Magnetism and Magnetic Effects of Current MCQ.pptx

MCQs in Magnetism and Magnetic effects of current. Based on Class XII Physics (TN State Board)(With explanation)

1. Magnetism and Magnetic Effects of Current MCQ Class-XII (TN State Board) (With explanation)

2. By Dr M. Arunachalam Head, Department of Physics ( Rtd .) Sri SRNM College, Sattur

3. 1. The magnetic field at the centre O of the following current loop is μ I/4rÄ (b) μ I/4r (c) μ I/2rÄ (d) μ I/2r Magnetic field at the centre of current carrying circular coil is B C = μ I/2r Ä Hence for a semi circular coil it is B S = (1/2) μ I/2r Ä = μ I/4r Ä (Vector quantity) Ans: a

4. 2. An electron moves in a straight line inside a charged parallel plate capacitor of uniform charge density σ. The time taken by the electron to cross the parallel plate capacitor undeflected when the plates of the capacitor are kept under constant magnetic field of induction ( B ) is (a) έ elB / σ b) έ lB /l σ (c) έ lB /e σ (d) έ lB / σ

5. Solution Here the electron moves undeflected. So the force due to the electric field and the force due to the magnetic field are equal. ie . qE = qvB ie . v = E/B. For a parallel plate capacitor, E = σ / ε Time taken, t = Distance/ velocity = l/v Substituting v in this equation, t = l/(E/B) = lB /E = lB /( σ / ε ) ie . t = ε lB / σ Ans: d

6. 3. A particle having mass m and charge q is accelerated through a potential difference V. Find the force experienced when it is kept under perpendicular magnetic field B . 2q 3 BV/ m q 3 B 2 V/2m c) 2q 3 B 2 V/m d) 2q 3 BV/ m 3 When a charged particle is accelerated by a potential V, KE of the particle = (½)mv 2 = qV ie . v 2 = 2qV/m ie v = 2q V/m  

7. Contd … Force due to the magnetic field B is F = Bqv Substituting the value of v we get, F = Bq 2q V/m ie . F = 2q 3 B 2 V/m Ans: C  

8. 4. A circular coil of radius 5 cm and 50 turns carries a current of 3A. The magnetic dipole moment of the coil is nearly 1.0 A m 2 (b) 1.2 A m 2 (c) 0.5 A m 2 (d) 0.8 A m 2 N- No. of turns = 50 i - Current = 3A r- Radus of the coil = 5cm Magnetic dipole moment p = NiA = Ni ie. p = 50x3x ) 2 = 50x3x ) = 11775 = 1.1775 A m 2 = 1.2 A m 2 Ans: b  

9. 5. A thin insulated wire forms a plane spiral of N = 100 tight turns carrying a current I = 8 m A (milli ampere). The radii of inside and outside turns are a = 50 mm and b = 100 mm respectively. The magnetic induction at the centre of the spiral is 5 μT b) 7 μT (c) 8 μT (d) 10 μT B = [( μ NI/2(b-a )]x ln(b/a) μ – Pemeability of free space = 4 10 -7 H/m N- No. of turns = 100  

10. Contd … I – Current = 10 -3 A a- Inner radius = 10 -3 m b- Outer radius = 10 -3 m ie . B = [( 4 10 -7 x100x 10 -3 ]x ln(100/50) 2(100-50 )]x 10 -3  

11. Contd … ie . B = ( 4 10 -7 x800) x ln(2) = 12.56x800x 10 -7 x0.6931/100 2x50 ie. B = 12.56x8x 10 -7 x0.6931 ie. B = 100.48x0.6931x 10 -7 ie . B = 69.3 x 10 -7 = 6.93 x 10 -6 T ie . B = 6.93 x 10 -6 T ie . B = 6.93 μT = 7 μT Ans: b  

12. 6. Three wires of equal lengths are bent in the form of loops. One of the loops is circle, another is a semi-circle and the third one is a square. They are placed in a uniform magnetic field and same electric current is passed through them. Which of the following loop configuration will experience greater torque ? Circle (b) Semi-circle (c) Square (d) All of them Torque = τ = BiAN ie . τ α A Here N = 1 ie . Torque is directly proportional to area (A) Of the three shapes circle will occupy more area. Hence circle loop will experience greater torque. Ans: a

13. 7. Two identical coils, each with N turns and radius R are placed coaxially at a distance R as shown in the figure. If I is the current passing through the loops in the same direction, then the magnetic field at a point P at a distance of R /2 from the centre of each coil is 8 μ N/ (b) 8 μ N/5 3/2 (c) 8 μ N/5 (d)4 μ N/  

14. Solution We know that, the magnetic field at a point on the axis of the coil is B = μ NiR 2 /2(R 2 +x 2 ) 3/2 . So the field will increase by two times. Also here x = R/2. Hence the equation becomes B = 2 μ NiR 2 /2(R 2 +(R/2) 2 ) 3/2 = μ NiR 2 /(4R 2 +R 2 )/4] 3/2 ie . B = μ NiR 2 x4 3/2 /(5R 2 ) 3/2 = μ NiR 2 x4 3/2 /(5) 3/2 xR 3  

15. Contd … ie . B = μ Nix4 3/2 /(5) 3/2 R ie . B = μ Nix8 /(5) 3/2 R ie . B = 8 μ N/5 3/2 Ans: b  

16. 8. A wire of length l carrying a current i along the Y direction is kept in a magnetic field given by B = β / i+j+k ) tesla . The magnitude of Lorentz force acting on the wire is   ( /3) β i l (b) ( /3) β i l (c) β i l (d) /2) β i l We know that, the force acting on a current carrying conductor is F = Bilsinθ . The wire carries current along the Y direction. So the j component of force is 0. (since θ is 0 and for other directions θ is 90) ie . F = Bilsinθ = β ( 1 2 +0+ 1 2 )il / = β (2 /3) il ie . F = = (2 /3) β il Ans: a  

17. 9. A bar magnet of length l and magnetic moment p m is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be p m (b) 3/ π p m (c) (2/ π ) p m (d) (1/2) p m Magnetic dipole moment = ml = p m Length of the arc l = ( 2 π r)x60/360 = π r/3 = l ie . r = 3l/ π New magnetic moment = mxr = mx 3l/ π = ( 3/ π ) ml ie . New magnetic moment = ( 3/ π ) p m Ans: b

18. 10. A non-conducting charged ring carrying a charge of q , mass m and radius r is rotated about its axis with constant angular speed ω. Find the ratio of its magnetic moment with angular momentum is (a) q/m (b)2q/m (c) q/2m (d)q/4m Magnetic moment is M = NiA ----- 1 Angular momentum is L = I ω = mr 2 ω ----- 2 N – No. of turns = 1 A- Area of cross section of the ring = π r 2 i = Current = q/T = q/(2 π / ω ) = q ω /2 π

19. Contd … Substituting values in equation 1 , we get M = 1x ( q ω /2 π ) π r 2 = Dividing equation 1 by equation 2 , we get, M/L = q ω r 2 /2 mr 2 ω = q/2m Ans: c

20. 11. The BH curve for a ferromagnetic material is shown in the figure. The material is placed inside a long solenoid which contains 1000 turns/cm. The current that should be passed in the solenonid to demagnetize the ferromagnet completely is ( a) 1.00 m A (b) 1.25 mA (c) 1.50 mA (d) 1.75 mA

21. We know that H = ni /l ie . i = Hl/n From the figure, the value of H when the material is completely demagnetized is H = -150Am -1 n = 1000 turns l = 1cm = 1x10 -2 m ie . i = 150x 1x10 -2 /1000 = 0.15x10 -2 A = 1.5x10 -3 A= 1.50 mA Current to be passed is i = 1.50 mA Ans: c

22. 12. Two short bar magnets have magnetic moments 1.20 Am 2 and 1.00 Am 2 respectively. They are kept on a horizontal table parallel to each other with their north poles pointing towards south. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centres is (Horizontal components of Earth’s magnetic induction is 3.6 × 10 –5 Wb m –2 ) (a) 3.60 × 10–5 Wb m–2 (b) 3.5 × 10–5 Wb m–2 (c) 2.56 × 10–4 Wb m–2 (d) 2.2 × 10–4 Wb m–2

23. Solution B net = B 1 + B 2 + B H B net = Net ( ie , Resultant) horizontal magnetic induction B 1 = Magnetic induction of I bar magnet = 1.20 Am 2 B 2 = Magnetic induction of II bar magnet = 1.00 Am 2 B H = Horizontal component of earth’s magnetic induction = 3.6 × 10 –5 Wb m –2

24. We know that, B = µ M/4 π x 3 B net = µ M 1 /4 π x 1 3 + µ M 2 /4 π x 2 3 + 3.6 × 10 –5 Here the bar magnets are separated by a distance 20cm. We find B net at the midpoint. So, x 1 = x 2 = 10cm = 10x 10 –2 m. = 0.1m = 10 –1 m ie B net = µ M 1 /4 π ( 10 –1 ) 3 + µ M 2 /4 π ( 10 –1 ) 3 + 3.6 × 10 –5

25. Contd … B net = [µ /4 π ( 10 –1 ) 3 ] ( M 1 + M 2 ) + 3.6 × 10 –5 B net = [4 π x 10 –7 /4 π x 10 –3 ] (1.2 + 1.0) + 3.6 × 10 –5 B net = ( 10 –4 x 2.2) + 3.6 × 10 –5 B net = 2.2x 10 –4 + 0. 36 × 10 –4 B net = 2.56x 10 –4 Wb/ m 2 Ans: c

26. 13. The vertical component of Earth’s magnetic field at a place is equal to the horizontal component. What is the value of angle of dip at this place? 30 o (b) 45 o (c) 60 o (d) 90 o When the vertical component of Earth’s magnetic field and the horizontal component of Earth’s magnetic field are equal, we have B V = B H ie . B cos θ = B sin θ . In this case, θ = 45 o Angle of dip = 45 o Ans: b

27. 14. A flat dielectric disc of radius R carries an excess charge on its surface. The surface charge density is σ. The disc rotates about an axis perpendicular to its plane passing through the centre with angular velocity ω. Find the magnitude of the torque on the disc if it is placed in a uniform magnetic field whose strength is B which is directed perpendicular to the axis of rotation 1/4 σώπ BR (b) 1/2 σώπ BR 2 (c) 1/4 σώπ BR 3 (d)1/4 σώπ BR 4 We can assume that the disc consists of a number of rings with varying radius from 0 to R.

28. Solution Consider one such ring of radius r and thickness dr Torque acting on that ring is dτ = iAB ---- 1 i - current = q/T = q/(2 π / ώ ) ie . i = q ώ /2 π ---- 2 q – Charge on the ring = σ x2 π rdr 2 π rdr – Surface area of the ring σ – Surface charge density B – Magnetic field induction A – Area of cross section of the ring = πr 2

29. Contd … Substituting the value of q in equation 2, we get, ie . i = σ x2 π rdr ώ /2 π ie . i = σώ rdr By equation 1, we have, Torque acting on that ring is dτ = iAB

30. Contd … Therefore, Torque acting on the disc is τ = ∫ dτ = ∫ iAB ---- 3 Substituting the values of I and A, we get τ = ∫ dτ = ∫ σώ rdr πr 2 B = σώπ B ∫ R r 3 dr ie . τ = σώπ BR 4 /4 Ans: d

31. 15. The potential energy of a magnetic dipole whose dipole moment is p m = (-0.5i + 0.4j) Am 2 kept in a uniform magnetic field B = 0.2i tesla is –0.1 J (b) –0.8 J (c) 0.1 J (d) 0.8 J Potential energy = - p m x B ie . Potential energy = - (-0.5i + 0.4j) x 0.2i Potential energy = + 0.1J Ans: c

32. Thank You