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NUMERICALS-ANSWERS QUESTIONS Copper has × 1028 conduction electrons per m3 A copper wire of length 1m and cross-sectional area × 10–6 m2, carrying a current and lying at right angle to a magnetic field of strength × 10–3 T, experience a force of 8.0 × 10–2 N Calculate the drift velocity of free electrons in the wire (CBSE Sample Paper) Two long, straight, parallel wires carry currents of 4.0 A and 5.0 A in the same direction They are separated by 3.0 cm Calculate the force per meter exerted by one wire on the other A proton enters a magnetic field of T with a velocity of 105 m/s at an angle of 60° with the field Calculate the magnitude of the force on this proton A beam of alpha particles and another of protons of the same velocity v, enter a uniform magnetic field at right angles to the field lines The particles describe circular paths What is the ratio of the radii of these two circles ? (AISSCE - Delhi 1993C) An Mev proton is moving transversely to a uniform magnetic field of 2.5 T Find the force on it Given mp = 1.6 × 10–27 kg A circular coil of radius 14 cm is made from a wire of 880 cm length If 1.0 A current is passed through this loop, calculate the strength of the magnetic field at its centre Calculate the distance between two long, straight, parallel wires, carrying currents of 10 A and A, if they repel each other with a force of 0.08 N/m A wire is bent in the form of a square of side 20 cm It carries a current of A Find the magnetic field at the centre of the square A circular coil of radius cm consists of 1000 turns The current through the coil is 5.0 A What is the magnitude of the magnetic field at a point on its axis at a distance of cm from its centre 10 A straight current-carrying conductor of length 20 cm is placed in a magnetic field of T making an angle of 60° with the field Calculate the force acting on it if the current flowing through it is A S Chand & Company Limited 11 A rectangular coil of area 100 cm2, having 1000 turns, carries a current of 1A It is suspended freely in a uniform magnetic field of 0.5 T Calculate (a) the maximum torque acting on the coil, and (b) the torque on the coil when the normal to the plane of the coil makes an angle of 30° with the direction of the field 12 A galvanometer coil has 50 turns, each of area × 10–4 m2 It is placed in a uniform magnetic field of 0.5 T If the torsion constant is 10–3 Nm/degree, calculate the current sensitivity of the galvanometer 13 A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of strength 0.10 T normal to the plane of the coil If the current in the coil is 5.0 A, find (i) total torque on the coil, (ii) total force on the coil, (iii) average force on each electron in the coil due to the magnetic field Given cross-sectional (AISSCE 1990 C) area = 10–5 m2 and electron density in copper = 1029 m–3 14 What is the the magnetic moment associated with a coil of turn, area of cross section 10–4 m2, carrying a current of A (AISSCE Delhi 1993) 15 An electron, after being accelerated through a potential difference of 100 V, enters a uniform magnetic field of 0.004 T perpendicular to the direction of motion Calculate the radius of the path described by the electron (e = 1.6 × 10–19 C and m = 9.1 × 10–31 kg) (AISSCE 1992) 16 An electron, moving with a speed of 108 m/s, enters a magnetic field of × 10–3 T in a direction perpendicular to the field Calcualte (i) the radius of the path, (ii) the frequency of revolution of the electron (AISSCE Delhi 1992) 17 A cyclotron in which the magnetic field is 4.5 T is used to accelerate deuterons What is the time period of revolution of deuterons and the freqency of the oscillating potential that must be applied across the dees of the cyclotron Given : mass of deuteron = 3.3 × 10–27 kg S Chand & Company Limited 18 A rectangular coil of area × 10–4 m2 and 60 turns is pivoted about one of its vertical sides The coil is in a radial horizontal field of 90 G What is the torsional constant of the hair springs connected to the coil if a current of 0.20 mA produces an angular deflection of 18° ? 19 A helium nucleus is completing one round of a circle of radius 0.8 m in s Show that the magnetic field at the centre of the circle is 1019 à0 tesla (e = 1.6 ì 1019 C) 20 Wires which connect a battery to a motor carry a current of 300 A What is the force per unit length between the wires if they are 1.5 cm apart ? Is the force attractive or repulsive ? 21 The electron in a hydrogen atom circles round the proton with speed of × 106 m/s in an orbit of radius 5.2 × 10–11 m What magnetic field does it produce at the proton ? 22 What is the magnitude of the force on a wire of length 0.04 m placed inside a solenoid near its centre making an angle of 30° with its axis? The wire carries a current of 12 A and the magnetic field due to the solenoid has magnitude of 0.25 T (AISSCE Delhi 1990C) 23 An electron is projected with a velocity of 105 m/s at right angle to a magnetic field of 0.019 G Calculate the radius of the circle described by the electron (C B S E Sample Paper) 24 An electron travels in a circular path of radius 20 cm in a magnetic field of × 10–3 T Calculate the speed of the electron What is the potential difference through which the electron must be accelerated to acquire this speed ? (AISSCE Delhi 1992) 25 A chamber is maintained at a uniform magnetic field of × 10–3 T An electron, moving with a speed of × 107 m/s, enters the chamber normal to the field Calculate (a) the radius of the circular path of the electron, and (b) the frequency of revolution of the electron S Chand & Company Limited 26 Two moving coil meters M1 and M2 have the following particulars : R1 = 10 Ω R2 = 14 Ω N1 = 30 N2 = 42 A1 = 3.6 × 10–3 m2 –3 A2 = 1.8 × 10 m B1 = 0.25 T B2 = 0.50 T (The spring constants are identical for the two meters) Determine the ratio of (a) current sensitivity (b) voltage sensitivity of M2 and M1 (NCERT Book) 27 An electron having energy 10 eV is revolving in a plane perpendicular to a uniform magnetic field of 1.0 × 10–4 Wb/m2 Calculate the following : (i) Radius of the path of the electron (ii) Cyclotron frequency (iii) Time period of revolution of the electron (iv) Work done by the magnetic force 28 A wire has × 1022 electrons per metre having average drift velocity × 10–5 m/s The wire is placed perpendicular to a magnetic field of 0.1 Wb/m2 Find (i) the current flowing through the wire, (ii) the force on an electron in the magnetic field due to its drift velocity (magnetic Lorentz force), (iii) force per unit length of the wire 29 There are 100 turns in a circular coil The effective radius of the coil is 5.0 cm, and a current of 0.10 ampere flows through it How much work will be required to rotate it in a magnetic field of 1.5 T from position θ = 0° to θ = 180°? 30 Two straight long parallel wires are kept cm apart Currents of 2.5 A and A flow through them Calculate the force acting per metre length of each wire and its nature if the currents in them flow in (i) the same direction, (ii) opposite directions S Chand & Company Limited 31 A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends A current of 5.0 A is set up in the rod through the wires (a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero ? (b) What will be the total tension in the wires if the direction of current is reversed, keeping the magnetic field same as before ? g = 9.8 m/s2 (NCERT Book) 32 A rectangular loop of sides 25 cm and 10 cm, carrying a current of 15 A, is placed with its longer side parallel to a long straight conductor 5.0 cm away from the nearer edge and in the same plane, carrying a current of 25 A What is the net force on the loop ? 33 A horizontal overhead power line carries a current of 90 A in east to west direction What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line ? (NCERT Book) 34 What is the force on a wire of length 4.0 cm placed inside a solenoid near its centre, making an angle of 60° with its axis ? The wire carries a current of 12 A and the magnetic field due to the solenoid has a magnitude of 0.25 T 35 What is the magnitude of the magnetic force per unit length on a wire carrying a current of A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T ? 36 If the current sensitivity of a moving coil galvanometer is increased by 20% its resistance also increases to 1.5 times How will the voltage sensitivity of the galvanometer be affected ? (AISSCE 1999) S Chand & Company Limited ANSWERS The magnetic force on the wire is given by F = BIL Here F = × 10–2 N B = × 10–3 T L=1m ∴ Now I = F BL = × 10 × 10 I = ne Avd ∴ Drift Velocity v d = I −2 −3 ×1 = = 16 A 16 neA × 10 × 1.6 × 10 −19 × × 10 −6 = 1.6 × 10–4 m/s Force per metre of each wire, F = = 4π × 10 −7 × 4×5 2π × × 10 28 µ i1 i2 2π r −2 −4 = 1.3 × 10 N Magnetic force on a particle is F = B q v sin θ Here q = 1.6 × 10–15 C B=1T v = 105 m/s θ = 60° ∴ F = × 1.6 × 10–19 × 105 × sin 60° = 1.39 × 10–14 N S Chand & Company Limited The radius of the circular path is given by mv r= Bq Since v and B are constants, we get rα rp But mα = × qα qp mp mα = 4m p and qα = q p So, rα = 4m p × qp = :1 rp q p m p Energy of proton = Me V = × 106 eV = × 106 × 1.6 × 10–19 J = × 1.6 × 10–13 J Now, E= v= 2 m v So, 2E = × × 1.6 × 10 1.6 × 10 = × 107 m/s Force on the proton F = Bq v Here m –13 −27 B = 2.5 T, q = 1.6 × 10–19 C and v = × 107 m/s S Chand & Company Limited F = 2.5 × 1.6 × 10–19 × × 107 = 16 × 10–12 = 1.6 × 10–11 N Total length of wire Number of turns in the coil, n = Perimeter of one loop 880 880 = = = 10 22 2π × 14 × 14 2× ∴ µ ni Magnetic field at the centre of the loop, B = Here i = A n = 10, r = 14 cm = 0.14 m 2r ∴ B= 4π × 10 −7 × 10 × × 0.14 = 4.5 × 10 −5 T Force per metre of each wire is F= Here So µ i1 i2 2π r F = 0.08 N, i1 = 10 A, i2 = 2A r= 4π × 10 −7 × 10 × 2π 0.08 = × 10–5 m S Chand & Company Limited The magnetic field due to a straight conductor is given by B= Here µ 0i 4πr (cos φ1 − cos φ2 ) i = 2A, r = 20 = 10 cm = 0.1 m φ1 = 45° and φ2 = 135° ∴ B= 4π × 10−7 × × (cos 45° − cos 135°) 4π 0.1 = 2 × 10 −6 T Total magnetic field at the centre = × 2 × 10 −6 = × 10 O φ1 = 45° φ = 135° Fig A.23 S Chand & Company Limited −6 T 9 The magnetic field due to a circular coil at an axial point at a distance x from the centre is given by µ ni r B= 2 3/ 2( r + x ) Here r = cm = × 10 x = cm = × 10 ∴ B= = 4π × 10 2π × 10 −7 −7 −2 −2 m, n = 1000, i = 5.0 A m −2 × 1000 × × (4 × 10 ) −2 −2 3/ [(4 × 10 ) + (3 × 10 ) ] × 1000 × × 16 × 10 × × × 10 −4 −6 −2 = 4.0 × 10 T 10 The force acting on a current carrying conductor in given by F = BIL sin θ Here B = T, l = 0.2 m, θ = 60°, I = A ∴ F = × × 0.2 × sin 60° = 1.4 N 11 (a) Maximum Torque acting on the coil, τmax = nBIA Here A = 100 × 10–4m2, n = 1000, i = 1A, B = 0.50 T ∴ τmax = 1000 × 0.5 × × 100 × 10–4 = 5.0 Nm (b) τ = nBIA sin θ = 1000 × 0.5 × × 100 × 10–4 × sin 30° = 2.5 Nm S Chand & Company Limited 12 Current sensitivity is given by nBA SI = k Here n = 50, B = 0.50 T, A = × 10–4 m2, k = 10–3 Nm deg ∴ SI = 50 × 0.50 × × 10 10 −4 −3 = degree/A 13 (i) Total torque acting on the coil is τ = nBIA sin θ Here n = 20, A = π r2 = π × (0.1)2 m2, B = 0.10 T, I = 5.0 A θ = 0º = 20 × 0.1 × × π × (0.1)2 × sin 0° = Zero Nm (ii) The total force on a plane current loop in a magnetic field is always zero (iii) Force on each electron = Bevd But vd = I/neA So, F = Be (I/neA) = F= 0.1 × 5.0 10 29 × 10 −5 BI nA = × 10 −25 N S Chand & Company Limited 14 Magnetic moment m = niA Here n = 1, i = 2A and A = 10–4 m2 M = × × 10–4 = × 10–4 A m2 15 The K.E of an electron is given by K = mv = eV ∴ 2eV or v = m Radius of the circular path, r = = Here m 2eV Be m m = 9.1 × 10 = −31 mv Be mV B e kg, B = 0.004 T, q = 1.6 × 10 −19 C, V = 100 V ∴ r= × 9.1 × 10 0.004 −31 1.6 × 10 × 100 −19 = 8.4 × 10 16 The radius of the circular path is given by r = Here v = 108 m/s, B = × 10–3 T, −3 m mv Bq m = × 10–31 kg S Chand & Company Limited e = 1.6 × 10–19 C ∴ r= × 10 × 10 −3 −31 × 10 × 1.6 × 10 The frequency of revolution, ν = = × 10 −3 × 1.6 × 10 × 3.14 × × 10 = 11.4 × 10 −19 −2 m Be 2π m −19 −31 = 1.4 × 10 Hz 17 Time period of revolution is given by T= Here 2π m Bq B = 4.5 T, q = 1.6 × 10 T= × 3.14 × 3.3 × 10 4.5 × 1.6 × 10 −19 C, m = 3.3 × 10 −27 −19 = 2.9 × 10 −8 −27 kg s Frequency of oscillating potential is ν= = 3.5 × 10 Hz T 18 When the coil is in equilibrium, nbIA = kα S Chand & Company Limited nBIA or k= Here n = 60, A = × 10 −4 α = 18°, i = 0.2 mA α ∴k = 60 × 90 × 10 −4 m , B = 90 × 10 × 0.2 × 10 −4 T, = 0.2 × 10 −3 × 10 −3 A −4 18 −11 −9 = 300 × 10 = × 10 Nm/degree 19 We know that a circulating charge is equivalent to a current loop If q is the magnitude of the charge and T is the time period of revolution, then the effective current is q I= T The magnetic field at the centre is B= = = µ0 I 2r = µ0q rT µ × 2e rT = µ 0e rT µ × 1.6 × 10 −19 0.8 ì 19 = 10 T 20 Assuming that the wires are very long as compared to the distance between them, the force per unit length is S Chand & Company Limited F L = µ i1 i2 2π r Here i1 = i2 = 300 A, ∴ F L = 4π × 10 r = 1.5 cm = 1.5 × 10 −7 × 2π = × × 10 −1 −2 m 300 × 300 1.5 × 10 −2 = 1.2 N/m The force is repulsive 21 v = × 106 m/s, r = 5.2 × 10–11 m, e = 1.6 × 10–19 C Number of revolutions completed per second by the electron is v n= 2π r I = ne Current = = v 2π r ×e × 10 × 3.14 × 5.2 × 10 −11 × 1.6 × 10 −19 = 9.8 × 10 Magnetic field produced at the proton, B = = 4π × 10 −7 × 9.8 × 10 11 × 5.2 × 10 –4 A µ0 I 2r −4 = 11.8 T S Chand & Company Limited 22 We know F = BIL sin θ Here L = 0.04 m, θ = 30°, i = 12 A B = 0.25 T ∴ F = 0.25 × 12 × 0.04 × sin 30° = 0.25 × 12 × 0.04 × = × 10–2 N 23 Radius r = mv Bq Here v = 105 m/s, B = 0.019 G, e = 1.6 × 10–19 C, m = × 10–31 kg −31 × 10 × 10 ∴ r= −4 −19 0.019 × 10 × 1.6 × 10 = 0.3 m mv 24 We know that r = Bq rBq or v= m Here r = 20 cm = 0.20 m, B = × 10 ∴ v= −3 0.20 × × 10 × 1.6 × 10 × 10 = 7.0 × 10 m/s −3 T −19 −31 S Chand & Company Limited If V is the required potential difference, then eV = mv −31 mv × 10 × (7.0 × 10 ) = × or V = − 19 2e 1.6 × 10 = 1.4 × 10 V 25 (a) Radius of the path, r = × 10 = −31 × × 10 −3 −19 mv Bq = 5.6 × 10 −2 × 10 × 1.6 × 10 (b) Frequency of revolution is ν= Be 2m π −3 = m × 10 × 1.6 × 10 −19 = 1.4 × 10 14 Hz × × 10 × 3.14 26 R1 = 10 Ω, N1 = 30, A1 = 3.6 × 10–3 m2, B = 0.25 T R2 = 14 Ω, N2 = 42, A2 = 1.8 × 10–3 m2, B = 0.50 T −31 Current sensitivity is given by S I = ⇒ (SI )2 ( S I )1 = N B2 A2 N1 B1 A1 = NBA k −3 42 × 0.50 × 1.8 × 10 30 × 0.25 × 3.6 × 10 = 1.4 −3 S Chand & Company Limited NBA (b) Voltage sensitivity is given by SV = ( SV ) (S I )2 or v= × 1.6 × 10 −18 × kR 10 SI R = × ( S I )1 14 = 27 E = 10 eV 1= 10 × 1.6 × 10–19 J, B = 1.0 × 10–4 T (i ) E = m v ( S V )1 = R1 = R2 2E m = × 10 −31 = 1.9 × 10 m/s r= mv Bq = × 10 × 10 −31 −4 × 1.9 × 10 × 1.6 × 10 −19 = 0.11 m = 11 cm (ii ) Cycloton frequency ν = = Bq 2m π 1.0 × 10−4 × 1.6 × 10−19 × × 10 −31 (iii ) Time period T = × 3.14 = = 2.8 × 10 Hz ν 2.8 × 10 = 3.6 × 10–7 s S Chand & Company Limited (iv) Work done by the magnetic force is zero as it always remains perpendicular to the direction of motion 28 Number of electrons per unit length = × 1022 metre vd = × 10–5 m/s, B = 0.1 T (i) Current I = e × vd × number of electrons per unit length = 1.6 × 10–19 × × 10–5 × × 1022 = 1.0 A (ii) Force on an electron = 0.1 × 1.6 × 10–19 × × 10–5 = 1.28 × 10–24 N (iii) Force on the wire, F = BIL Force per unit length = F L = BI = 0.1 × 1.0 = 0.1 N/m 29 A current loop behaves as a magnetic dipole of moment m = nIA work down W = mB (cos θ1 – cos θ2) = nIAB (cos 0° – cos 180°) = nIAB (since cos 180° = –1.) = × 100 × 0.10 × 3.14 × (5 × 10–2)2 × 1.5 = 0.24 J 30 r = × 10–2 m, i1 = 2.5 A, i2 = A S Chand & Company Limited Magnitude of the force per metre of each wire, = 4π × 10 2π −7 × F L = µ i1 i2 2π r 2.5 × × 10 −2 = × 10–5 N/m When the currents are in the same direction the force is attractive and when they are in opposite directions, the force is repulsive 31 l = 0.45 m m = 60 × 10–3 kg, I = A (a) The tension in the wires will be zero if the weight of the rod is balanced by the force due to the magnetic field, i.e., mg = BIL mg or B= IL 60 × 10−3 × 9.8 = 0.26 T × 0.45 (b) When the direction of current is reversed, then the total tension in the wires is T = mg + BIL = 2mg (since mg = BIL) = = × 60 × 10–3 × 9.8 = 1.18 N S Chand & Company Limited 32 Force on AD due to the wire PQ is L 2π r 4π × 10 −7 × 2π 25 × 15 × 10 −2 –4 = 3.75 × 10 N, towards left Force on BC is F2 = 4π × 10 F1 × 0.25 F2 15 A −7 2π × 25 cm = 10 cm 25 A F1 = µ i1 i2 P 25 × 15 0.15 cm × 0.25 −4 = 1.25 × 10 N, towards right Forces on sides AB and CD are equal and opposite, so they cancel out The net force on the loop is D Q C Fig A.24 F = F1 – F = (3.75 – 1.25) × 10–4 N = 2.50 × 10–4 N, towards the conductor PQ Note : In case the direction of current in loop ABCD is reversed the direction of force will be away from PQ S Chand & Company Limited 33 i = 90 A, We have B= r = 1.5 m µ 0i 2πr = 4π × 10 −7 × 90 2π × 1.5 = 1.2 × 10–5 T, towards south 34 l = 4.0 cm = 4.0 × 10–2 m, θ = 60° I = 12 A, B = 0.25 T We have F = BIL sin θ = 0.25 × 12 × × 10–2 × sin 60° = 0.25 × 12 × × 10–2 × = 0.110 N = 1.04 × 10–1 N 35 I = 8A, θ = 30°, B = 0.15 T We have F = BIL sin θ or F L = BI sin θ = 0.15 × × sin 30° = 0.15 × = 0.6 N/m S Chand & Company Limited 36 Current Sensitivity S I = θ I S Voltage Sensitivity S = θ = θ = I V V IR New Current Sensitivity = S I + New resistance R 20 100 S I = 1.2 S I = 1.5 R New Voltage Sensitivity S′ = V 1.2 S I 1.5R = 0.8 SV Thus, the voltage sensitivity decreases S – SV′ Percentage decrease = V × 100 SV = 0.2 × 100 = 20% S Chand & Company Limited