction of a Uniform Magnetic Field of 0.15T ?
Answer:Given that,
Current in the wire, I=8A
Magnitude of the uniform magnetic field, B=0.15T
Angle between the wire and magnetic field, θ=30°
We have the expression for magnetic force per unit length on the wire as,
F=BIsinθ
Substituting the given values, we get,
F=0.15×8×1×sin 30°
⇒ F=0.6N m-1
Thus, the magnetic force per unit length on the wire is found to be 0.6N m-1
15. A 3.0cm Wire Carrying a Current of 10A is Placed Inside a Solenoid Perpendicular to Its Axis. The Magnetic Field Inside the Solenoid Is Given to Be 0.27T. What is the Magnetic Force on the Wire?
Answer:We are given the following,
Length of the wire, l=3 cm=0.03m
Current flowing in the wire, I=10A
Magnetic field, B=0.27T
Angle between the current and magnetic field, θ=90°
(Since the magnetic field produced by a solenoid is along its axis and current carrying wire is kept perpendicular to the axis)
The magnetic force exerted on the wire is given as,
F=BIlsinθ
Substituting the given values,
F= 0.27×10×0.03sin 90°
⇒ F= 8.1×10-2 N
Clearly, the magnetic force on the wire is found to be 8.1×10-2 N. The direction of the force can be obtained from Fleming’s left-hand rule.
16. (a) How does the electric field vary with distance?
(b) What is an electric flux?
Answer:
(a) The electric field decreases as the distance between the charged object and the point of measurement increases. This is because the electric field is a direct result of the electric charge at that point. As the distance between the two points increases, the charge decreases, and so does the electric field.
(b) Electric flux is a measure of the amount of electric field passing through a given area. It is measured in volts per meter (V/m).
Electric flux is related to the electric field by the following equation:
Φ = E A
Where Φ is the electric flux, E is the electric field, and A is the area. This equation states that the electric flux is equal to the electric field multiplied by the area.
17. A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60º with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
Solution : Given:
Number of turns in the coil, n = 30
Radius of coil, r = 8 cm
Current through the coil, I = 6.0 A
Strength of magnetic field = 1.0 T
Angle between the direction of field and normal to coil, θ = 60°
We can understand that the counter torque required to prevent the coil from rotating is equal to the torque being applied by the magnetic field.
Torque on the coil due to magnetic field is given by,
T = n × B × I × A × sinθ …(1)
Where,
n = number of turns
B = Strength of magnetic field
I = Current through the coil
A = Area of cross-section of coil
A = πr2 = 3.14 × (0.08 × 0.08) = 0.0201m2 …(2)
θ = Angle between normal to cross-section of coil and magnetic field
Now, by putting the values in equation (1) we get,
⇒ T = 30 × 6.0T × 1A × 0.0201m2 × sin 60°
T = 3.133 Nm
Hence, the counter torque required to prevent the coil from rotating is 3.133 Nm.
18. Answer the Following Questions:
(a) A Magnetic Field That Varies in Magnitude from Point to Point but Has a Constant Direction (east to West) Is Set up in a Chamber. A Charged Particle Enters the Chamber and Travels Undeflected Along a Straight Path With Constant Speed. What Can You Say About the Initial Velocity of the Particle?
Ans: The initial velocity of the particle could either be parallel or be anti-parallel to the magnetic field. So, it travels along a straight path without undergoing any deflection in the field.
(b) A Charged Particle Enters an Environment of a Strong and Non-Uniform Magnetic Field Varying from Point to Point Both in Magnitude and Direction, and Comes Out of it Following a Complicated Trajectory. Would Its Final Speed Equal the Initial Speed If it Suffered No Collisions With the Environment?
Ans: Yes, the final speed of the charged particle would be equal to its initial speed as the magnetic force has the potential to change the direction of velocity even though not its magnitude.
(c) An Electron Travelling West to East Enters a Chamber Having a Uniform Electrostatic Field in the North to South Direction. Specify the Direction in Which a Uniform Magnetic Field Should Be Set up to Prevent the Electron from Deflecting from Its Straight-Line Path.
Ans: An electron traveling from West to East enters a chamber having a uniform electrostatic field along the North-South direction.
This moving electron stays undeflected when the electric force acting on it is equal and opposite of the magnetic field.
The magnetic force would stay directed towards the South. Also, according to Fleming’s left-hand rule, the magnetic field must be applied in a vertically downward direction.
19. A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude B inside the solenoid near its center.
Answer:
Given l = 80 cm = 0.8 m
Total number of turns, N = 5 × 400 = 2000
∴ No. of turns per unit length
n = Nl = 20000.8 = 2500
D = 1.8 cm, I = 8.0 A
∴ Magnetic field inside the solenoid near its center
B = 4π × 10-7 × 2500 × 8
= 2.5 × 10-2 T
20. A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal to the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
Answer:
n = 30, r = 8 cm
∴ A = π (8)2 cm2,
I = 6A,
B = 1T,
θ = 60°
Torque = nBIA sin θ
= 30 1682100100sin 60。
= 3.133 Nm.
21. What is Biot-Savart Law ?
Answer: Biot-Savart Law :- Consider an infinitesimal element dl of the conductor. The magnetic field dB due to this element is to be determined at a point P which is at a distance r from it. Let θ be the angle between dl and the position vector r. The direction of dl is the same as the direction of the current.
According to Biot-Savart law, the magnitude of the magnetic field dB is proportional to the current I, and the element length dl is inversely proportional to the square of the distance r. Its direction is perpendicular to the plane containing dl and r. Thus in vector notation,
dB∝Idl sin r2
dB = 04 Idl sin r2
Where 04 is a constant of proportionality. The above expression holds when the medium is a vacuum. The proportionality constant in the S.I. unit has a value, 04 =10-7 T-m/A.
We call 0 the permeability of free space.
22. What is the Moving Coil Galvanometer ?
Answer: The galvanometer consists of a coil, with many turns free to rotate about a fixed axis, in a uniform radial magnetic field. The coil is wrapped on a cylindrical soft iron core which enables the field to remain radial in all positions of the coil.
When a current flows through the coil, a torque acts on it.
=MB
=NIABsinθ
Here the symbols have their usual meaning. A spring Sp provides a counter torque Kφ that balances the magnetic torque NIAB, resulting in a steady angular deflection φ. In equilibrium
K = NIAB
where K is a torsional constant of the spring, that is restoring torque per unit twist. The deflection is indicated on the scale by a pointer attached to the spring. We have
= NABKI
The quantity in brackets is a constant for a given galvanometer.
23. Find the Torque on a Current Loop in a Uniform Magnetic Field.
Answer: A current loop ABCD is placed in a uniform magnetic field B. The dimensions are AB = l and BC = b. The plane of the loop makes an angle θ with a magnetic field. M is magnetic dipole moment. The angle between M and B is θ.
The four straight sections AB, BC, CD and DA experience forces due to the magnetic field. F represents the magnitude of the force on AB and CD while F′ represents the magnitude of forces on AD and BC. The directions are shown in the figure.
F1 = F2 = i l B sin90
F1 = F2 = i l B
Clearly, the net force is zero. But the forces on AB and CD constitute a couple. The torque due to them is
= F × perpendicular distance between the lines of action.
⇒ = F × bsinθ
= i l B bsinθ
= i(l b) B sinθ
= iA B sinθ
= MB sinθ
= M x B
24. What is Magnetic Force?
Answer:
Magnetic Force :- It is observed that when the charge is at rest it experiences almost no force. However, if the charge q is given a velocity v in the direction of the current, it is deflected towards the wire. Hence we conclude that the magnetic field exerts a force on a moving charged particle. The combination of the electric and magnetic force on a point charge is known as Lorentz Force.
Consider a point charge q moving with velocity v located at position vector r at a given time t. If an electric field E and a magnetic field B exist at that point, then force on the electric charge q is given by
F=q[E+v×B]
This force was first given by H. A. Lorentz, hence it is called the Lorentz force.
25. What is Ampere’s Circuital Law? Find Magnetic Field Due to a Straight Infinite Current-Carrying Wire.
Answer: Ampere’s circuital law :- Ampere’s circuital law states that the line integral of a steady magnetic field over a closed loop is equal to μ0 times the total current (Ie) passing through the surface bounded by the loop i.e.,
∫ B.dl=0Ie
where Ie is enclosed current
Magnetic Field Due to a Straight Infinite Current-Carrying Wire :-
Consider a straight infinite current-carrying wire. We draw an Amperian loop, a circle of radius r with wire as its axis. The Amperian loop exploits the symmetry of the situation as the magnetic field is tangential at every point in the loop. The magnetic field has equal magnitude at all the points.
Now using the Ampere's Law
∫ B.dl =0I
∫ Bdl.cos 0 =0I
B ∫dl=0I
B(2r) =0I
B = 0I2r
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