20- Magnetic effect of current & Magnetic force of on charge_current

TOPIC MAGNETIC EFFECT OF CURRENT & MAGNETIC FORCE OF ON CHARGE 20 SECTION - I : STRAIGHT OBJECTIVE TYPE [k.M-I: lh/ks oLrqfu"B izdkj 20.1 The negatively and uniformly charged nonconducting disc as shown in the figure is rotated clockwise with great angular speed. The direction of the magnetic field at point A in the plane of the disc is : _.k RedrFk ,dleku:ilsvkosf'kr,ddqpkydpdrhfp=k uqlkjnf{k.kkorZfn'k esacgqrvf/kddks.kh;pky ls?kwejghgsApdrhdsryesafLFkrfcUnqAijpqEcdh;{ks=kdhfn'k gksxh& (A*) into the page (B) out of the page (C) up the page (D) down the page (A*)istesavUnjdhvksj (B) istlsckgjdhvksj (C) istes Åij dh vksj (D)istesauhpsdhvksj Sol. (A) Disc behaves like made up of coils arranged in a plane in which current is flowing in anti-clockwise direction. Hence, the field at A is directed into the page. Hence (A). pdrhblizdkjO;ogkjdjrhgStSls,dryesaO;ofLFkrdq.Mfy;ksalscuhgSftlesa/k jkokekorZfn'k esaizokfgr gksrhgSA blizdkj Aij{ks=k istds vUnjdh vksjgSA blizdkj (A). 20.2 A particle is moving with velocity →  ˆi  3ˆj and it produces an electric field at a point given by → E  2k It will produce magnetic field at that point equal to (all quantities are in S.. units) osx v  ˆi  3ˆj lsxfrekud.kfdlhfcUnqijfo|qr{ks=k →  2kˆ mRiUudjrkgSAd.k}kjkblhfcUnqijmRiUu pqEcdh;{ks=kgksxk (lHkhjkf'k;k¡varjk"Vªh;ekudi)fresagS) (A*) 6ˆi  2ˆj c 2 (B) 6ˆi  2ˆj c2 (C) zero 'kwU; (D) can not be determined from the given data fn;s x;s vk¡dM+ksa ls Kkr ugha fd;k tk ldrk Sol. (A) → → 1 → → → → = 0 q v  r and → = qr  → =   →  → v  E = B 4 r 3 E 4 0 r 3 B 0 0 (v E) c2 20.3 Two observers moving with different velocities see that a point charge produces same magnetic field at the same point A . Their relative velocity must be parallel to r , where r is the position vector of point A with respect to point charge. This statement is : fHkUufHkUuosxksalspyrsgq,nksizs{kdfdlhfcUnqorvkos'k}kjklekufcUnqAijlekupqEcdh;{ks=kmRiUudjrk gqvkns[krsgSaAmudkvkisf{kdosx r dslekukUrjgksukpkfg,]tgk¡ r , fcUnqAdkfcUnqorvkos'kdslkis{kfLFkfr lfn'kgSaA;gdFkugSA (A*) true lR; (B) false vlR; (C) nothing can be said dqN ugha dg ldrs → (D) true only if the charge is moving perpendicular to the r → rHkhlR;tcvkos'k r Sol. (A) Since dsyEcor~xfrekugSA  → → → = 0 q v  r , → → must be same B 4 r 3 v  r where → = velocity of charge with respect to observer Let A and B are the observers then →  → → = → → → (vC v A )  r (v C  vB )  r or →  → → or → → → (v A vB )  r  0 (v A  vB ) || r  → → pwafd → = 0 q v  r , → → leku gksuk t:jh gSA B 4 r 3 v  r tgk¡→ =izs{kddslkis{kvkos'kdkosxAoBizs{kdgSaA rks → (vC → v A ) → → r (v C → → vB ) r ;k →  →  →  0 ;k →  → → (v A vB ) r (v A vB ) || r 20.4 In the figure shown A B C D E F A was a square loop of side 𝑙 , but is folded in two equal parts so that half of it lies in x z plane and the other half lies in the y z plane. The origin ' O ' is centre of the frame also. The loop carries current ' i '. The magnetic field at the centre is: fn[kk;s x;s fp=k AB CD EF A, Hkqtk 𝑙 dk ,d oxkZdkj ywi Fkk ijUrq bls nks leku Hkkxks esa bl izdkj eksM+k x;k gS fd bldk vk/kk Hkkx xz ry esa rFkk vk/kk Hkkx yz ry esa jgrk gSA Qzse dk dsUnz Hkh ewyfcUnq O gSA ywi esa /kkjk dk eku i gS rks dsUnz ij pqEcdh; {kS=k gksxk%& (A) 0 i  ˆi  ˆj  (B) 0 i   ˆi  ˆj  (C*)  ˆi  ˆj  (D) 0 i  ˆi  ˆj    4  𝑙    𝑙     Sol. (C) Due to FABC the magnetic field at O is along y – axis and due to CDEF the magnetic field is along x–axis . Hence the field will be of the form A ˆi  ˆj  Calculating field due to FABC : due to AB : → 0i = (sin 45º + sin45º) ˆj = ˆj BAB 4 𝑙  2 2𝑙   Due to BC : → 0i sin 0º sin 45ºˆj B = 4  𝑙  = 0i ˆj BC l  2 2 2𝑙   Similarly due to FA : → BFA = 0i ˆi Hence → = 0i  1  1  2  ˆi BPABC 𝑙  2 2  2  → BFABC = 20i 𝑙 ˆj  Similarly due to CDEF: → BCDEF = Sol. (C) 20i ˆi  𝑙 →  Bnet = 20i ˆi  ˆj  𝑙 FABCdsdkj.kOijpqEcdh;{ks=ky–v{kdsvuqfn'kgS,oaCDEFdsdkj.kpq-{ks=kx–v{kdsvuqfn'kgS& blizdkj{ks=kA ˆi  ˆj  ds:iesagksxkA FABC ds dkj.k {ks=k dh x.kuk : ABdsdkj.k → 0i = (sin 45º + sin45º) ˆj = 0i ˆj BAB 4 𝑙  2 2𝑙   BC ds dkj.k : → BAB 0i sin 0º sin 45º =  𝑙  = 0i 4l   2  blh izdkjFAdsdkj.k : → BAB = 0i ˆi blh izdkj → = 0i  1  1  2  ˆi BPABC 𝑙  2 2  2  → BPABC = 20i 𝑙 ˆj  blh izdkj CDEF ds dkj.k : → BCDEP = 20i ˆi  𝑙 →  Bnet = 20i ˆi  ˆj  𝑙    20.5 If the magnetic field at 'P' in the given figure can be written as K tan  2  then K is   ;fn 'P' ij pqEcdh; {ks=k K (tan(/2) ls fy[kk tk;sa] rks K gSA (A) Sol. (B) 0 4d (B*) 0 2d (C) 0  d (D) 20 d Let us compute the magnetic field due to any one segment : B = 0i (cos 00  cos(180  )) = 0i 1 cos  = 0i tan  4(dsin) Resultant field will be : 4(dsin) 4d 2 B = 2B = 0i tan    k = 0i net 2d 2 2d fdlh,d[k.Mdsdkj.kpq-{ks-dhx.kukdjrsgS& B = 0i (cos 00  cos(180  )) = 0i 1 cos  = 0i tan  4(dsin) ifj.keh{ks=kgksxk: 4(dsin) 4d 2 B = 2B = 0i tan    k = 0i net 2d 2 2d 20.6 The magnetic field at the origin due to the current flowing in the wire as shown in figure below is fp=kkuqlkj rkj esa çokfgr /kkjk ds dkj.k ewy fcUnq ij pqEcdh; {ks=k gS & (A)  0 (ˆi  kˆ) 8a (B) 0 (ˆi  kˆ) 2a (C*) 0 (ˆi  kˆ) 8a (D) 0  (ˆi  kˆ) Sol. (C) B = 0 B = 0 = 0 [cos 45(ˆi )  cos 45kˆ] = 0 (ˆi  kˆ) 8a 20.7 An infinitely long wire carrying current  is along Y axis such that its one end is at point A (0, b) while the wire extends upto + . The magnitude of magnetic field strength at point (a, 0) is ,d vuUryEckbZdsrkj esa/kkjkI,Y–v{k dsvuqfn'kcgjghgSrFkkftldk ,d fljkfcUnqA(0,b)ijgStcfd nwljkfljk+rdvxzljgSAfcUnq(a,0)ijpqEcdh;{ks=kdkifjek.kgS 0  b  0  b  0  b  (A) 1  (B*) 1  (C)   4a  a2  b2  4a  a2  b2  4a  a2  b2  (D) None of these buesa ls dksbZ ugha Sol. (B)  i 0 i 0 i  b  B = 0 (sin 90+sin(– )) = (1 – sin ) = 1   4 a 4 a 4 a  a2  b2  20.8 Two infinitely long linear conductors are arranged perpendicular to each other and are in mutually perpendicular planes as shown in figure. If  = 2A along the y-axis and  = 3A along negative z-axis and AP = AB = 1 cm. The value of magnetic field strength B at P is - nksvuUryEckbZdsjs[kh;lqpkydijLij,d&nwljsdsyEcor~la;ksftrgSrFk fp=kesafn[kk;svuqlkjijLij yEcor~ ry esa gSA ;fn y v{k ds vuqfn'k 1 = 2ArFkk _.kkRed z-v{k ds vuqfn'k 2 = 3AgS rFkk AP=AB = 1 cm gSrksfcUnq'P'ijpqEcdh;{ks=kBgS& (A) (3 × 10–5 T) ˆj + (– 4 × 10–5 T) kˆ (B*) (3 × 10–5 T) ˆj + (4 × 10–5 T) kˆ (C) (4 × 10–5 T) ˆj + (3 × 10–5 T) kˆ (D) (– 3 × 10–5 T) ˆj + (4 × 10–5 T) kˆ Sol. (B) Magnetic field strength at P due to  → B1 = 01 kˆ 2(AP) 4  107  2 ˆ = 2 1 102 k 1 = (4 × 10–5T) kˆ Magnetic field strength at P due the  → → B2 = 02 ˆj 2(BP) 4  107  3 ˆ = 2  2  102 j = (3 × 10–5T) ˆj Hence, B = (3 × 10–5 T) ˆj + (4 × 10–5T) kˆ 1 ds dkj.k P ijpq-{ks- → B1 = → 01 kˆ 2(AP) 02 ˆ 4  107  2 ˆ = 2 1 102 k 4  107  3 ˆ = (4 × 10–5T) kˆ 2dsdkj.k P ijpq-{ks- → B2 = j 2(BP) = 2  2  10 2 j = (3 × 10–5T) ˆj bl izdkj, B = (3 × 10–5 T) ˆj + (4 × 10–5T) kˆ 20.9 A steady current is set up in a cubic network composed of wires of equal resistance and length d as shown in figure. What is the magnetic field at the centre of cube P due to the cubic network ? ,dfLFkj/k jkdkscjkcjizfrjks/krFk yEckbZddsrkjksalsla;ksftr?kukdkjifjiFkesafp=kuqlkjizokfgrdjrsgSaA ?kukdkj ifjiFk ds dkj.k ?ku ds dsUnz 'P' ij pqEcdh; {ks=k D;k gSA 0 2 0 3 0   (A) Sol. (C) 4 d (B) 4 2 d (C*) 0 (D) 4 d By symmetry, the magnetic field at the centre P is zero. lefer klsdsUnzPijpq-{ks-'kwU;gSA 20.10 Figure shows an amperian path ABCDA. Part ABC is in vertical plane PSTU while part CDA is in horizontal plane PQRS. Direction of circulation along the path is shown by an arrow near point B and at D. o → B.d𝑙 for this path according to Ampere’s law will be : fp=k,d,sfEifj;uiFkABCDAdksçnf'kZrdjrkgSAHkkxABCÅ/okZ/kjryPSTUesagStcfdHk xCDA{kSfrtry PQRSesagSAiFkdsvuqfn'kpyusdhfn'k fcUnqBoDijrhj}kjkçnf'kZrgSAbliFkdsfy,,sfEi;jdsfu;eds vuqlkj o → gksxk: (A) (i1 – i2 + i3) 0 (B) (– i1 + i2) 0 (C) i3 0 (D*) (i1 + i2) 0 Sol. (D) → → B.d𝑙 = 0 (i1 + i3 + i2 – i3) = 0 (i1 + i2) [since for the given direction of circumlation i3 entering at PSTU ABCDA is positive while i3 at PQRS is negative] [pwafdlapj.knhxbZfn'kkdsfy,PSTUijizosf'kri3/kukRedgStcfd PQRSiji3a_.kkRedgSA] Alternate solution oSdfYid gy → → B.d𝑙 ABCDA → → = B.d𝑙 ABCA → → + B.d𝑙 CDAC = µ0 i1 + µ0 i2 = µ0 (i1 + i2) 20.11 A coaxial cable is made up of two conductors. The inner conductor is solid and is of radius R1 & the outer conductor is hollow of inner radius R2 and outer radius R3. The space between the conductors is filled with air. The inner and outer conductors are carrying currents of equal magnitudes and in opposite directions. Then the variation of magnetic field with distance from the axis is best plotted as: ,d lev{kh; /kkjkokgh rkj dks nks pkydksa ls cuk;k x;k gSA vUnj okyk pk y d Bk s l g S r Fk k bl dh f =k T; k R1 gS ,oa ckgj okyk pkyd [kks[kyk gS] ftldh vkUrfjd f=kT;k R2 o ckg~; f=kT;k R3 gSA nksuksa pkydkssas ds e/ ; LFkku esa ok;q Hkjh gSA vkUrfjd o cká pkyd esa leku ifjek.k o foijhr fn'kk esa /kkjk xeu dj jgh gSA rks v{k ls nwjh ds lkFk pqEcdh; {kS=k esa ifjorZu dk loZJs"B oØ gksxk & (A) (B) (C*) (D) Sol. (C) From ampere's law, the field at the axis is zero. From x = zero to R1 , the field increases linearly as the current enclosed increases. From x = R1 to R2 and from x = R2 to R3, the field decreases hyperbolicaly but with different slopes as the media are different. Hence the required graph is ,sfEi;jdsfu;elsv{kij{ks=k'kwU;gSA x= 'kwU;lsR1rd,{ks=kjs[kh;:ilsc<+rkD;ksafdcUn/k jkc<+rhgSA x = R1 ls R2 rd o x =R2 ls R3 rd] {ks=k vk;rh; vfrijoy; ds :i esa ?kVrk gSA ijUrq vyx&vyx a We can consider the given cylinder as a combination of two cylinders. One of radius 'R' carrying current  in one direction and other of radius R carrying current 2 I 3 in both directions. At point A : B = 0 (I / 3) + 0 = 2(R / 2) 0  4 I / 3   R   0 I 3R 0 I At point B : B =  2   + 0 = Sol. (C), (D) csyuds fy, 0 ir 2  R   2  3R B = 2 R2 ; r < a 0 i = 2 r ; r > a gefn;sx;scsyudksnkscsyuksadkla;kstuekuldrsgSA,d'R'f=kT;kdkgSftl s/k jk,dfn'kkesaizokfgr R I gSonwljk 2 f=kT;kdkgSftls 3 /kjknksuksafn'kvksaesaizokfgrgSA fcUnq A ij : B = 0 (I / 3) + 0 = 2(R / 2) 0  4 I / 3   R   0 I 3R 0 I fcUnq Bij B =  2   + 0 = 2  R   2  3R SECTION - III : ASSERTION AND REASON TYPE [k.M- III: dFku vkSj dkj.k izdkj 20.41 STATEMENT-1 : A charged particle undergoes uniform circular motion in a uniform magnetic field. The only force acting on the particle is that exerted by the uniform magnetic field. If now the speed of the same particle is somehow doubled keeping its charge and external magnetic field constant, then the centripetal force on the particle becomes four times. STATEMENT-2 : The magnitude of centripetal force on a particle of mass m moving in a circle of radius R mv2 with uniform speed v is R : oDrO;-1: ,dvkosf'krd.k,dlekupqEcdh;{ks=kesa,dlekuo`Ùkh;xfrdjrkgSAd.kijdk;Zdjusokyk,dgh cygStksfd,dlekupqEcdh;{ks=k}kjkyxk;ktkrkgSAvc;fnmlhd.kdhpkyfdlhizdkjnqxquhgkstkrhgStcfd bldkvkos'krFk ckg~;pqEcdh;{ks=kfu;rgSrksd.kijvfHkdsUnzh;cypkjxqukgkstkrkgSA oDrO;-2: Rf=kT;kdso`Ùkesa,dlekupkyvlsnzO;ekum ds,dd.kijvfHkdsUnzh;cydkifjek.k mv2 R gksrkgSA (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D*) Statement-1 is False, Statement-2 is True (A) oDrO;-1lR; gS] oDrO;-2lR; gS ;oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k gSA (B)oDrO;-1lR; gS] oDrO;-2lR; gS ; oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k ugha gSA (C) oDrO;-1lR; gS] oDrO;-2vlR; gS ; (D*) oDrO;-1vlR; gS] oDrO;-2lR; gS Sol. (D) The magnitude of magnetic force on charged particle undergoing uniform circular motion in uniform magnetic field is F=qvB  If v is doubled keeping q and B constant, the force F just doubles. Hence statement 1 is false. Sol. (D) ,dlekupqEcdh;{ks=kesa,dlekuo`Ùkh;xfrdjrsgq,,dvkosf'krd.kijpqEcdh;cydkifjek.kgS F=qvB ;fnqrFk Bdksfu;rj[krsgq;s]vdksnqxukfd;ktkrkgS]rkscyFnqxukgkstkrkgSA blfy;s oDrO;-1vlR;gSA 20.42 Statement 1 : A current carrying closed loop remains in equilibrium in a uniform and constant magnetic field parallel to its axis. Consider forces only due to this magnetic field . Statement 2 : Torque on a current carrying closed loop due to a magnetic field is maximum when the plane of the coil is parallel to the direction of the magnetic field. oDrO;&1: ,d/k jkokghcUnywi],dlekufu;rpqEcdh;{ks=ktksywidsv{kdslekUrjgSesalkE;oLFk esagSAdsoy pqEcdh;{ks=kdsdkj.kcyksadksghfyft,A oDrO;&2: ,d/k jkokghcUnywi¼dq.Myh½ijpqEcdh;{ks=kdsdkj.kcyk?kw.kZvf/kdregksxk;fnywidkrypqEcdh; {ks=k dh fn'kk ds lekUrj gSA (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B*) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True (A) oDrO;-1lR; gS] oDrO;-2lR; gS ;oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k gSA (B*)oDrO;-1lR; gS] oDrO;-2lR; gS ; oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k ugha gSA (C) oDrO;-1lR; gS] oDrO;-2vlR; gS ; (D) oDrO;-1vlR; gS] oDrO;-2lR; gS Sol. (B) Both statements are correct, but statement 2 is not a correct explanation of statement 1. (B) nksuksoDrO;lgh gS fdUrqdFku 2,dFku1dklgh Li"Vhdj.kughagSA 20.43 STATEMENT-1 : A solenoid tend to contract (along its length) when a current is passed through it. STATEMENT-2 : If current in two coaxial circular rings of equal radii is in same sense( as seen by an observer on axis away from both the rings), the rings attract each other. Further the given current carrying rings attract each other because parallel current attract oDrO;-1: tc,d/k jkdksifjukfydklsxqtkjktkrkgSrksifjukfydkviuhyEckbZdsvuqfn'kfldqM+usdhizo`fÙk j[krhgSA oDrO;-2: ;fnnkslek{kh;lekuf=kT;kokyho`Ùk dkjoy;ksaesa/k jk¼nksuksaoy;ksdsv{kijnwjfLFkrizs{kdds fy,½,dghleku(sensefn'k )gSrksoy;s,dnwljsdksvkdf"kZrdjsxhrFk nhxbZ/k jkokghoy;s,dnwljsdks vkdf"kZrdjrhgSD;ksafdlekUrj/k jk,vkdf"kZrdjrhgSA (A*) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 oDrO;-1 lR; gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dk lgh Li"Vhdj.k gSA (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 oDrO;-1 lR;gS, oDrO;-2 lR; gS; oDrO;-2 oDrO;-1 dklgh Li"Vhdj.k ugha gSA (C) Statement-1 is True, Statement-2 is False oDrO; -1 lR; gS, oDrO;-2 vlR; gSA (D) Statement-1 is False, Statement-2 is True. oDrO; -1 vlR; gS, oDrO;-2 lR; gSA Sol. (A) Solenoid tends to contract because the current in all the circular turns is in same direction. Hence all turns( can be assumed as a ring) attract each other. Sol. (A) lksysMksbM dh fldqM+us dh izo`fÙk gksxh D;ksafd lHkh pDdjksa esa /kkjk leku fn'kk esa gSA vr% lHkh pDdjksa dks oy;ksa dh rjg eku ldrs gSa tks fd ,d nwljs dks vkdf"kZr djsaxsA 20.44 Statement 1 : A direct uniformly distributed current flows through a solid long metallic cylinder along its length. It produces magnetic field only outside the cylinder . Statement 2 : A thin long cylindrical tube carrying uniformly distributed current along its circumference does not produce a magnetic field inside it. Moreover, a solid cylinder can be supposed to be made up of many thin cylindrical tubes. oDrO;&1 : ,d fn"V (direct) /kkjk tks fd ,d Bksl /kkfRod yEcs csyu esa ls bldh yEckbZ ds vuqfn'k cg jgh gS rFkk ;g ,d leku :i ls blesa forfjr gSA ;g csyu ds dsoy ckgj gh pqEcdh; {ks=k mRiUu djrh gSA oDrO;&2 : ,d yEch iryh csyukdkj [kks[kyh V~;wc esa bldh yEckbZ ds vuqfn'k ,d leku :i ls forfjr /kkjk cg jgh gSA ;g V~;wc vius vUnj fLFkr fcUnq ij pqEcdh; {ks=k mRiUu ugha djrh gS rFkk ,d Bksl csyu dks dbZ lkjs irys csyukdkj V~;wcksa ls cuk gqvk eku ldrs gSA (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D*) Statement-1 is False, Statement-2 is True (A) oDrO;-1lR; gS] oDrO;-2lR; gS ;oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k gSA (B)oDrO;-1lR; gS] oDrO;-2lR; gS ; oDrO;-2,oDrO;-1 dk lgh Li"Vhdj.k ugha gSA (C) oDrO;-1lR; gS] oDrO;-2vlR; gS ; (D*) oDrO;-1vlR; gS] oDrO;-2lR; gS Sol. (D) The current through solid metallic cylinder also produces magnetic field inside the cylinder. Hence statement-1 is false (D) ,d/k fRodBksl/k jkokghcsyudsdkj.kcsyudsvUnjHkhpqEcdh;{ks=kmRiUugksrkgSvr%dFku-1vlR;gSA SECTION - IV : COMPREHENSION TYPE [k.M- IV:cks/ku izdkj Paragraph for Question Nos. 11 to 12 iz'u 11 ls12 dsfy, vuqPNsn A small particle of mass m = 1kg and charge of 1C enters perpendicularly in a triangular region of uniform magnetic field of strength 2T as shown in figure : fp=kuqlkj2TrhozrkdspqEcdh;{ks=kdsf=kHkqtkdkj{ks=kesanzO;ekum=1kgrFk vkos'k1CdkNksVkd.kfp=kuqlkjyEcor~ izfo"V gksrk gS : 20.45 Calculate maximum velocity of the particle with which it should enter so that it complete a half–circle in magnetic region : pqEcdh;{ks=kesav/kZo`Ùkiw.kZdjldusdsfy,d.kdkvfHk"Bvf/kdreosxdkekuifjdfyrdhft,: (A) 2 m/s (B) 2.5 m/s (C*) 3 m/s (D) 4 m/s Sol. In triangle PMC f=kHkqt PMC esa MP cos53º = MC 3 R 5 = 4  R 12 = 8R R= 3 m (Risthemaximum radiusof half–circle) (v/kZo`Ùk dh vf/kdre f=kT;k RgS) Rmax = mumax qB max = 3 m/s. 20.46 In previous question, if particle enters perpendicularly with velocity 48 m/s in magnetic region. Then, how much time will it spend in magnetic region : ;fnfiNysiz'uesa]d.kpqEcdh;{ks=kesa48m/sosxlsyEcor~izfo"VgksrkgS]rkspqEcdh;{ks=kesa;gfdrukle;O;rhr djsxk: 11 7 13 17 (A) 360 sec. (B*) 360 sec. (C) 360 sec. (D) 360 sec. mu Sol. R = qB = 24 m Let, ekuk MPQ =  Bygeometry, T;kferh ls CPO = (37 – ) InCPOesa, OC sin(CPO) 20 sin(37º) 5 OP = sin(PCO) 24 = sin(180º37º ) 5  6 sin(37º) = 3 sin(37º – ) = 1 2  =  = qB m 7 180 rad.  = 2 rad/sec. 7 t = 360 sec. Comprehension vuqPNsn # 2 Curves in the graph shown give, as functions of radial distance r, the magnitude B of the magnetic field inside and outside four long wires a, b, c and d, carrying currents that are uniformly distributed across the cross sections of the wires. Overlapping portions of the plots are indicated by double labels. ,dleku:ilsvuqçLFkdkVesaforfjr/k jkçokghpkjyEcsrkjksaa,b,crFk d dsckgjrFkkvUnjpqEcdh;{ks=k ds ifjek.k ,oa f=kT;h; nwjh r ds chp oØ n'kkZ;s x;s gSA oØ ds vfrO;kih {ks=k] nks js[kkvkssa }kjk n'kkZ;s x;s gSA B a,b a b c c,d r 20.47 Which wire has the greatest radius ? fdlrkjdhf=kT;klokZf/kdgS\ (A) a (B) b (C*) c (D) d Sol. Inside the cylinder csyudsvanj B.2r =  .  R2 r 2 B = 0 .r 2R2 ..............(1) outside the cylinder csyudsckgj B.2r =    B = 0 2r ..............(2) Inside cylinder B  r and outside B  1 r csyu ds vanj BrrFkk ckgj B 1 So from surface nature of mag field changes. pqEcdh;{kS=kdh lrgdh izdfrcnyrh gSA Hence it is clear from the graph that wire 'c' has greatest radius. vr%xzkQlsLi"BgSfdrkj'c'dhf=kT;klokZf/kdgSA 20.48 Which wire has the greatest magnitude of the magnetic field on the surface ? fdlrkjdhlrgijpqEcdh;{ks=kdkifjek.klokZf/kdgS& (A*) a (B) b (C) c (D) d Sol. Magnitude of mag field is maximum at the surface of wire 'a'. rkj adhlrgijpqEcdh;{kS=kdkifjek.kvf/kdregSA 20.49 The current density in wire a is rkj'a' esa/k jk?kuRogS& (A*) greater than in wire c. (B) less than in wire c. (C) equl to that in wire c. (D) not comparable to that of in wire c due to lack of information. (A*)'c'rkj ls T;knk gSA (B)'c' rkj ls de gSA (C) 'c'rkj ds cjkcj gSA. (D)lwpukvi;kZIrgksusdsdkj.kclsrqyukughadhtkldrhA Sol. Inside the wire rkjdsvanj B(r) = 0 .  .r 2 R2 dB dr = 0 .  2 R2 i.e.slope vFkkZr~ r3 = r2 > r1 The ions fall at different positions 1, 2, 3 and 4, as shown. Correctly match the ions with respective falling positions. ,diqatftlesaA,B,CoDpkjizdkjdsvk;ugSAfp=kesan'k Z;savuqlkjle:ipqEcdh;{ks=kesaizos'kdjrkgSA {kS=kdkxt dsrsydsyEcor ,dijUrqbldhmfprfn'kkughanhxbZgSAlHkhvk;uiqatesaleku pkyls pyrs gSAuhpsnhxbZlkj.khvk;uksadsnzO;ekuovkos'kdkscrkrhgSA v k; u nzO; eku v kos' k A 2m + e B 4m – e C 2m – e D m + e v k; u i aqt r4 > r3 = r2 > r1 vk;ufofHkUufLFkfr;ksa1,2,3o4ijfp=k uqlkjfxjrsgSaAvk;uksadsmudsfxjusdhlghfLFkfr;ksadslkFklqesfyr djokb;saA (A) a (p) 1 (B) b (q) 2 (C) c (r) 3 (D) d (s) 4 Sol. R = mv/qB RB > RA andvkSj RA = RC (in oppositesense foifjr fn'kk esa) and vkSjRD issmallest U;wure gSA (A) s (B) p (C) q (D) r 20.54 Three wires are carrying same constant current i in different directions. Four loops enclosing the wires in different manners are shown. The direction of d→ is shown in the figure : rhurkjksaesalekufu;r/kjkifHkUufn'kvksaesaizokfgrgSArkjksads?ksjsesafHkUuizdkjlspkjywin'kZ;sx;sgSA → d𝑙 dhfn'k fp=kesafn[k ;svuqlkjgS& Loop-1 Loop-2 Loop-3 Loop-4 → → (A) Along closed Loop-1 (p)  B .d𝑙 cUnywi1dsvuqfn'k = µ0 i → → (B) Along closed Loop-2 (q)  B .d𝑙 cUnywi2dsvuqfn'k = –µ0 i → → (C) Along closed Loop-3 (r)  B .d𝑙 = 0 cUnywi3dsvuqfn'k (D) Along closed Loop-4 (s) net work done by the magnetic force to move a unit charge along the loop is zero. cUnywi4dsvuqfn'k ,dkadvkos'kdkscUnywidsvuqfn'kxfrdjkusesapqEcdh;cy }kjk ifj.keh dk;Z'kwU;gS& → → (t)  B .d𝑙 =  (2i) Ans. (A)  (q, s) (B)  (p, s) (C)  (q, s) (D)  (p, s) Sol. Work done by magnetic force on the charge = 0 in any part of its motion.  'S' is matching for all parts (i), (ii), (iii), (iv) For loop 1 For loop 2  = – i + i – i = – i in  = i – i + i = i in   → → B.d𝑙 → → B.d𝑙 = 0(i) = 0 (i) For loop 3 For loop 4  = –i – i +i = –i in  = + i + i – i = +i in → → B.d𝑙 → → B.d𝑙 = 0(i) = 0 (i) (Note : That current will be taken as positive which produces lines of magnetic field in the same sense in which d𝑙 is taken) Sol. xfrds fdlh Hkh Hkkxds fy,pqEcdh; cy}kjk ,d vkos'k ijfd;k x;k dk;Z =0  'S' izR;sd Hkkx(i), (ii), (iii), (iv) ls esy [kkrk gSA → → ywi1 ds fy, in = – i + i – i = – i  B.d𝑙 = 0(i) → → ywi2 ds fy, in = i – i + i = i  B.d𝑙 = 0 (i) → → ywi 3 ds fy, in = –i – i +i = –i B.d𝑙 = 0(i) → → ywi4 ds fy, in = + i + i – i = +i B.d𝑙 = 0 (i) (Note:og/k jk/kukRedyhtk,xhtksmlhfn'k esapqEcdh;{ks=kdhjs[k ,amRiUudjsxhftlesa d𝑙 fy;kx;kgSA) 20.55 Column-II gives four situations in which three or four semi infinite current carrying wires are placed in xy-plane as shown. The magnitude and direction of current is shown in each figure. Column-I gives statements regarding the x and y components of magnetic field at a point P whose coordinates are P (0, 0, d). Match the statements in column-I with the corresponding figures in column-II LrEHk-IIesapkjfLFkfr;k¡nhxbZgSftlesarhu;kpkjv/kZvuUr/k jkokghrkjxy-ryesafLFkrgStSlkfdfn[k ;k x;kgSA/k jkdhfn'k rFk ifjek.kizR;sdfp=kesafn[k ;sx;sgSaALrEHk-IesafcUnqPftldsfunsZ'k ad(0,0,d)gSa]ij pqEcdh;{ks=kdsxrFk y?kVdksalslEcfU/krdFkufn;sx;sgSaALrEHk-Iesafn;sx;sdFkuksadksLrEHk-IIesafn;sx;slaxr fp=k salslqesfyrdhft,rFk viusmÙkjdksnhxbZOMResa4×4eSfVªDlesamfprcqycqyksadksdkykdjn'k Zb;sA Column I Column II (A) The x component of magnetic field at (p) point P is zero in (B) The z component of magnetic field at (q) point P is zero in (C) The magnitude of magnetic field at (r) point P is 0 i 4d in (D) The magnitude of magnetic field at (s) point P is less than 0 i 2d in (t) LrEHkI LrEHkII (A) fcUnqP ijpqEcdh;{ks=kdkx-?kVd (p) 'kwU;gSA (B) fcUnqPijpqEcdh;{ks=kdkz-?kVd (q) 'kwU;gSA (C) fcUnqPijpqEcdh;{ks=kdkifjek.k (r) 0 i 4d gSA (D) fcUnqPijpqEcdh;{ks=kdkifjek.k (s) 0 i 2d ls de gSA (t) Ans. (A) p,q, r, t (B) p, q, r, s, t (C) r (D) p, q, r, s, t Sol. The magnetic field is along negative y-direction in case A,B and C Hence z-component of magnetic field is zero in all cases. The magnetic field at P is 0 i for case (r) 4 d The magnetic field at P is less than 0 i for all cases. 2 d Sol. lHkh pkjksa fLFkfr;ksa esa pqEcdh; {ks=k _.kkRed y-fn'kk ds vuqfn'k gSA blfy, lHkh fLFkfr;ksa esa pqEcdh; {ks=k dk z-?kVd'kwU;gSA fLFkfr(r)dsfy,PijpqEcdh;{ks=k 0 i gSA lHkhfLFkfr;ksadsfy,PijpqEcdh;{ks=k 0 i lsdegSA 20.56 A square loop of uniform conducting wire is as shown in figure. A current I ( in amperes) enters the loop from one end and exits the loop from opposite end as shown in figure. The length of one side of square loop is 𝑙 metre. The wire has uniform cross section area and uniform linear mass density. In four situations of column I, the loop is subjected to four different magnetic field. Under the conditions of coulmn I, match the column I with corresponding results of column II ( Bo in column I is a positive nonzero constant) Column I Column II → (A)  B ˆi in tesla (p) magnitude of net force on loop is (B)  B ˆj in tesla (q) magnitude of net force on loop is zero BoI 𝑙 newton (C) →  B (ˆi  ˆj) in tesla (r) magnitude of net torque on loop about its centre is zero (D) →  B kˆ in tesla (s) magnitude of net force on loop is BoI 𝑙 newton (t) magnitude of force on wire along y axis is B0I 𝑙/2 along x axis ,dleku pkyd rkj dk ,d oxkZdkj ywi fp=kkuqlkj fn[kk;k x;k gSA ,d /kkjk I (,fEi;j esa) ywi ds ,d fljs ls izos'k djrh gS rFkk ywi ds foijhr fljs ls fp=kkuqlkj fudyrh gSa oxkZdkj ywi dh ,d Hkqtk dh yEckbZ 𝑙 ehVj gSA rkj ,dleku vuqizLFk dkV rFkk ,dleku js[kh; nzO;eku ?kuRo dk gSA LrEHk I dh pkj fLFkfr;ksa esa] ywi dks pkj fHkUu&fHkUu pqEcdh; {ks=k esa j[kk tkrk gSA LrEHk I dh fLFkfr;ksa ds vUrxZr] LrEHk II ds laxr ifj.kkeksa ls lqesfyr dhft,A ¼LrEHk I esa Bo ,d v'kwU; fu;rkad gS½ LrEHkI LrEHkI (A) →  B ˆi Vslykesa (p) ywiij usV cy dk ifjek.k 2 B I𝑙 U;wVugSA B o o (B) →  B ˆj Vslykesa (q)ywiijusVcydkifjek.k'kwU;gSA → (C) B  Bo (ˆi  ˆj) Vslykesa (r)ywiijbldsdsUnzdsifjr%usVcyvk?kw.kZdkifjek.k'kwU;gSA → (D)  B kˆ Vslyk esa (s) ywi ij usV cy dk ifjek.k B I 𝑙 B o U;wVu gSA vuqfn'k B0I 𝑙/2 gSA o (t) y v{k ds vuqfn'k j[ks rkj ij cy x v{k ds Ans. A – r,s B – r,s C – q,r D – p,r, t Sol. (A) Because the magnetic field is parallel to x-axis, the force on wire parallel to x-axis is zero. The force on each wire parallel to y-axis is B i 𝑙. Hence net force on loop is B i𝑙. Since force on each wire parallel o 2 o to y-axis passes through centre of the loop net torque about centre of the loop is zero. (B) Because the magnetic field is parallel to y-axis, the force on wire parallel to y-axis is zero. The force on each wire parallel to x-axis is B i 𝑙. Hence net force on loop is B i𝑙. Since force on each wire parallel o 2 o to x-axis passes through centre of the loop, net torque about centre of the loop is zero. (C) Since net displacement of current from entry point in the loop to exit point in the loop is along the diagonal of the loop.The direction of external uniform magnetic field is also along the same diagonal. Hence net force on the loop is zero. Since force on each wire on the loop passes through centre of the loop net torque about centre of the loop is zero. (D) The net displacement of current from entry point in the loop to exit point in the loop is along the diagonal ( of length 𝑙.) of the loop.The direction of external uniform magnetic field is also perpendicu- lar to the same diagonal. Hence magnitude of net force on the loop is Boi( 𝑙 ). Since force on each wire on the loop passes through centre of the loop net torque about centre of the loop is zero. Sol. (A)pwafdpqEcdh;{ks=kx-v{kdslekarjgS]rksx-v{kdslekarjrkjijcy'kwU;gksxkAizR;sdrkjijy-v{kds lekUrjcyB i 𝑙gSAblfy;sywiijusVcyB i𝑙gksxkApwafdy-v{kdslekarjizR;sdrkjijcy]ywidsdsUnz o 2 o lsxqtjrkgS]rksywidsdsUnzdsifjr%cyk?kw.kZ'kwU;gksxkA (B) pw¡fd pqEcdh; {ks=k y-v{k ds lekarj gS] blfy;s y-v{k ds lekarj izR;sd rkj ij cy 'kwU; gSA x-v{k ds lekUrj izR;sd rkj ij cy Boi𝑙/2 gSA blfy;s ywi ij usV cy Boi𝑙 gSA pwafd x-v{k ds lekarj izR;sd rkj ij cy ywi ds dsUnz ls xqtjrk gS] blfy;s ywi ds dsUnz ds ifjr% usV cyk?kw.kZ 'kwU; gksxkA (C)pwafdywiesaizos'kfcUnqlsfudklfcUnqrd/k jkdkusVfoLFk iuywidsfod.kZdsvuqfn'kgSAcká,dleku pqEcdh;{ks=kdhfn'k Hkhmlhfod.kZdsvuqfn'kgSAblfy;sywiijusVcy'kwU;gSApwafdywiesaizR;sdrkjijcy ywidsdsUnzlsxqtjrkgS]blfy;sywidsdsUnzdsifjr%usVcyk?kw.kZ'kwU;gksxkA (D)ywiesaizos'kfcUnqlsfudklfcUnqrd/k jkdkusVfoLFk iuywids( 2 𝑙yEckbZds)fod.kZdsvuqfn'kgSAcká ,dlekupqEcdh;{ks=kdhfn'k Hkhmlhfod.kZdsyEcor~gksxhAblfy;sywiijusVcydkifjek.kBoi( 2 𝑙)gSA pwafdywiesaizR;sdrkjijcyywidsdsUnzlsxqtjrkgSrksywidsdsUnzdsifjr%usVcyk?kw.kZ'kwU;gksxkA 20.57 A particle enters a space where exists uniform magnetic field B  Bx i  By j  Bzk & uniform electric field E  E i  E j  E k with initial velocity →  u i  u j  u k . Depending on the values x y z u x y z of various components the particle selects a path. Match the entries of column A with the entries of column B. The components other than specified in column A in each entry are non-zero. Neglect gravity. → ,dd.kizkjfEHkdosx u  ux i  uy j  uzk . ls,sls{ks=kesaizos'kdjrkgSftlesa,dlekupqEcdh;{ks=k B  Bx i  By j  Bzk ,oa,dlekufo|qr{ks=k E  Ex i  Ey j  Ezk fo|ekugSAfofHkUu?kVdksadsekuksa ds vk/k jijd.k,diFkpqurkgSALrEHkB dhizfof"B;ksadkfeykuLrEHkA dhizfof"B;ksadslkFkdhft;saA LrEHkAdh izR;sdizfof"Besavkusokys?kVdkasdsvfrfjDrlHkh?kVdv'kwU;gSA ¼xq:Rokd"kZ.kux.;ekus½ Column -I Column -II (A) By = Bz = Ex = Ez = 0; u = 0 (p) circle (B) E = 0, uxBx + uyBy   u B (q) helix with uniform pitch → → and constant radius u LrEHk-I LrEHk-II (A) By = Bz = Ex = Ez = 0 u = 0 (p) o`Ùk (B) E = 0, u B + uyBy   u B (q) ,dleku fip¼dq.Mfyuh;kpwM+hvUrjky½ofu;r → → f=kT;kdhdq.MyhuqekiFk¼helical iFk½ u Sol. (A) r, (B) q,t (C) t, (D) s → → → → F  qE  q(V B) If u  0 , → B  Bxi and → E  Ey j then charge particle will start to move in y-direction due to electric field and as it aquires velocity it will experience force due to magnetic field and will move in a cycloid path. Similarly, one can find path for other cases. (A) If By = Bz = Ex = Ez = 0 ; u = 0 then path is cycloid. (B) If E = 0, uxBx + uyBy   u B then path is helix with uniform pitch and constant radius or straight line. → → (C) If ux B  0, ux E  0 then path is straight line. (D) If →  B, B||E then path is helix with variable pitch and constant radius. (A) r, (B) q,t (C) t, (D) s → → → → F  qE  q(V B) ;fn →  , →  B ˆi rFkk →  E ˆj u 0 B x E y rcvkos'ky-fn'k esafo|qr{ks=kdsdkj.kxfrdjukizkjEHkdjsxkArFkkosxdksizkIrdjusijblijpqEcdh; {ks=kdsdkj.kHkhcyyxsxkod.kpØh;iFkesaxfrdjsxkAblhrjgvU;fLFkfr;ksadsfy,HkhiFkdhx.kuk dj ldrs gSA (A) ;fn By = Bz = Ex = Ez = 0 ; u = 0 rc iFk pØh; gSA (B) ;fn E = 0, uxBx+ uyBy  uzBz rc iFk dq.Myhuqek ds lkFk ,dleku pwM+h vUrjky rFkk fu;e f=kT;k oljyjs[kkgksxhA → → (C) ;fn ux B  0, ux E  0 rc ljy js[kk gksxhA (D) ;fn →  B, B||E rc iFk dq.Myhuqek ds lkFk vleku pwM+h vUrjky rFkk fu;e f=kT;kA SECTION - VI : INTEGER TYPE [k.M - VI : iw.kkZad izdkj SHORT SUBJECTIVE y?kqÙkjkRediz'u 20.58 A uniformly charged ring of radius 10 cm rotates at a frequency of 104 rps about its axis. Find the ratio of energy density of electric field to the energy density of the magnetic field at a point on the axis at distance x 20 cm from the centre is 10 109 then x is . ,dle:ivkosf'kr10cmf=kT;kdhoy;bldsv{kdslkis{k104rpsdhvko`fÙkls?kw.kZudjrhgSAv{kijdsUnz x ls20cmnwjhij,dfcUnqijfo|qr{ks=kdkÅtkZ?kuRoopqEcdh;{ks=kdsÅtkZ?kuRodkvuqikr 10 109gSrks xgksxkA Ans. 91 Sol. Electric field at P is Pijfo|qr{ks=kgS& Q x E = 4 0 (x 2  r 2 )3 / 2 0 Magnetic field at P is B = 4 2 ir 2 (x2  r 2 )3 / 2 0 = 4 2 Q2 f 2 r 2 (x2  r 2 )3 / 2 f = frequency of revolution. Electric energy density = 1  E2 2 B2 ; Magnetic energy density 2 Electric energy density 1 o E 2 2 = 2 = E = x 2 c 2 0 9 = × 1010 = 9.1 × 109 magnetic energy density B 2 2 0 0 c 2B 2 2 ir 2 42 f 2 r 4 2 0 2 Q2 f 2 r 2 P ij pqEcdh; {ks=k gS B= 4 (x2  r 2 )3 / 2 = 4 (x2  r 2 )3 / 2 f=?kw.kZuxfrdhvko`fÙk 1 2 B2 fo|qr ÅtkZ ?kuRo = 2 oE ; pqEcdh; ÅtkZ ?kuRo 2 0 fo|qr ÅtkZ?kuR pqEcdh; Åkt?ZkuR = Ans. 9.1 × 109 1 o E 2 2 B 2 = 2 0 E2 c 2B 2 = x 2 c 2 42 f 2 r 4 9 = 2 × 1010 = 9.1 × 109 → 20.59 A neutral particle is at rest in a uniform magnetic field B . At t = 0, particle decays into two particles each of mass ' m ' and one of them having charge ' q '. Both of these move off in separate paths lying in plane → perpendicular to B. At later time, the particles collide. if this time of collision is xm /qB then x is (neglecting the interaction force). → ,dmnklhud.kle:ipqEcdh;{ks=k B esafojkeijgSAt=0ijd.knksHk xksaesafoHk ftr(decay) gkstkrkgSAizR;sd dknzO;ekumomuesals,ddkvkos'k'q'gSAnksuksaBdsyEcorryesavyx&vyxekxZijpyrsgSAdqNle;ckn d.k Vdjk tkrkgSA vUr%fØ;k cyux.; ekurs gq, VDdj dk ;gle; xm/qBgksrks x Kkr djksA [ Ans.:  m ] q B Sol. Since, total charge is zero initially thus the two particle will be of opposite charges. Initially the neutral particle is at rest, so both will have same speed. As both particle move in opposite directions, magnetic force on them will be in the same direction and of same magnitude. Using R = mV , both will be moving in the circle of same radius. So they will meet at qB t = R  m Ans. V qB pwafdizkjEHkesadqyvkos'k'kwU;gSblizdkjnksd.kfoijhrvkos'kdsgksaxsAizkjEHkesamnklhud.kfojkeijgS]A blfy,nksuksadhpkylekugksxhAD;ksafdnksuksad.kfoijhrfn'k esaxfrdjrsgSa]muijpqEcdh;cylekufn'k esaolekuifjek.kdkgksxkA R=mV dkmi;ksxdjrsgq,]nksuksalekuf=kT;kdso`ÙkesaxfrdjjgsgksaxsAblfy,osfcUnqQijfeysaxsAvFk Zr~ qB izkjfEHkd fcUnqPdsO;klr%foijhrAblfy,fy;kx;kle;gksxk& t = R  m Ans. V qB 20.60 As shown in the figure, three sided frame is pivoted at P and Q and hangs vertically. Its sides are of same length and have a linear density of 3 kg/m. A current of 10 3 Amp is sent through the frame, which is in a uniform magnetic field of 2T directed upwards as shown. Then angle in degree through which the frame will be deflected in equilibrium is : (Take g = 10 m/s2) fp=kesafn[k ;svuqlkj]rhuHkqtkvksaokyk,dÝsePrFk QijdhyfdrgSrFk m/oZ>wyrkgSAbldhizR;sdHkqtk dhyEckbZlekugSrFk bldkjs[kh;?kuRo fd-xzk-@eh-gSA10 3 ,Eih;jdh/k jkÝseesaizokfgrdhtkrhgS rFk fn[k ;svuqlkj2Tdk,dlekupqEcdh;{ks=kftldhfn'k mijdhvksjgS]mifLFkrgSAlkE;oLFk esaogdks.k fMxzh esa ftlls Ýse fopfyr gksxk gS & (g = 10 eh-@lS-2 ekus) Ans. 45° Sol. Torque of magnetic force about PQ (1) PQdslkis{kpqEcdh;cy dkcyk?kw.kZ (1) m = ( L B) L cos =  L2 B cos (2) Torque of gravitational force about PQ PQ dslkis{kxq:Roh;cydkcyk?kw.kZ g = [(L) g L sin + 2 ( L) g (1/2) L sin] = 2  L2 g sin (1) m = g  tan  = IB 2 g = = 1   = 45°