PYQ Chapter 4 Physics || Moving Charges & Magnetism || Most expected question || CBSE Class12

[@Shubham-jp4ri]

00:07 Importance of practicing previous year questions 02:19 Understanding the Fleming's Left Hand Rule for determining the direction of magnetic force on a moving charge. 06:54 Installing a shunt to increase range up to 10 amps for an ammeter 09:05 Using shunt to measure resistance of ammeter 13:23 Key concept: Current carrying wire in magnetic field 16:05 Understanding the interaction of a wire with a magnetic field 20:10 Velocity changes but speed remains constant in circular motion of charges. 22:01 Circular motion is determined by the value of speed and distance. 26:34 Charged particle's acceleration and kinetic energy with potential difference 28:34 The formula for radius in the context of potential difference and charge 33:19 Magnetic field and current enclosed in Ampere loop 35:36 Calculation of magnetic field for different distances from the source 40:07 Discussion on magnetic field and motion of charges in magnetic field 41:56 Electron motion in a magnetic field 45:44 Find the radius of the alpha particle path with given mass, charge, and kinetic energy formula. 47:47 Discussing the calculation of charge and magnetic field 52:19 Understanding the mass and charge of protons and neutrons 54:20 Proton and neutron charges, kinetic energy ratios, and circular path radii compared. 58:51 Alpha particles have twice the mass and charge of a proton, affecting kinetic energy and magnetic field. 1:01:03 Particles with an angle of 30° experience forces and motion components 1:05:15 Comparison of paths described by alpha particle and proton in a magnetic field 1:07:29 Calculating mass ratio of proton to alpha particle 1:12:08 Calculating frequency and mass ratio of electron and proton 1:14:39 Frequency of proton and electron comparison 1:19:14 Conversion of geometric properties in physics problems 1:21:31 Understanding the magnetic field at the center of a current-carrying circular loop 1:26:26 Finding the value of √8 and 2√2 using ratio of b1 and b2. 1:28:56 Understanding the magnetic field and current flow between two identical loops. 1:33:17 Calculating the net magnetic field in a specific scenario 1:35:29 Understanding the net magnetic field and direction calculation. 1:40:14 Understanding the direction of the magnetic field due to current-carrying coils 1:42:18 Determining magnetic moment direction and magnetic field formula 1:46:37 Solving questions related to wound solenoids in physics 1:48:48 Understanding the magnetic field between parallel wires with equal current 1:53:18 Understanding the variation of magnetic field with distance. 1:55:43 Calculation of the magnetic field due to a current-carrying wire on a proton's motion. 2:00:52 Analysis of magnetic torque on a suspended coil 2:03:19 Current value is 0.96 Newton meter 2:07:42 The video discusses the force between parallel wires carrying current. 2:09:48 Moving charges create magnetic fields around them 2:14:31 The net force on a square loop due to current can be zero under certain conditions. 2:16:44 Magnetic field exerts force on the wire

[@vani_.7702]

Thanks

[@sudheksharavichandran5802]

Thank you

[@43.sumandey32]

bro konsi website use ki tune?

[@Niharikavt]

23:42 (a)speed of the particle 1:35:20 S.I Unit of permeability of free space (mu not)=Wb A^-1 m^-1 and T A^-1 m 1:40:06 Bnet=4 pi ×10^-7/R 2:07:16 option C should be correct as in reason , it is written that shunt is added in series but to increase the range , shunt will be added in parallel

[@1mflexin_]

43:57 mein helical path follow krega na ?

[@kmbro-i7u]

​@@1mflexin_nahi circular hoga

[@ADIL-bk2rv]

🤗

[@Supriya-e8b]

b net will be 6pie *10^-7/R

[@Shagun.662]

Just wanted to know if any question (similar to those in this video) came in board exam of 2023-24???

[@ranisharma7194]

23:28(a) speed of the particle. 1:10:54 ,r= root2 r

[@erasparky]

2:21:07 GUYS ANSWER WRONG HAI SIR KA CORECT ANSWER IS :-- 5.33*10^-7 DROP A LIKE SO SBKA ANSWER GLT NA AYE 👇🏻👇🏻👇🏻👇🏻 SIR NAI L GLT LELIA HAI SO BE AWRE

[@sanjeevsirrajivsirfanclub2751]

Ratta marta sir

[@ranisharma7194]

2:07:12 reason is false Because , to increase the range, resistance is to be lowered and to lower the resistance shunt should be connected on parallel.

[@Dkmishra6002]

Same

[@Tengu732]

Same

[@thefresherfitness6804]

2:18:47 The value of i1 is 2A not 1A 😢 And the correct answer is 5.33×10^-7 ❤

[@study-zq9oy]

THNX BROTHER BASS YAHI MERA ANS THA LEKIN MUJHA APNA ANS PAR DOUBT HONA LAGA THA

[@thefresherfitness6804]

@@study-zq9oy Your welcome bro..

[@NavrajSingh-rx2ms]

✌️

[@harmanbrar5028]

23:20 a is ans 1:11:05 root2 r 1:35:09 kgms-2A-2 and TmA-1 1:39:56 munot / R and pie / 6 2:05:40 Voltmeters should have very high resistance. As ammeter has very low resistance, thus very high resistance must be connected in series with the ammeter to use it as voltmeter 2:07:18 c

[@trivendrasingh683]

root 2r kese aaya hh ...?

[@danyasharma26]

​@@trivendrasingh683Initially, r= √2mqV/qB For new , Potential diff = 2V r' = √2mq2V /qB r'= √2 r

[@AnishAryan027]

2:06:21 Sir you have made a mistake , shunt is connected parallelly with ammeter to increase its range. Therefore ASSERATION IS CORRECT BUT THE REASON IS WRONG. Ans should be option C

[@SuzumeChan-kr7fs]

thought of the same

[@tanishqpanchal6275]

I also noticed this

[@adishreepanda8531]

Yes

[@navneetrocks8255]

25:26 que ka ans D hoga hoga sir. Jb dono wire ek doosre ko attract krengi to are decrease hoga instead of increasing

[@sunnyysinhaa]

W

[@BeautyGupta-yn4oe]

Right

[@nobitakumar6080]

but current to opposite hana bhai to vo to repel karengi

[@SAMRIDDHI-tk7tw]

yeahh right

[@mrBlk02025]

b option hoga.....both are true but R not correct expln of A

[@AnishAryan027]

23:52 Option A 1:11:08 R`= root 2 R 1:40:05 B = u0/R , THETHA = 30 DEGREE

[@omsawant3899]

Bro there we can direct compare naa R= mv/qb R’=2mv/qb Equating it we will get R’= 2R Then why we following that root 2mk wala step I know my steps is wrong doubt is why and why we using that root mkv wala step

[@sagunrath2728]

Bhai wo velocity wala v hai and root ke andar potential wala v

[@danyasharma26]

​@@sagunrath2728 haa

[@ranisharma7194]

2:04:43( a)low resistance in parallel When high resistance is connected in series with the galvanometer it works as a voltmeter when a low resistance is connected in parallel with galvanometer it works as a ammeter

[@tapaswinidash6718]

But here voltmeter hin use karna hai...par voltmeter ka kam ammeter ko karna hai.... So the right ans will be high resistance in series.... Cause ammeter has low resistance and voltmeter has high resistance...so ammeter ko voltmeter ke jagah use karenge so resistance badhane ke liye hame toh series mein high resistance hin add karna padega nah

[@raghavmittal6506]

2:06:26 me option c ayega bcz to increase range we have to use shunt in parallel with it Nice video sir

[@chiragvlogs5769]

Yes, u are right

[@pratyasaha9876]

Yes

[@gaganrana-qo6rm]

Sahi kha bhai merko bhi c option lga but ohir comment section me apne jaiso ko dhundne aya hu 🫡💀

[@RobinShaju]

​@@gaganrana-qo6rm yaa

[@BlueYelloww-m8g]

26:16 option D IS RIGHT

[@srushtidarji6690]

1:35:22 *SI unit* kg•m•s^-2•A^-2 N/A^2 T•m•A^-1 1:40:07 u°/R 2:21:00 ans *wrong* he *5.33×10^-7 N* aayega Aapne F(AB) me *i1* ki value *2A* ki jagah *1A* rakhi he

[@AnishAryan027]

2:21:12 Net force on the loop will be = 5.33 x 10-7 . Correct krlo guys aaplog aapna sir se mistake hogyi hai F(AB) Calculate krne me

[@Yellow_flash50]

Same

[@BusinessclassGaming]

Thanku bhai for that😶I was really confused about it

[@I_am_the_Upgrade]

1:11:00 The new radius will be r' = sqrt(2)*r where, r is the initial radius of the proton.

[@DineshKumar-uu2vw]

Yess

[@areebakhan141]

1:40:09 ans: Bnet= μ⁰/R Direction= tan^-1(1/√3)

[@mohit4562]

2:07:16 wrong ans ❌ Shunt in parallel 2:09:00 sir waha unit n/m hogi

[@yashrajsrivastava7751]

Sir 30 Important Derivations wali video jaldi banao

[@l.s.mmurmu8709]

Yes sir

[@namansingh5977]

Yes sir please thoda sa samay nikal kar bana digyea please sir

[@maazahmadshariff3084]

Yes sir

[@trendingmoviehub1584]

Yes sir

[@apnedesilokgeet2412]

Abhi ban rhi h 😅,sir live h

[@harshsingh9655]

1:40:09 Bnet =uo/R AND direction in XZ plane....

[@AmritaSoni-df9zq]

23:42 option A Speed of the partical

[@AdityaYadav-jd8ly]

2:07:17 Option c is correct, but sir had done (a) option correct Here assertion is correct, but reason is false. As to increase range of ammeter , we should combine shunt to parellel of ammeter.

[@aarushidhankar2250]

1:13:43, sir charge of p = negative of charge of e, isn't it then how you canceled it without - ve sign and in q 53, 2:06:56- you added shunt in parallel so the assertion is correct but the reason is wrong

[@_atbw_]

2:21:00 F = 5.33 x 10^-7 N

[@prakashpharmapowayan2795]

How?

[@mathverse-n5d]

25:26 Sir iska answer D hona chahiye kyonki a statement false hai .

[@DakshMishra69]

Sahi h bro

[@Saimehak2906]

2:07:18- Sir here the optic C would be correct bcz we add shunt in parallel to the galvanometer to make it ammeter(not in series)

[@rashil8703]

true

[@AbhinayKumarSingh-cf4zc]

Right

[@theoggy8630]

1:11:08 answer is r' = root 2 r

[@sjoshi6825]

23:31 a option 24:22 a option 1:10:55 root 2 r 1:40:03 12.56 guna 10^-7

[@Kunal_Kumar1401]

1:40:00 Bnet = 4pie × 10-7 /R tesla.. Dir_ñ = pie /6

[@BinitBarnwal]

2:14 c) ❌ 6:58 b) 9:37 😊 10:22 b) 18:12 c ) 20:09 unattempt 23:20 a) 25:42 d) a is false ❌ (real ans is A) 27:15 a) ❌ 50:10 I Used r = √2mk/qB (k=qV) 😊 1:11:06 r' = √2 r 1:11:42 as electron mass is very less as compared to protons so electron will move with higher frequency in circular path keeping the magnetic field constant. f = 2πm/qB 1:35:20 Forgot 😢

[@vectoranonymous4908]

Bahut errors hai sir. But I really appreciate your efforts Sir. Radhe Radhe 🙏

[@UnknownGuy_22]

Please batado errors

[@bliss-sb7zp]

2:06:04 sir in reason it is written that shunt should be connected in series but shouldn't we shunt in parallel to Inc the range?

[@navyakunwar1532]

1:11:08 √2 r 1:35:28 (mue not)si unit= Wb A^-1 m^-1 & T A^-1 1:40:05 Bnet= Munot/R Tan@= 1/√3 @= 30°

[@chuck5369]

23:35 Answer- (a) speed of the particle

[@Musicvy73]

1:37:11 H/m and N/A

[@Musicvy73]

1:40:10. Muenot/R

[@ashatripathi5836]

Kaise ?

[@HarshGoyal-l2l]

2:04:45 the answer should be low resistance in parallel which is shunt resistance

[@thundermemes9119]

Yes u r correct he did it parallel but failed to read it properly

[@mihir2085]

C hi hoga uska answer theek se padho question

[@SbSk-c9w]

Sir at 2:21:13 last question, I1 is 2 but you substituted wrongly by mistake.. ANSWER SHOULD BE 5.333×10-⁷N

[@Mehz12]

Yes bro you are right 👍

[@khus_raho_or_rahne_do]

23:28 (a) 1:10:54 r'=√2 r 1:15:51 frequency mass ke inversely proportional h thus e- is ans

[@adivaaaaaa]

1:40:07 B=u0 /R and theta=30 degree

[@harshlikeshiphop7840]

These PYQs series is just like the booster dose for students❤❤❤ Thanks...🎉

[@harshagrawal4735]

2:19:30 in the last question there is a mistake ,sir you have done the wrong solution the correct answer should be 5.33×10^-7 . The overall video was very helpful.😊

[@yashvardhansinghyash3224]

Sahi baat hai bro current ki value 2 amp nhi li❤ koi nhi hojati sabse galti❤

[@harshsingh9655]

2:07:18 SIR IN SERIES DIYA HUA HAI TOH Reason false ho jaayega na

[@KrishnapalsinghRajput-yw2fy]

You are right bro by mistake ho gya sir se

[@nevergiveup....9592]

Hnn ji

[@Aliya-nk5ce]

1:37:22 SI untitof uo is TmA-1 and Wb A-1 m-1

[@thefresherfitness6804]

Nice broo❤

[@HarshGoyal-l2l]

2:07:19 the answer should be option 'c' bcs reason is wrong as shunt should be connected in parallel with it.

[@MuskanSingh-cb5dy]

Same here

[@sarladevi3980]

1:55:00 For x=d2 magnetic field B=0. For xd2 magnetic field B is directed downward. Graph is incorrect? ..

[@sarladevi3980]

D2 = d/2

[@dhanasrikanisetti2263]

23:35 option (a)

[@vectoranonymous4908]

Yes

[@IamAdi45264]

2:07:06 q no 53 me b option hoga na sir kyuki ammeter me shunt parallel me lagta h

[@AnkitSharma-ex8vk]

Exactly

[@AJBeastFun]

1:40:00 Ans 4pi x10^-7/R

[@Mayateriram01]

2:07:13 option C hoga kyuki usme likha h added in series to it

[@thomashemsworth1545]

2:07:16 sir shunt parallel meh lagta hai series meh nahi Reason is wrong assertion is correct

[@ayushsinghkaintura7422]

Hn Bhai sahi bol rahe ho

[@AdityaYadav-jd8ly]

21:40 Answer will be option (a) Speed of particle

[@amanuniyal27]

2:19:58 sir B¹ me current I¹ hai jo 2 amphere h jiski vajaha se ans 16/3 ×10^-7 ayega

[@Pradeepthakur_g1]

1:11:09 r root2

[@PankajSardhana-mj1fc]

Root two r

[@Mehz12]

(a)Speed of the particle

[@thundermemes9119]

02:15:12 Can't we do direct by formula taking distance between them as 2L , why to do minus? Any specific reason for that?

[@RobinShaju]

Direct formula possible Minus is done because both force to opposite side , so net force will be LARGE force minus SMALL force

[@Abhay--1]

2:07:00 option c❤❤❤

[@devilarmy367]

a) sppped of particle 23:40 ❤❤❤❤❤❤❤

[@hasanmazhar902]

2:20:55 wrong answer

[@PrachiSingh-j2i]

23:37 speed of particle {A}

[@AjayKumar-mq5ir]

2:07:15 per option C hoga

[@Nsjdbdj89]

23:29 option A: speed of the particle

[@Musicvy73]

1:11:5 R2=√2r

[@creativityx1000]

1:10:57 r'=√2 r

[@parasytegaming2181]

thank you sir so much my confidence in phy is growing only because of you thank you so much sir

[@joydeepdeka7946]

1:10:00 r'= √2r

[@harshsingh9655]

23:45 speed of the particle

[@DreamGirl-f2r]

Yess correct

[@abhinishtsingh9767]

23:30 - opt a speed

[@RobinShaju]

1:09:02 root 2 r

[@pratyasaha9876]

23:51 Opt-a) Speed of a particle

[@harshsingh9655]

1:35:26 SI UNIT OF uo IS TmA-1

[@creativityx1000]

23:39 ---- answer is (A)speed

[@MoneyHeistShorts09]

01:11:08 SIR IT'S ANSWER WILL BE √2r = ∆r Means change in r' = √2r

[@tanishk565]

2:07:12 yha par sir assertion to shi h but reason false h q ki ammeter me hum shunt Parralel me add karte h??????

[@ManiKaushik5912]

Yes bro... ye sir ne dhyan nh diya

[@dolorous_me]

23:41 speed of the particle 😊😊

[@GuptakïPlaylist]

Assertion is false 25:50 as per figure

[@RobinShaju]

2:07:12 option C right? Shunt should be in parallel not in series? ??? 🙂

[@ManiKaushik5912]

Q 10 me P side se Current ja rha h Or Q ki taraf a rha h jbki dono wires me current ki direction same h.. how?? Could anybody answer??

[@sushmitachauhan1153]

Yes bro same doubt ... attract krna chaiye jisse area km ho jayega

[@ManiKaushik5912]

@@sushmitachauhan1153 yes bro

[@Yashika723]

I also have same doubt

[@NehaSingh-ok2ft]

Exactly .. Same doubt

[@12THBOARDPHODNAHAI]

23:45 A very easy when you teach sir❤❤

[@neonblade534]

23:30 option a)

[@PrashantKumar-te4xq]

2:06:52 answer will be a is true Ris false 😊

[@AryanGupta-ws2un]

14:20 18:27 28:24 40:50 58:24 1:03:57 1:07:08 1:53:40 2:14:10

[@LeviAckerman-yt8hj]

27:28 Important Question

[@shubham.359]

1:55:04 mei graph galat hai kyuki ye graph d/2 ke about symmetric hoga, agr ek wire se dur ja rhe honge toh durse wire ke pas bhi toh aa rhe honge toh inhone jo graph wo toh pura galat hai

[@neerajverma619]

Sir question number 53 me apne dhyan nhi Diya syd Usme likha h ki...shunt in series ...but apne parallel smj ke dono ko correct bol diya. But reason glt h usme...ak bar chk kr lijiye please

[@SbSk-c9w]

SI units of mu naught - Testla metre per ampere( Tm/A) and newton per ampere square(N/A²)

[@nevergiveup....9592]

26:25 sir current wah jo bent shape main aisa kyun jayega...ye v to ho skta hai na same direction main jae🤔

[@shagunsom7023]

2:21:11 answer is wrong. Because the current in straight wire is 2A but you in solution take it as 1A

[@Sakshi_2007]

1:35:15 Tm/A Weber/A Am i correct??

[@rishikapanjeta]

Sir aapne ques. 32 me magnetic field ki direction alag li h aur next ques. me current ki direction previous quest. jaisi he h phir bhi dono questions me alag alag directions kyu h ??

[@ayushiraikwar8801]

Thnquh so much sir ..For Everything u r doing for us ☺

[@kuljitsingh-qp4oe]

I have a doubt sir. In ques 39 1:46:08 direction of B should be on right side bcz current flow clockwise in 1. Please its my humble

[@SahilVerma-sh5op]

Same doubt bro 😢

[@CriNageinfo]

2:07:06 Sir Reason galat hai waha likha hai shunt series mai add hoga.

[@kirti118]

4 march ke liya dar lagta hai

[@Tanmaybs]

Or hme 21 feb ke liye

[@safgour496]

Best explanation 😊😊 شکریہ استادے محترم 😊😊

[@kuljitsingh-qp4oe]

2:07:03 sir reason should be false bcz shunt is connected parallel to it. Please reply to my confusion

[@aygamingstar2058]

Sir at 2:12:02 you wrote that F on NO =F on MP -- F on No . Please check 😢😢

[@akd.bhaiya]

Sir que 35(b) mein doosri taraf B bhi to hai to ye A se chl ke ghatna cyahiye phir ek point pe 0 hona chahiye aur phir badhna chahiye....................❤

[@nehachouhan80]

Speed of the particle ques 9 24:14

[@DHRUVYADAV-yb3qk]

23:37 answer is option (a)