If there is no current flowing in the conductors that are parallel to each other, the magnetic forces do not obey Newton's third law.
The spiral route is formed by the fields E and B in their current configuration.
mv2/R = evB eB/m = v/R = ꞷ B = F/ev = [MA-1T-2] [ꞷ] = [eB/m]=[v/R] = [T-1]
The quarter's vector sum of the magnetic field at the origin is given as \({{\vec{B}}_{net}}=\frac{1}{4}\left( \frac{{{\mu }_{0}}I}{2R} \right)(\widehat{i}+\widehat{j}+\widehat{k})\)
The time period of the radio frequency is halved when the frequency is doubled, resulting in a half revolution of the charges.
Because velocity varies depending on frame of reference, the net acceleration might have a different value in different frames of reference.
\(dW=\vec{F}\cdot dl=0\) \(\vec{F}\cdot \vec{v}=0\)
In a cyclotron, the particle follows a circular path, with magnetic force acting as a centripetal force. mv2/R = evB eB/m = v/R = ꞷ B = F/ev = [MLT-2]/[AT][LT-1] = [MA-1T-2] [ꞷ] = [eB/m] = [v/R] =...
a) E = 0, B ≠ 0 b) E ≠ 0, B ≠ 0 d) E = 0, B = 0
b) the magnetic forces on both the particles cause equal accelerations c) both particles gain or lose energy at the same rate d) the motion of the centre of mass (CM) is determined by B alone
b) the value of
c) there may be a point on C where B and dl are perpendicular d) B vanishes everywhere on Cb) the value of \(\oint\limits_{c}{B.dl\) is independent of sense of C c) there may be a point on C where B and dl are perpendicular
b) some charges inside the wire move to the surface as a result of B d) if the wire moves under the influence of B, no work is done by the magnetic force on the ions, assumed fixed within the wire
a) independent of which orbit it is in b) negative
d) zero
a) undergoes acceleration all the time
d) the electron will continue to move with uniform velocity along the axis of the solenoid
a) the magnitude of magnetic moment now diminishes
a) B ┴ v
d) the charge to mass ratio satisfy: (e/m)1 + (e/m)2 = 0
The circular coil C1 has a radius of R, a length of L, and a number of turns per unit length of n1 = L/2R. The square C2 has a side, a perimeter, and a number of turns per unit length of n2 = L/4a....
The declination is zero, P is in the plane, S is in the north, and P is in the plane. The declination for point P is 0 since it is in the plane S created by the dipole axis and the earth's axis....
a) |B| is minimum at the magnetic equator. b) Angle of dip is zero when θ = π/2 c) When dip angle is ±45o θ = tan-1 is the locus.
χm = I/H = intensity of magnetisation/magnetising force χ is dimensionless as I and H has the same units χ = 10-4
Magnetic field = 0M/4(1/a2-1/R2) along the z-axis b) On the circular arc, the magnetic field at point A is = 0m/4R2. c) (d) The magnetic moment is 0
The amperian loop is denoted by the letter C. We can find the angle between the two points by solving the above equation. cos is negative because it is greater than 90o.
(a) 1 mole ${{C}_{2}}{{H}_{6}}$ contains two moles of C- atoms. ∴∴ No. of moles of C- atoms in 3 moles of ${{C}_{2}}{{H}_{6}}$ = 2 * 3 = 6 (b) 1 mole ${{C}_{2}}{{H}_{6}}$ contains six moles of H-...
Average atomic mass of Cl. = $[(\text{Fractional abundance of }\!\!~\!\!\text{ }_{{}}^{35}Cl)(\text{molar mass of }\!\!~\!\!\text{ }_{{}}^{35}Cl)+(\text{fractional abundance of }\!\!~\!\!\text{...
Mass percent of Fe = 69.9% Mass percent of O = 30.1% No. of moles of Fe present in oxide =$ \frac{69.90}{55.85}$ = 1.25 No. of moles of O present in oxide =$\frac{30.1}{16.0}$ =1.88 Ratio of Fe to...
1 mole of $CuSO_{4}$ contains 1 mole of Cu. Molar mass of $CuSO_{4}$ = (63.5) + (32.00) + 4(16.00) = 63.5 + 32.00 + 64.00 = 159.5 grams 159.5 grams of $CuSO_{4}$ contains 63.5 grams of Cu....
Mass percent of HNO3 in sample is 69 % Thus, 100 g of HNO3 contains 69 g of HNO3 by mass. Molar mass of HNO3 = { 1 + 14 + 3(16)} g.mol^{-1}g.mol−1 = 1 + 14 + 48 = 63g mol^{-1}=63gmol−1 Now,...
required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1.0.375 Maqueous solution of $CH_{3}COONa$ = 1000 mL of solution containing 0.375 moles of $CH_{3}COONa$ Therefore, no. of moles of $CH_{3}COONa$ in 500 mL = 0.1875 mole Molar mass of sodium acetate...
(i) 1 mole of carbon is burnt in air. $C+{{O}_{2}}\to C{{O}_{2}}$ 1 mole of carbon reacts with 1 mole of O2 to form one mole of CO2. Amount of $CO_{2}$ produced = 44 g (ii) 1 mole of carbon...
Percentage of Fe by mass = 69.9% [As said previously] By mass, 30.1 percent of O2 is present. [As said previously] Relative moles of Fe in iron oxide: = (69.9)x(55.85}/55.8569.9 = 1.25 Relative...
Now for Na2SO4. Molar mass of Na2SO4 = [(2 x 23.0) + (32.066) + 4(16.00)] =142.066 g Therefore, mass percent of the sodium element: = 32.379 = 32.4% Mass percent of the sulphur element: = 22.57...
(ii)
(iii)
(i)Molecular weight of methane= (1 x Atomic weight of carbon) + (4 x Atomic weight of hydrogen) = [1(12.011 u) +4 (1.008u)] = 12.011u + 4.032 u = 16.043 u (ii)Molecular weight of water= (2 x...
Option (iii) is correct. Explanation: The lowest boiling point among isomeric pentanes is of 2,2-dimethylpentane, and further on branching, its boiling point decreases
Option (i) is correct Explanation: In nitration, benzene is treated with a nitrating mixture, which consists of conc. $HNO_3$ and $H_2SO_4$, and $H_2SO_4$ aids in the production of $NO_2 +$....
Option (i) is correct
Option (i) is correct. Compounds with the following features are aromatic: planarity and complete delocalization of the electrons in the ring. The ring contains (4n+2) electrons, where n is an...
(i) is d (ii) is a (iii) is b (iv) is c
(i) is d (ii) is c (iii) is b (iv) is a
(i) is b (ii) is c (iii) is a
(i) is d (ii) is a (iii) is e (iv) is c (v) is b
At about 140°C, ethanol is processed with sulphuric acid to produce hydrogen sulphate. The response is
Huckel's rule states that it must meet the (4n+) rule. Aromatic compounds include B, C, D, and F.
Aromatic compounds: A, E and F Non-Aromatic : B, C, D and G
In comparison to the 1° free radical, the 3° free radical is stabilised by 9 hyperconjugation structures, giving it greater stability.
When a combination of 1-iodo-2-methylpropane and 2-iodopropane is treated with salt, it produces three compounds as a result of intermolecular and intramolecular reactions:
Number of hydrogen reactivity = number of mono-chlorinated compounds 1° H = 9 1 = 9 mono-chlorinated products 2° H = 2 3.8 = 7.6 mono-chlorinated products 3° H = 1 5 = 5 mono-chlorinated products...
Nucleophiles I (vi), (vii), and (viii) Electrophiles are (ii), (iii), (iv), and (v).
Step 1: Peroxide homolysis to produce free radicals Step 2: Bromine free radical formation ∙C6H5 + H-BR → C6H6 + Br Step 3: hydrogen bromide reacts with an alkyl radical.∙
I We can cycle acetylene using the intermolecular condensation technique and then treat it at high temperatures in a red hot iron tube. (ii) After being converted to benzene, the aliphatic molecule...
A nitrogen atom is linked to two extremely electronegative oxygen atoms in the Nitro group. This causes a net reduction in electron density around the nitrogen atom, giving nitrogen a positive charge.
Halogens have a ns2p5 outer configuration, indicating that they may take one electron and are near to completing their octate; halogens have a significant affinity for attracting one electron,...
I Bromine undergoes electrophilic substitution with Br2 in the presence of anhydrous FeBr3 to generate bromobenzene. We get a p-nitrobromobenzene after treating it with conc. HNO3 and conc. H2SO4 at...
Friedel Crafts alkylation with a Lewis acid is demonstrated in this process. The development of the carbocation will occur initially, followed by the production of a more stable secondary...
The ascending order of reactivity is HI >HBr>HCl. The rising order of halogen reactivity corresponds to the rise in bond energy.
Because there is less C – H bond pair repulsion and the atoms are at their furthest distance from one other, the staggered arrangement is more stable than an eclipse.
In ethane, the single bond is a – bond, which is a coaxial overlap of orbitals that allows the C C bond to rotate on its axis. However, due to the torsional strain that the bond receives as a result...
The negative charge that has generated on one carbon attacks the proton from NH3 and causes another sodium atom to lose its electron, causing the atom to create a second negative charge. The second...
To produce a more stable saturated product, alkenes undergo an addition reaction. Hybridization shifts from sp2 to sp3 in this reaction. A substitution process maintains the resonance stability of...
Option (ii) and (iii) are the answers
Option (i) and (iii) are the answers.
Option (i) and (iii) are the answers.
Option (i) and (iii) are the answers.
The solutions are options (i) and (iii).
Option (i) and (iv) are the answers.
The solutions are options (iii) and (iv).
The solutions are options (iii) and (iv).
The solutions are options (iii) and (iv).
Option (iii) is the answer.
The solution is option (iv).
The solution is option (ii).
The solution is Option (iii)
Option (iv) is the answer.
Option I is the correct response.
Option I is the correct response.
Option (ii) is the correct response.
Option (i) is the solution
the answer is Option (iv)
Option (iii) is the answer.
Option (i) is the answer.
Option (iv) is the answer.
Option (ii) is the answer.
Option (i) is correct.
Option (ii) is correct.
correct is Option (iii)
(i) are a and d (ii) is c (iii) is a (iv) is b
(i) is e (ii) is d (iii) is a (iv) is b (v) is c
(i) is d (ii) is e (iii) is a (iv) is c (v) is b
(i) are c and d (ii) are e and d (iii) is b (iv) is a
The laxative effect is the loosening of faeces and increased bowel movement induced by laxative drugs. Laxatives in high dosages might induce diarrhoea. This might be due to an excessive amount of...
The ozone layer is heavier than oxygen and is thermodynamically unstable, yet the interaction between ozone production and dissociation has a dynamic equilibrium. As a result, they avoid settling...
The action of ultraviolet light on atmospheric oxygen forms the ozone layer in the stratosphere. UV rays cause molecular oxygen to divide or dissociate into two oxygen atoms. O 2 (g) → O (g) + O (g)...
Nitrogen oxides, volatile organic compounds (VOC), ozone, and PAN (peroxyacetyl nitrate) are all components of photochemical smog. In the presence of sunshine, nitrogen oxide in the smog dissociates...
The uncatalyzed oxidation of sulphur dioxide to sulphur trioxide is sluggish, but it may be oxidised quickly in the presence of a catalyst in the environment. There are substances in the atmosphere...
This sort of breathing difficulty can be caused by toxic nitrogen and sulphur oxides in the environment. These oxides are produced in industry by the oxidation of fossil fuels such as coal. N2 (g) +...
Algae breakdown by bacteria continually generates a nasty odour, and algae contaminate and make water unpleasant. The quantity of dissolved oxygen in the water drops as well, which might cause...
The quantity of oxygen required by bacteria to breakdown organic materials is known as biochemical oxygen demand (BOD). The higher the BOD of water, the more contaminated it is. The water with a...
Water absorbs oxygen from the atmosphere when it comes into direct contact with atmospheric air. During the day, green aquatic plants photosynthesise. Photosynthesis does not occur at night, but the...
Biodegradable pollutants are ones that can be degraded by bacteria or other natural elements such as fruits, sewage, and so on. Bacteria have a difficult time decomposing non-biodegradable...
The dumping of household garbage in the lake can give nutrients for algae and aquatic plants to develop fast. The breakdown of these, which is aided by bacteria, can also result in a bad odour. The...
Inadequate industrial and household waste management in a city can result in significant harm to both living and non-living objects. 1. If household trash is not properly disposed of, sewage pipes...
The action of ultraviolet radiation causes chlorofluorocarbons to dissociate, releasing the free chlorine radicle. U V radiations + CF2Cl2 + Cl + CFC Now that this Chlorine radicle has formed, it is...
The process of eutrophication is when the amount of dissolved oxygen in the water decreases. It's a process in which nutrient-rich water bodies support a dense plant population, which then kills...
When ozone is removed from the atmosphere, UV radiation comes into direct contact with living organisms, causing harm such as skin cancer and a variety of other dangerous illnesses. The ozone in the...
Acid rain is caused by human activities that release sulphur and nitrogen oxides into the atmosphere. 2H2SO4 = 2SO2 (g) + O2 (g) + 2H2O (l) (aq) 4HNO3 (aq) = 4NO2 (g) + O2 (g) + 2H2O (l). Acid rain...
Greenhouse gases such as carbon dioxide, methane, ozone, chlorofluorocarbon compounds (CFCs), and water vapour are all responsible for the greenhouse effect, which leads to global warming.
The solutions are (i), (ii).
The solutions are (ii), (iii), and (iv).
The solutions are option (i) and (ii)
The solutions are option (iii) and (iv)
The solution is option (iV).
The solution is option (iii).
The solution is option (iii).
The solution is option (i).
The solution is option (ii).
The solution is option (ii).
The solution is option (iii).
The solution is option (i).
The solution is option (i).
The solution is option (i).
The solution is option (iii).
Option (i) is the correct answer
T stands for the time period. The moment of inertia is me. The magnet's mass is m. B stands for magnetic field. T = 2I/MB M' = M/2 magnetic dipole moment T' = T/2 is the time period.
pE sin θ = μB sin θ pE = μB E = cB pcB = μB p = μ/c
As the system is in equilibrium, the net torque and net force will be equal to zero.
P is the point at a distance r from O and OP, then the magnetic field is given as: dS is the elementary area of the surface P, then dS = r2 (r2 sin θ d θr) Solving the above we...
I The superconducting material will repel the bar magnet. ii) The magnetic moment will be directed from left to right.
The temperature has little effect on the temperature dependence of susceptibility for a diamagnetism. The temperature affects the temperature dependence of susceptibility for paramagnetism and...
nitrogen Density = 28 g/ 22400 cc copper Density = 8 g/ 22400 cc Ratio = 16 × 10-4 Diamagnetic susceptibility = density of nitrogen/density of copper = 1.6 × 10-4
Intensity = 106 A/m l = 0.1 m M = IM/l IM = Ml = 105 A
A comparison of a proton's and an electron's spinning is made by comparing their magnetic dipole moment, which is given as μp = eh/4πmp μe = eh/4πme μp/μe = me/mp = 1/1837 >> 1 μp << μe...
a) electrostatic field lines can end on charges and conductors have free charges c) lines of B cannot end on any material and perfect shielding is not possible d) shells of high permeability...
a) the H field in the solenoid is unchanged but the B field decreases drastically d) the magnetisation in the core diminishes by a factor of about 108
a) atomic currents d) intrinsic spin of electron
a) lines of B are necessarily continuous across S d) lines of H cannot all be continuous across S
b) 2/3 Am-1
b) In the case of magnetic fields, instance (ii) violates Gauss's law.
d) all of the domains are exactly matched
a) The declination ranges from 11.3 degrees West to 11.3 degrees East.
c) is zero; otherwise, a field dropping as 1/r3 at great distances outside the toroid would exist.
The solenoid's changing magnetic field is represented as: onI = B1(t) (t) The second coil's magnetic flux is 2 = onI(t).b2 As a result of the solenoid's changing magnetic field, the induced emf in...
v = mgR/B02π2λ2l4
The angle between A and B = 0o Therefore, the emf is Bvd. -LdI(t)/dt + Bvd = IR LdI(t)/dt + RI (t) = Bvd Solving the equation, we get I = Bvd/R [1-e-Rt/2]
It = /R is the current induced in the loop. BA = 1/R.d/dt It = 1/R.d/dt It = 1/R.d/d vBd/R = it The angle formed by B and A is equal to zero. At t = o, the switch S is closed. Q(t) = Cv is the...
The angle formed by B and PQ is 90 dϕ = B.dA dϕ = B v d cos θ -ε = B v d cos θ I = -Bvd/R cos θ Using Newton's second law to solve the preceding problem, we get v as v = α g sin θ [1 – e...
When the magnetic field is lowered in t, the magnetic flux across the conducting ring drops to zero from its maximum. E2b = induced emf According to Faraday's law of emf, The induced emf is equal to...
If the instantaneous current is t, then I(t) = 1/R d/dt I(t) If q is the charge that passes during time t, dQ/dt = I(t) 1/R d/dt = dQ/dt When we integrate the equation, we obtain Q = 0L1L2/2R log...
The strip has a width of dr and a length of l, and it is located inside the rectangular box at a distance of r from the current-carrying conductor's surface. The magnetic field along the length of...
I If the spinning conductor's location is considered to be in the time period t = 0 to t = 4 We obtain current I = 1/2 Bl2/l sec2 t cos t = Bl/2 cos t = Bl/2 cos t = Bl/2 cos t = Bl/2 cos t = Bl/2...
m = B.A = BA cos m The area vector is A, while the magnetic field vector is B. e1 = -dB(t)/dt lx e1 e2 = B(t) lv (t) The total emf in the circuit is equal to the emf owing to field change plus the...
Allow wire AB to travel with velocity v at time t = 0. x(t) = vt at time t AB = e1 = Blv Motional emf across (Bo sin t)vd = e1 (-j) d(B)/dt = e2 e2 = -B0 cos tx (t)d e2 = -B0 cos tx (t)d e2 = -B0...
The current flowing through the coil is denoted by Ia. Mutual induction between A and B is known as Mab. The number of turns in coil A is Na. The number of turns in coil B is Nb. an is the flux...
We may deduce from the graph that when the rate of change of magnetic flux reaches its highest, the electromagnetic force, which is proportional to the rate of change of current, reaches its...
F is the force acting on PQ's free charge particle. The motional emf is calculated by multiplying E along the PQ by the effective length of the PQ. As a result, the induced current will be vBd/R,...
The magnetic flux lines that pass through are identical to those that flow through the surface. The magnetic field lines in an area A with magnetic flux B are represented by = B1dA1 + B2dA2. As a...
The magnetic field is directed along the z-axis. B.A = BA cos = B.A cos = B.A cos = B.A cos = B.A cos Using the electromagnetic induction law of Faraday, R sin t = Boa2/R sin t = I = Boa2/R sin t =...
The magnetic flux across the pipe changes when a cylindrical bar magnet with a radius of 0.8 cm is dropped through it, causing eddy currents to form. The existence of eddy current causes the magnet...
We already know that current was flowing through the solenoid, causing it to act like a magnet with the S pole on the upper side. As a result, the ring has no induced current. When the current is...
The metal ring leaps up when the current is quickly turned on because the magnetic flux across the ring is enhanced.
The magnetic flux rises when an iron core is introduced into the solenoid. According to Lenz's law, as the flux increases, the current flow through the coil decreases.
When the spiral coil is extended and there are gaps between consecutive components, the current rises. Reactance must drop in order for current to rise.
There will be no current induced since there is no change in the magnet or the circuit area. In addition, there has been no change in the angle
(b) the magnetic field is in the same plane as the circular coil and it may or may not vary. (c) the magnetic field has a perpendicular (to the plane of the coil) component whose magnitude is...
(a) increases when they are brought nearer. (d) is the same as M21 of coil 2 with respect to coil 1.
(a) the coil is in a time-varying magnetic field. (b) the coil moving in a time-varying magnetic field. (c) the coil moving in a constant magnetic field.
(a) The plate is receiving a direct current. (b) it is exposed to a magnetic field that changes over time. (c) it is put in a magnetic field that varies in space but not in time.
(b) l decreases and A increases.
(a) clockwise steady current
(d) In A, there is a counterclockwise current that is constant.
(b) There is no current flowing through ammeter A.
(b) 2 Bo L2 Wb. is the correct answer
(c) 4 Bo L2 Wb. is the correct answer
(i) is a, b and d (ii) is b (iii) is b (iv) are c and d
(i) is c (ii) is e (iii) is a (iv) is b (v) is d
(i) is c (ii) is f (iii) is b (iv) is a (v) is d (vi) is e
(i) is e (ii) is d (iii) is a (iv) is b (v) is c
Compound A's two resonant structures are not equal, but compound B, which has a negative charge on the oxygen atom, has two equivalent structures, making it more stable.
Steam distillation can be used to separate the mixture. The boiling points of alcohol and hydrocarbon are almost same. Steam distillation is used to purify temperature-sensitive materials in...
Because all of the atoms in structure I have a complete octet, while the carbon atom with a positive charge does not have a complete octet in structure II, structure I will be more stable.
Conjugation is only feasible when an atom possesses any of the double bond's charge or alternate positions. Because of this, CH3OH will not exist as a resonance hybrid here.
1. In the inductive effect, the electron is only sent through the sigma bond, but in the resonance effect, the electron is transmitted through both the sigma and pi bonds. 2. It is feasible when the...
(i) Because the bromine atom destabilises the positive charge on a carbon atom, C+H3 will be more stable. Bromine possesses a lone pair of electrons and is an electron-withdrawing group. (ii) CCl3...
(i)The compound's name is 3-ethyl-4-methyl-5-heptane-2-one. (ii) The chemical is called 1-nitro-cyclohexane-2-ene.
3NaCNS + FeSO4 + dilute sulphuric→ Fe(CNS)3 (Blood red colour) + 3Na+ Ramesh's organic molecule has both Nitrogen and Sulphur, resulting in the Blood-red colour of Fe(CNS)3, whereas Manish and...
The following is the stability order: (III) > (II) > (I) > IV This is due to the fact that (III) is a tertiary carbocation, (II) is a secondary carbocation, and (I) and (IV) are primary...
Triphenylcarbocation is a tertiary carbocation with a positive charge on the carbon atom that is stabilised by resonance thanks to the three phenyl groups. The stability of the system improves as a...
Inductive effect, hyperconjugation, resonance, and other factors influence the stability of carbocation. Because structure A has a carbocation that is both primary and allylic, whereas...
Structure C is more stable than structure A because all atoms' octets are complete in structure C, but in structure A, a C-atom with a positive charge does not have 8 electrons in its valence...
Because the liquid component decomposes near its boiling point, indicating that it is heat-sensitive, we purify it using "Steam distillation." For temperature-sensitive materials, this is done.
Because nitrogen cannot be transformed into ammonium sulphate at the conditions employed in the Kjeldahl technique, it cannot be used to estimate nitrogen levels in DNA and RNA. This test for DNA...
a. The main(parent) chain should be the longest carbon chain feasible, including all carbons containing functional groups. With both functional groups in the main chain, the aforementioned option is...
The functional groups (thioether/sulphide) are the similar in both structures, but the atoms in the main chain are arranged differently. As a result, they can show chain isomerism.
CH3—CH2—CH2—δ-CH2—δ+Mg+ X- Because the electronegativity differential in the C-Mg bond is strongly polarised, carbon is more electronegative than magnesium. Because C is more electronegative, its...
There are three forms of carbon hybridization in organic compounds: sp, sp2, and sp3. The electronegativity of carbon increases as the ‘s' character increases because ‘s' orbitals are closer to the...