Check out the latest Numericals MCQ,Subjective Questions on CBSE/BSEB Class 12 Physics Chapter 4 Moving Charges and Magnetism. These questions are extremely helpful for Bihar Board Exams. NCERT Exemplar questions are also included.
Question 1.
A’ flat circular coil of average radius 3 cm has 100 turns. It is placed in the plane of paper and a current of 3 A is passed in it in the anticlockwise direction as seen from above. Calculate the magnetic induction at:
(a) the centre of the coil.
(b) at a point on the axis of the coil at a distance of 4 cm from its centre.
Answer:
Here, a = 3 cm = 3 x 10-2 m
N = 100
I = 3A
x = 4 cm = 0.04 m
(a) B at centre = ?
Using the formula, B = \(\frac{\mu_{0} \mathrm{NII}}{2 \mathrm{a}}\), we get
(b) Using formula, B = \(\frac{\mu_{0} \mathrm{NIa}^{2}}{2\left(a^{2}+x^{2}\right)^{\frac{3}{2}}}\), we get
and it acts along the axis of coil in upward direction.
Question 2.
Calculate the magnetic field at the centre of a coil in the form of a square of side 4 cm carrying a current of 5 A.
Answer:
Here, I = 5A flowing in the clockwise direction.
AB = CD = BC = AD = 4 cm = 0.04 m.
O is the centre of square where two diagonals intersect at right angles to each other. So ∠BOC = 90° or ∠BOM = ∠MOC = 45°.
OM = a = perpendicular distance of the conductor BC from O,
where B is to be calculated. Also OM = a = \({\frac{4}{2}}\)cm = 2 x 10-2 m.
Let B1 be the field at O due to the conductor BC.
Using this Formula
If B be the total magnetic field at O due to the square, then
B = 4 B1 = 4 x 3.54 x 10-5T
= 1.42 x 10-4 T
and it acts perpendicular to the plane of the loop and directed downwards.
Question 3.
An electron is moving at 10s ms-1 in a direction parallel to a current of 5 A flowing through an infinitely long straight wire separated by a perpendicular distance of 10 cm in air. Calculate the magnitude of the force experienced by the electron.
Answer:
Here, q = 1.6 x 10-19 C, υ = 106 ms-1 I = 5A,
Perpendicular distance = a – 10 cm = 0.10 m.
F = force on the electron = ?
Using formula, F = quB, where B = \(\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{a}}\), we get
F = qυ. \(\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{a}}\)
= 1.6 x 10-19 x 106 x \(\frac{4 \pi \times 10^{-7}}{2 \pi} \times \frac{5}{0.10}\)
= 1.6 x 10-18 N
Question 4.
A wire of length 20 cm and mass 50 mg lies in a direction 300 east of south. The earth’s magnetic field at this point is horizontal and has a magnitude of 8.0 x 103 T. What current must be passed through the wire so that it may float in air ? (g = 10 ms-2).
Answer:
Here, m mass of wire = 50 mg =50 x 10-6 kg.
g = 10 ms-2
B = 8.0 x 10-3T.
l = 20 cm = 0.20 m.
θ = 30°
Force acting on the current carrying wire in the magnetic field is
F = BIl sin θ
and acts vertically upward. The wire will float in air if the force on the wire due to \(\overrightarrow{\mathrm{B}}\) is balanced by the weight of the wire.
Question 5.
Compare the current sensitivity and voltage sensitivity in the following moving coil galvanometers:
MeterA:N = 30, A =1.5 x 10-3m2,B = 0.25T,R = 20 Ω
MeterB:N =35,A = 2.O x 10-3m2,B = 0.25T,R = 30 Ω
Springs in the two meters have the same torsional constants.
Answer:
Current sensitivity, Is = \(\frac {NBK}{k}\)
Question 6.
A galvanometerof resistance 50 gives full scale deflectioi for a current of 0.05 A. Calculate the length of shunt wire required t convert the galvanometer into an ammeter of range 0.5 A. The diamete of shunt wire is 4 mm and its resistivity is 5 x 10-7 Ωm.
Answer:
Here, Ig = 0.05 A
I = 0.5 A
G = 50 Ω
S = ?
Using the formula,
l = length of shunt =?
Also we know that
S = R = ρ \(\frac {1}{A}\) = ρ = \(\frac{l}{\pi r^{2}}\)
or l = \(\frac{\mathrm{S} \pi \mathrm{r}^{2}}{\rho}\) ……..(i)
Here, D = 4mm =4 x 10-3 m
p = 5 x 107 Ω m
∴ r = \(\frac {D}{2}\) = 2 x 103m
Question 7.
A galvanometer of 24Ω resistance can carry a full load of 500 μA. If it is shunted by a resistance of 3Ω, how much current can this system carry now without damage?
Solution :
Here, Ig = 500 μA = 500 x 10-6 A
G = 24 Θ
S = 3Ω
I = ?
Using formula, S = \(\frac{\mathrm{I}_{\mathrm{g}} \mathrm{G}}{\mathrm{I}-\mathrm{I}_{\mathrm{g}}}\), we get
I – Ig = \(\frac{\mathrm{I}_{\mathrm{g}} \mathrm{G}}{\mathrm{S}}\)
Question 8.
The current to be measured is arranged to go through two long parallel wires of equal lengths in opposite directions, one of which is linked to the point of current balance. The resulting repulsive form on the wire is balanced by placing a mass in the scale pan handing on the other side of the pivot. In one measurement, the mass in the scale pan 30g, the length of wires is 50 cm each and the separation between
them is 10 mm. What is the value of the current being measured? Take g = 9.8 ms-2 and assume that the arms of the balance are equal.
Solution:
Let F’ = fore experienced per unit length
I1 = I2 = I (say),
r = separation between wires
F = \(\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{I}_{1} \mathrm{I}_{2}}{\mathrm{r}}\)
If F be the total force on wire of length l attached to the scale pan
then
When the pan is blanced, then
Question 9.
A coil of radius 20 cm produces a magnetic field of 0.5 G at its centre when a current of 0.2 A flows through it. Find the number of turns the coil must have.
Answer:
Here, r = radius of coil = 20 cm = 0.20 m
B = magnetic field produced at the centre of the coil = 0.5 G , = 0.5 x 10-4 T
I = Current flowing in the coil = 0.2 A
n = number of turns in the coil = ?
Using formula,
Question 10.
An electron of kinetic energy 25 keV moves perpendicular to the direction of uniform magnetic field of 0.2 millitesla. Calculate the time period of rotation of the electron in the magnetic field.
Solution:
Here, B = magnetic field = 0.2 mT
= 0.2 x 10-3
K.E. of Electron = \(\frac {1}{2}\) mu2 = 25 keV
= 25 x 103 eV = 25 x 103 x 16 x 1019J
T = time period of rotation of electron in the magnetic field = ?
Using the relation,
Question 11.
A beam of protons with velocity 4 x 105 ms-1 enters a uniform magnetic field of 0.3 T at an angle of 60° to the magnetic field. Find the radius of the helical path taken by the proton beam. Also find the pitch of helix. Mass of Proton = 1.67 x 10″ 27 kg.
Solution:
Here, θ = 60°
m = 1.67 x 10-27kg
B = 0.3 T ‘
υ = 4 x 105 ms-1
e = 1.6 x 10-19 c
r = radius helical path = ?
p = pitch of helix = ?
Using formula, r = \(\frac{m v_{\perp}}{e B}\) = \(\frac{\mathrm{m} v \sin \theta}{\mathrm{eB}}\), we get
Question 12.
A rectangular loop of side 25 cm and 10 cm carrying a current of 15A is placed with its longer side parallel to a long straight conductor 2.0 cm apart carrying a current of 25A. What is the net force on the loop?
Solution :
Let XY be a straight wire and ABCD be a rectangular loop placed 2 cm a way from it as shown in figure.
Here, AB = CD = 25 cm 0.25 m
BC = DA = 10 cm = 0.10 m
Let r1 and r2 be the distances of the arms AB and CD from XY
∴ r1 = 2 cm = 0.02 m
and r2 = 2 + 10 = 12 cm – 0.12 m
I1 = 15 A flowing in ABCD
I2 = 25A flowing in XY
The arm AB will get attracted while CD will get repelled and forces on BC and AD will cancel each other.
if \(\overrightarrow{\mathrm{F}_{\mathrm{1}}}\) and \(\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{I}_{1} \mathrm{I}_{2}}{\mathrm{r}_{1}}\) x leanth of AB
= 10-7 x \(\frac{2 \times 15 \times 25}{0.02}\) x 0.25
= 9.375 x 10-4N towards XY
and \(\overrightarrow{\mathrm{F}_{\mathrm{2}}}\) = \(\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{I}_{1} \mathrm{I}_{2}}{\mathrm{r}_{2}}\) x length of CD
1.5625 x 10-4 N (a way from XY) i.e replusive
if \(\overrightarrow{\mathrm{F}}\) be the net force on the loop then
\(\overrightarrow{\mathrm{F}}\) = \(\overrightarrow{\mathrm{F}_{\mathrm{1}}}\) – \(\overrightarrow{\mathrm{F}_{\mathrm{2}}}\)
= 9.375 x 10-4 – 1.5625 x 10-4 towards XY
= 7.8125 x 10-4towards XY
Question 13.
A voltmeter reads 5.0V at full scale deflection and is graded according to its resistance per volt at full scale deflection as 5000 ΩV-1. How will you convert it into a voltmeter that reads 20V at full scale deflection? Will it still be graded as 5000 ΩV-1. Will you prefer this voltmeter to one that is graded as 2000 ΩV-1?
Solution:
Here, resistance per volt at full scale deflection = 5000 ΩV-1
∴ Currant for maximum deflection is given by
Ig = \(\frac{1}{5000}\) A = 2 x 10-4 A
(∴ Ig = resistance/volt at full scale deflection)
Reading of volt meter at full scale deflection = 5 V
If Rv be the voltmeter resistance, then
Rv = \(\frac{V}{I_{g}}=\frac{5}{2 \times 10^{-4}}\) = 2.5 x 104 Ω
New range of voltmeter, V’ = 20V
Let R be the resistance that should be connected in series so that voltmeter may read upto 20 V.
R + Rv = \(\frac{\mathrm{V}^{\prime}}{\mathrm{I}_{\mathrm{g}}}\)
R = \(\frac{\mathrm{V}^{\prime}}{\mathrm{I}_{\mathrm{g}}}\) – Rv
= \(\frac{20}{2 \times 10^{-4}}\) – 2.5 x 104
= (10 – 2.5) x 10+4
If R’v be the resistance of the new voltmeter then
R’v = R + Rv
= 7.5 x 104 + 2.5 x 104
= 10 x 104 = 104 Ω
∴ Resistance per volt of the new voltmeter is
= \(\frac{20}{2 \times 10^{-4}}\)= 5000 ΩV-1
Yes, new voltmeter will be still graded as 5000 ΩV-1.
A voltameter graded as 5000 ΩV-1 is better than one graded as 2000 ΩV-1. It is because, higher the resistance per volt lesser is the current it draws from the circuit and hence better it is.
Question 14.
A test charge of 1.6 x 10-19C is moving with velocity \(\overrightarrow{\mathrm{υ}}\) = (2\(\hat{\mathrm{i}}\) + 3\(\hat{\mathrm{j}}\)) ms-1 in a magnetic field \(\overrightarrow{\mathrm{B}}\) =(l\(\hat{\mathrm{i}}\) + 3j\(\hat{\mathrm{j}}\)) wb m2. Find the force
acting on the test charge.
Solution :
Here,
Using Formula,
i.e. no force acts on the test charge as it moves parallel to \(\overrightarrow{\mathrm{B}}\)
Question 15.
A beam of electrons is moving with a velocity of 3 x 106 ms-1 and carries a current of 1 pA.
(a) How many electrons per second pass the given point?
(b) How many electrons are in lm of the beam?
(c) What is the total force on all the electrons in lm of the beam, if it passes through a field of 0.1 NA-1 m-1?
(d) What is the force on a single electron, if the electrons experience the same force?
Solution :
Here, υ = 3 x 106 ms-1,
1 = 1 μA = 10-6 A
(a) The number of electrons passing a given point per second is
(b) Since the electrons traverse a distance of 3 x 106 m in Is, the number of electrons in lm of the beam is given by:
∴ Force on lm of the beam of electrons is given by
F = B I l = 0.1 x 10-6 x 1 = 10-7 N
(d) Since lm contains 2.08 x 106 electrons
If f be the force on one electron, then F = n’f
Question 16.
In a cyclotron to accelerate proton, the magnetic field is 15000 gauss and radius of the dees is 15 cm. Calculate the energy of the proton beam and the frequency of the oscillator. Tabe e/m for proton to be 9580.84 e.m.u g’1 and mass of proton = 1.67 x 1024 g.
Solution :
Here, m = 1.67 x 10-24 g
e/m = 9580.84 e.m.u. g-1
B – 15000 G
r = radius of dees = 15 cm
Emax = energy of the proton ?
v = frequency = ?
Using formula,
Also using formula, v
Question 17.
A current of 1.0 A is flowing in the sides of an equilateral triangle of side 4.5 x 10-2 m. Find the magnetic field at the control of the triangle, μ = 4π x 10-7 Vs A-1 m-1.
Solution :
Here, l = side of the equilateral
triangle = 4.5 x 10-2 m
I = Current flowing in the triangle .
If \(\overrightarrow{\mathrm{B}_{\mathrm{1}}}\) be the magnetic field at G due to be current carrying conductor BC, then
As the field due to the three sides is in the same direction (i.e. perpendicular to the plane of triangle in upward direction), so the resultant field at a is
B = 3B1 = 3 x \(\frac {4}{5}\) x 10-4 = 4 x 10-5 T
Question 18.
A coil in the shape of an equilateral triangle of side 0.02 m is suspended from the vertex such that it is hanging in a vertical plane between the pole pieces of a permanent magnet producing a horizontal ,, magnetic field of 5 x 10-2t. Find the couple acting on the coil when a current of 0.1 A is passed through it and the magnetic field is parallel to its plane.
Answer:
Couple or Torque acting on the coil is given by x = BNAI sin θ
Here, A = area of equilateral triangle = \(\overrightarrow{\mathrm{B}_{\mathrm{1}}}\) a2
a = 0.02 m = side of equilateral triangle
B = 5 x 10-2 T, θ = 90°
∴ τ = 5 x 10-2 x 1 x (0.02)2x 1
= 8.66 x 10-7Nm,
Question 19.
An electron moves along a circular path of radius 2 x 10-10 m with a uniform speed of 3 x 106 ms-1. Find the magnetic field produced at the centre of the circular path. Also calculate the magnetic moment of the current loop.
Solution :
An electron moving in a circular path is equivalent to a circular current loop with current
I = ne
when n = no. of resolution/s = \(\frac {υ}{ 2πr }\)
1 = \(\frac {υ}{ 2πr }\) e
υ = 3 x 106,
r = 2 x 10-10m, e = 1.6 x 1019 C
Let M = Magnetic moment of the current loop =?
… M = IA = 0.38 x 10-3 x πr2
= 0.38 x 10-3 x 3.14 x (2 x 10-10)2
= 4.77x 10-23 Am2.
Question 20.
Given a uniform magnetic field of 100G in an east to west direction, and a 44 cm long wire with a current carrying capacity of at most 10A. What is the shape and orientation of the loop made of this wire which yields maximum turning effect on the loop? What is the magnitude of the maximum torque?
Solustion :
Here B = magnetic field
= 100G = 100 x 10-4 T = 10-2 T
length of wire – l = 44 cm = 0.44 m ‘
I = current carrying capacity of the wire = 10A
We know that the torque on a current loop will be maximum if it has maximum area. For a given perimeter, a circle encloses maximum area. So the wire should be bent into a circular loop having radius V (say) given by
2πr = 0.04
0r r = \(\frac {0.04}{ 2 x 22}\) x 7 = 0.07 m
area of the circular loop
A = πr2 = \(\frac {22}{7}\) x (0.07)2 = 154 x 10-4 m2
We know that torque on the coil is given by
τ = BIA sin θ
For τ to be maximum τmax sin θ = 1,
τmax = BIA
= 10-2 x 10 x 154 x 10-4
= 1.54 x 10-3Nm.
Orientation of loop: From equation (1)
τ = BIA sin θ
where θ is the smaller angle between \(\overrightarrow{\mathrm{E}}\) and \(\overrightarrow{\mathrm{B}}\)
For τmax , sin θ = 1 or θ = 90 which means that the plane of the coil is parallel to the direction of the magnetic field.
Bihar Board 10th/All video Youtube Channel :- Vidyanjali Point
Numerical Problems Class 12 Physics Moving Charges and Magnetism Numerical Problems Class 12 Physics Moving Charges and Magnetism Numerical Problems Class 12 Physics Moving Charges and Magnetism Numerical Problems Class 12 Physics Moving Charges and Magnetism Numerical Problems Class 12 Physics Moving Charges and Magnetism Numerical Problems Class 12 Physics Moving Charges and Magnetism Numerical Problems Class 12 Physics Moving Charges and Magnetism Numerical Problems Class 12 Physics Moving Charges and Magnetism Numerical Problems Class 12 Physics Moving Charges and Magnetism Numerical Problems Class 12 Physics Moving Charges and Magnetism Numerical Problems Class 12 Physics Moving Charges and Magnetism Numerical Problems Class 12 Physics Moving Charges and Magnetism Numerical Problems Class 12 Physics Moving Charges and Magnetism Numerical Problems Class 12 Physics Moving Charges and Magnetism
1xbet Зеркало Рабочее и Сегодня%3F Рабочий Домен Букмекерской Конторы 1хбет1xbet Зеркало Букмекерской Конторы На сегодня…
1xbet Зеркало Рабочее На день%3F Рабочий Домен Букмекерской Конторы 1хбетАктуальное Зеркало 1xbet Рабочее На сегодня…
1xbet ᐉ Ставки На Спорт Онлайн ᐉ Букмекерская Контора 1хбет ᐉ 1xbet Com ᐉ Ma-1xbet…
Bihar Board Class 11 Physics Solutions Chapter 1 भौतिक जगत BSEB Bihar Board Class 11…
NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols Phenols and Ethers in hindi (ऐल्कोहॉल,…
NCERT Solutions for Class 12 Chemistry Chemical Kinetics in Hindi (रासायनिक बलगतिकी) एनसीईआरटी कक्षा 12…