Title 04. MOVING CHARGES AND MAGNETISM(H).pdf ✅

NEET REVISION 14. () : Explana on If we divide the total loop into four parts then field = sum of field due to individual parts. 15. () : Explana on Dividing 16. () : Explana on (A) (due to circular coil) Net magne c field at (B) Net magne c field at (C) - (S) (C) Net magne c field at (D) - (R) (D) (A)-(R), (B)-(P), (C)-(S), (D)-(Q). B = BI + BII + BIII + BIV BII = BIV = 0 B = BI + BIII = ( ) + ( ) = μoI 2R 3 4 μoI 4R 1 4 7μoI 16R 10mT = (i1 − i2) μ0 2π(r/2) 30mT = (i1 + i2) μ0 2π(r/2) : i1i2 = 2 (A) − (R) B1 B1 = ⋅ = ⊗ μ0 4π 2πI r μ0I 2r O Bnet = B1 − B2 = − = [1 − ] μ0I 2r μ0I 2πr μ0I 2r 1 π Bnet = ( ) ⊗ μ0I 2r π − 1 π (B) − (P) B1 = ⋅ ⊙ μ0 4π I r B2 = ⋅ = ⊙ μ0 4π πI r μ0I 4r B3 = ⊙ μ0 4π I r O Bnet = B1 + B2 + B3 = [ + 1 + ] μ0 4 I r 1 π 1 π = [ + 1] μ0I 4r 2 π Bnet ( ) μ0I 4r 2 + π π O B1 = 0, B2 = ⋅ = ⊗ μ0 4π πI r μ0I 4r B3 = ⊗ μ0 4π I r Bnet = B2 + B3 + B1 [1 + ] μ0I 4r 1 π = ⊗ μ0I(π + 1) 4πr B1 = B3 = 0 Bnet = B2 = ⋅ μ0 4π πI r ⇒ Bnet = μ0I 4r
NEET REVISION 17. () : Explana on Magne c at the centre of current carrying circular coil is given by No. of turns Similarly, 18. () : Explana on Resistance of galvanometer coil, and current Range of current is So, the shunt resistance is 19. () : Explana on 20. () : Explana on Magne c field due and is zero. Magne c field due 21. () : Explana on Here, Magne c field, 22. () : Explana on The magne c dipole moment of the current carry‐ ing coil is given by The torque ac ng on the coil is 23. () : Explana on Magne c field inside the long cylindrical conduc‐ tor, Magne c field outside the cylinder at a distance from the axis Here, 24. () : Explana on For to be zero at , Thus will be zero at point at a distance from wire carrying current. B = μ0Ni R = N1 = 9 B1 = = μ0N1i 2R 9μ0i 2R B2 = = μ0N2i 2r μ0i 2R = = 9 ⇒ B1 = 9B2 B1 B2 9μ0 i 2R μ0 i 2R ∴ B2 = B1 9 R = 10Ω = 1 mA 0 − 100 mA. S = IgR I−Ig = 1×10−3×10 100×10 −3−1×10 −3 = = 0.1 Ω 10 99 → B 0 = → B PQ + → B QR + → B RS + → B SP = → 0 + ( ) k^ + → 0 + ( ) (−k^) = [ − ] ^k = [ − ] μ0I 4πr1 π 3 μ0I 4πr2 π 3 μ0I 12 1 r1 1 r2 4π × 10 −7 × 10 12 1 4 × 10 −2 1 7 × 10 −2 = [ ] = 0.11 × 10 −4 T ∴ B→ 0 = 1.12 × 10 −5 T 4π × 10 −7 × 10 12 × 10 −2 7 − 4 4 × 7 AB CD BC = μoIα 4πR v = 6.6 × 10 15rps = 6.6 × 10 15 Hz q = e = 1.6 × 10 −19C, r = 0.53 × 10 −10 m B = = (∵ I = qv) μ0I 2r μ0qv 2r = 4π×10−7×1.6×10−19×6.6×10 15 2×0.53×10−10 = 12.5 T M ̄ = NiAn^ = 100 × 0.5(0.08) × 0.044 ^i = 16 × 10 −2 (^i)Am2 τ = M × B = MB(^i × ^j) = 16 × 10 −2 × 0.05 k^ √2 = 5.66 × 10 −5 (Nm)^k Bin = μ0 4π 2Ir R2 r ′ Bout = μ0 4π 2I r ′ ∴ = ; Bout = Bin Bin Bout rr ′ R2 R 2 rr ′ Bout = 10 T, r ′ = R + 4R = 5R r = R − = R 4 3R 4 Bout = 10 × = T R2 ( )(5R) 3R 4 8 3 B P B1 = B2 = μ0(12.5) 2π(6 − x) μ0(2.5) 2πx = 5 1 6 − x x 5x = 6 − x x = 1 cm B P 1cm 2.5A