Physics-12 Past Paper Questions PDF

## Charges and Magnetism ### 66 **(v)** When a charged particle moving along the combination of electric field and magnetic field acting mutually perpendicular to each other, then the nature of the path is a straight line. ### Q. 14. Write Fleming's left hand rule. A particle of having charge 1.0...

## Charges and Magnetism ### 66 **(v)** When a charged particle moving along the combination of electric field and magnetic field acting mutually perpendicular to each other, then the nature of the path is a straight line. ### Q. 14. Write Fleming's left hand rule. A particle of having charge 1.0 x 10^-9 C enters in a magnetic field of B = *(4i + 3j)* tesla with velocity = *(-75i + 100j)* m/s. Find the magnitude and direction of magnetic force exerting on the particle. [2023] **Ans.** Fleming's Left Hand Rule: Refer to Short Answer Type-I Questions Q. No. 10. **Numerical:** Given, B = *(4i + 3j)* tesla, v = *(-75i + 100j)* m/s, q = 1 x 10^-9 C, F =? We know that, F = q * ( v x B) F = 1 x 10^-9 * ( -75, 100, 0 | 4, 3, 0 ) F = 1 x 10^-9 * ( (0 - 0), - (0 - 0), + (-225 - 400) ) F = 1 x 10^-9 * (0 - 0 - 625) F = 1 x 10^-9 * (-625k) F = -6.25 x 10^-7 N ...Magnitude of magnetic force act on a particle will be - 6.25 x 10^-7 N and in negative direction of z-axis. ### Q. 15. What is a solenoid? Show the magnetic field lines produced by a current carrying solenoid. Length of a solenoid is 80 cm and 2000 turns are in it. Current in the solenoid is 10 amperes. Find the magnitude of magnetic field inside the solenoid near its centre. [2024] **Ans.** The Solenoid: A solenoid is an electromagnet in which the insulated wire is wound closely in helical form. Fig (a). It is used to produce controlled magnetic fields. Generally the length of a solenoid is... ### 65 It produced resultant magnetic field is zero at the point P, then B2 - B1 = 0 μ0*i/(2π* 5 x 10^-2) - μ0*i/(2π*15 * 10^-2) = 0 i = 30 A ### Q. 12. An electron enters in the region of 0.3 tesla magnetic field at an angle 60° with the speed of 4 x 10^5 m/s. Find the radius of helical path and pitch (distance between two consecutive spirals) of the electron beam. [2023] **Ans.** Given, B = 0.3 tesla, v = 4 x 10^5 m/s, θ = 60°, r = ? We know that, r = mv/(Bqsinθ) r = (9 x 10^-31 x 4 x 10^5) / (0.3 x 1.6 x 10^-19) r = (3 x 10^-31 x 10^7) / (4 x 10^-19) r = 3/4 x 10^-31 x 10^7 x 10^19 r = (√3/2 ) x 10^-31 x 10^26 = (√3/2) x 10^-5 r = (√3/2) x 10^-5 meter ### Q. 13. Write the nature of path of a charged particle moving (i) Along the electric field. (ii) Perpendicular to the electric field. (iii) Perpendicular to the magnetic field. (iv) Along the magnetic field. (v) Along the combination of electric field and magnetic field acting mutually perpendicular to each other. [2023] **Ans.** (i) When a charged particle moving along the electric field, then the acting force on it will be in the direction of electric field. Hence, the nature of path will be straight line (F = q. Ė). (ii) When a charged particle moving perpendicular to the electric field then the acting force on it will be in perpendicular direction of electric field. Hence, the nature of path will be parabolic. (iii) When a charged particle moving perpendicular to the magnetic field, then the acting force on it will be perpendicular to the direction of velocity. Hence, the nature of path will be circular. (iv) If a charged particle moves along the magnetic field, then the force acting on it will be zero. Hence, the particle moves in a straight line, (F = qvBsinθ, θ = 0). (v) If a charged particle moves along mutually perpendicular electric field and magnetic field, then the nature of path will be helical. ### 68 ...The torque tends to align the dipole in the direction of the field. Work has to be done in rotating the dipole against the action of the magnetic torque. This work done is stored as potential energy of the dipole... Now, a small amount of work is done in rotating dipole through a small angle dθ: dW = -M B dθ sin θ ... ### Q.6 The intensity of the magnetic field B due to a current carrying circular loop of radius 12 cm at its centre is 0.50 x 10^-4 tesla. Find the magnetic field due to this loop at a point of the axis at a distance of 5.0 cm from the centre. [2015] **Ans.** Magnetic field at the centre of a circular loop, B0 = μ0*i/(2R) = 0.50 x 10^-4 = μo*i / (2R) Magnetic field at a point on the axis for this loop B = μ0*i*R^2 / (2*(R^2 + x^2)^3/2) Divide eq (ii) by eq (i), B = B0 * R^3 / ((R^2 + x^2)^3/2) B = 0.50 x 10^-4 * (12)^2 / ((12)^2 + (5)^2)^3/2 B = 3.9 x 10^-5 T ### Q. 7. A small bar magnet is placed in an uniform external magnetic field of 4.5 tesla such that its axis is 30° angle from the direction of the field. If a torque of 4.5 x 10^-2 newton-meter acts on it, then find the potential energy of the bar magnet. [2022] **Ans.** Given, B = 4.5 tesla, θ = 30°, τ = 4.5 x 10^-2 newton-meter We know, τ = MBsinθ 4.5 x 10^-2 = M x 4.5 x sin30° 4.5 x 10^-2 = M x 4.5 x 1/2 M = 15 x 10^-2 A-m² ### 57 With the help of Biot-Savart’s Law, determine the expression for magnetic field at the centre of current carrying coil. [2016, 19, 20, 24] **Ans.** Biot-Savart’s Law: Every current carrying conductor produces magnetic field in its surrounding region (Oersted’s experiment). It is convenient to assume that this field is produced by contribution from different segments of the conductor which are called current elements. A current element is denoted by the symbol *I dl* and it has the same direction as that of the current *I*. After performing number of experiments two french physicists Jean-Baptiste Biot and Felix Savart in 1820 abtained an expression for magnetic field due to a small current element, which is known as Biot-Savart law. If current is anti-clockwise in the coil, then the direction of magnetic field will be in upward direction and is perpendicular to the field. If the current in the coil is in anti-clockwise direction, then the magnetic field produced at the centre of coil will be in downward direction and will be perpendicular to the plane of the coil. ### Q.3. Why is the coil in a moving coil galvanometer wound on an aluminium frame? A galvanometer with a coil of resistance 12.0 Ω shows a full scale deflection for a current of 2.5 mA. How will you convert it into an ammeter of range 0 to 7.5 A? [2022] **Ans.** The coil of the moving galvanometer is wound over an aluminium frame to provide electromagnetic damping because aluminium is dimagnetic and it does not interact with external magnetic field. Here, G = 12 Ω, I = 2.5 mA = 2.5 x 10^-3 A, I = 7.5 A S = (I * G) / (I - I ) S = (2.5 x 10^-3 * 12) / (7.5 - 2.5 x 10^-3) S = 0.004 Ω ### Q. 4. Obtain formula for the potential energy of a magnetic dipole placed in an external magnetic field. [2017, 20, 24] **Ans.** Magnetic dipole moment: Magnetic dipole moment is the torque acting on a dipole held perpendicular to a uniform magnetic field of unit strength. In figure, a uniform magnetic field B is represented by equidistant parallel lines. NS is a bar magnet of length 2l and strength of each pole m. The magnet is held at an angle θ with the direction of B. Force on N-pole = *mB, along B* Force on S-pole = *mB, opposite to B* where, m = strength of each pole and B = strength of magnetic field. When a magnetic dipole of moment M is held at an angle θ with the direction of a uniform magnetic field... ### 56 Where N is the number of turns and A is area of loop. Here, N = 1, i = 8 x 10^-4 A and A = πr^2 A = 3.14 (0.5 x 10^-10)^2 M = N i A M = 1 x 8 x 10^-4 x 3.14 (0.5× 10^-10)^2 M = 6.28 x 10^-24 A-m^2 ### Q. 9. An electron of energy 45 eV is revolving in a circular path in magnetic field of intensity 9 x 10^5 Wb/m^2. Find the radius of circular path. [2023] **Ans.** Given, K = 45 eV = 45 x 1.6 x 10^-19 J = 72 x 10^-19 J, m = 9.0 x 10^-31 kg, e = 1.6 x 10^-19 C, B = 9 x 10^5 Wb/m^2 We know that, r = (√2mK)/(eB) So, r = √(2 x 9 x 10^-31 x 72 x 10^-19) / (1.6 x 10^-19 x 9 x 10^5) r = √(36 x 36 x 10^-50) / (14.4 x 10^-24) r = √(36 * 10^-25) / (14.4 x 10^-24) r = 6 x 10^-13 / (14.4 x 10^-24) r = 0.25 meter ### Q. 10. What is Fleming’s left hand rule? [2023] **Ans.** If thumb, forefinger and middle finger of left hand are adjusted or arranged such that thumb is perpendicular to both fingers and forefinger represents direction of B, and middle finger represents the direction of velocity v of the particle (see fig.), then thumb represents the direction of magnetic force Facting on positively charged particle. This law is known as Fleming’s left hand rule. ### Q. 1. A proton, deutron and an alpha particle with same spacing entered parallel in magnetic field. (i) Comparing these kinetic energies. (ii) If radius of circular path of proton is 10 cm, then what will be the radius of path of deutron and a-particle? (iii) Find the ratio of periodic time of proton and a-particle. [2023] **Ans.** (i) Kinetic energy of charge Q and potential V volt: K = QV joule Kinetic energy of proton, Kp = eV (charge q = e) Kinetic energy of deutron, Kd = eV Kinetic energy of a-particle, Ka = 2eV Kp: Kd: Ka = 1: 1: 2 (ii) In magnetic field B, for circular path of radius r, charged particle is moving with velocity v (mass m, charge q) r^2 = mv^2 / (q^2B^2) r = (mv^2 / (q^2B^2))^1/2 = (mv^2/qB)^1/2 For proton, mass m, charge e and kinetic energy 2mKp rp = (2mKp/e^2B^2)^1/2 For deutron mass 2m, charge e and kinetic energy 4mKd rd = (4mKd/e^2B^2)^1/2 For a-particle, mass 4m, charge 2e and kinetic energy 2Ka ra = (2Ka / (2e)^2B^2)^1/2 = (2mKa/e^2B^2)^1/2 From equations (1) and (2), 2Kd = 2Kp Kp: Kd = 1: 1 rp = (2m Kp/e^2B^2)^1/2 = √2 x 10 = 1.414 x 10 = 14.14 cm From eqns. (1) and (3) 2Ka = 4Kp Kp: Ka = 1: 2 ra = (2m Ka/e^2B^2)^1/2 = √4 x 10 = 1.414 x 10 = 14.14 cm (iii) Periodic time, T = 2πm / (Bq) Tp = 2π mp / (Bqp) Ta = 2π ma / (Bqa) Divide eq (1) by eq (2), Ta / Tp = (2π ma / (Bqa)) / (2π mp / (Bqp)) Ta / Tp = (ma / qa) / (mp / qp) Ta / Tp = (2m / q) / (m / q) Ta / Tp = 2 / 1 Ta : Tp = 2: 1 ### Q. 2. Deduce an expression for magnetic field at centre of current carrying coll. [2017, 18, 22, 23] ### 55 Total resistance of ammeter, G’ = (G * S) / (G + S) G = (20 x 0.012) / (20 + 0.012) G = 12 x 10^-3 Ω ### Q. 5. A galvanometer is of 500 resistance and current for full scale deflection is 0.05 A. What length of the shunt will be required so as to convert the galvanometer into an ammeter of 5A range? Area of cross-section of wire 2.97 x 10^-2 cm². Specific resistivity of the wire material = 5 x 10^-7 Ω-m. [2019, 24] **Ans.** Given, resistance of galvanometer, R1 = 50 Ω, Current, I1 = 0.05 A, Ammeter current, I = 5 A, Cross-section of wire = 2.97 x 10^-2 cm^2 = 2.97 x 10^-2 x 10^-4 m^2 = 2.97 x 10^-6 m^2 , And specific resistivity of the wire = 5 x 10^-7 Ω-m S = (R1 * I1) / (I - I1) S = (50 x 0.05) / (5 - 0.05) S = 2.5/4.95 = 50/99 Ω Also, S = ρl/A l = (S * A) / ρ l = (50 x 2.97 x 10^-6) / (5 x 10^-7) l = (50 x 2.97) / (5 x 99) l = ≈ 3 m Hence, length of the shunt required so as to convert the galvanometer into an ammeter of 5A range will be ≈3m. ### Q. 3. Define magnetic dipole moment. Determine the expression for torque on magnetic dipole held in a uniform magnetic field. [2023, 24] **Ans.** Suppose a bar magnet of magnetic length 2l is placed in a uniform magnetic field B and the pole strength of the magnet is m. See given fig. Let the angle between the magnet and the magnetic field is θ. τ = *mB x 2l sinθ* = MB sin θ Here, M = *m(2l)*, is the magnetic moment of the bar magnet. In vector form: The torque in vector form is τ = *M x B* The direction of the torque given in equation (1) is perpendicular to the plane containing M and B, given by right handed screw rule. Some times, the magnetic dipole moment is defined AS: Let B=1 and θ=90°, then using equation (2), we have τ = *MB* [ sin 90° = 1 ] Therefore, the magnetic dipole moment is defined as the torque acting on the dipole, placed in unit magnetic field perpendicular to it. Its unit is ampere-metre and dimensional is [AL²]. ### Q. 7. When a coil, magnetic torque of 3 x 10^-8 A-m² is placed in a magnetic field like that its surface is parallel to the area then torque of force applied on it is 75 x 10^-9 N-m. Find the value of magnetic field. If area of the coil is 1.5 cm² and number of turns in it is 15, then what is the value of current flowing in it? [2018, 23] **Ans.** Given, M = 3 x 10^-8 A-m^2, N = 15, A = 1.5 cm^2 = 1.5 x 10^-4 m^2, τmax = 7.5 x 10^-9 N-m, B = ?, i = ? When surface of coil is parallel, then torque of force is maximum. From τmax = MB Magnetic field, B = τmax/M = (7.5 x 10^-9) / (3 x 10^-8) B = 0.25 N/A-m Now, from M = NIA i = M / (NA) = (3 x 10^-8) / (15 x 1.5 x 10^-4) = 1.33 x 10^-5 I = 1.33 x 10^-5 A ### Q. 8 An electron is moving around the nucleus of an atom in a circular path of radius 0.5 Å with a frequency of 5 x 10^15 round/sec. Find the magnetic torque of an atom. [2018, 20, 22] **Ans.** Moving electron in a class is equivalent to a current loop in which value of current, i = electron charge x number of revolutions in one second i = 1.6 x 10^-19 x 5 x 10^15 i = 8 x 10^-4 ampere Magnetic torque of equivalent current loop, M τ = i A B ### 58 ...The work done is stored as potential energy of the dipole... Let us assume that, at θ = 90°... U = W = MB (cos θ - cos 90°) U = MB cosθ In vector notation, we may rewrite this equation as: U = - MB ### Q. 5. With the help of Biot-Savart’s law determine the expression for magnetic field on the axis of a current loop. [2016, 19, 20, 24] **Ans.** Suppose the radius of circular coil is a and current *i* is passing through it. B = μ0*i * a^2 / (2*(a² + x²)3/2) ### Q. 6. The intensity of the magnetic field *B* due to a current carrying circular loop of radius 12 cm at its centre is 0.50 x 10^-4 tesla. Find the magnetic field due to this loop at a point of the axis at a distance of 5.0 cm from the centre. [2015] **Ans.** Magnetic field at the centre of a circular loop, B = μ0 * i / (2R) = 0.50 x 10^-4 = μ0*i / (2R) Magnetic field at a point on the axis for this loop B = (μ0 * i * R^2) / (2 * (R^2 + x^2)3/2) Divide eq. (ii) by eq. (1), B = B0 * R^3 / ((R^2 + x^2)^3/2) B = 0.50 x 10^-4 * (12)^2/((12)^2 + (5)^2)^3/2 B = 3.9 x 10^-5 T ### Q. 7. A small bar magnet is placed in an uniform external magnetic field of 4.5 tesla such that its axis is 30° angle from the direction of the field. If a torque of 4.5 x 10^-2 newton-meter acts on it, then find the potential energy of the bar magnet. [2022] **Ans.** Given, B = 4.5 tesla, θ = 30°, τ = 4.5 x 10^-2 newton-meter We know, τ = MBsinθ 4.5 x 10^-2 = M x 4.5 x sin30° 4.5 x 10^-2 = M x 4.5 x 1/2 M = 15 x 10^-2 A-m^2 ### 57 With the help of Biot-Savart’s Law, determine the expression for magnetic field at the centre of current carrying coil. [2016, 19, 20, 24] **Ans.** Biot-Savart’s Law: Every current carrying conductor produces magnetic field in its surrounding region (Oersted’s experiment). It is convenient to assume that this field is produced by contribution from different segments of the conductor which are called current elements. A current element is denoted by the symbol *I dl* and it has the same direction as that of the current *I*. After performing number of experiments two french physicists Jean-Baptiste Biot and Felix Savart in 1820 abtained an expression for magnetic field due to a small current element, which is known as Biot-Savart law. ### Q.3. Why is the coil in a moving coil galvanometer wound on an aluminium frame? A galvanometer with a coil of resistance 12.0 Ω shows a full scale deflection for a current of 2.5 mA. How will you convert it into an ammeter of range 0 to 7.5 A? [2022] **Ans.** The coil of the moving galvanometer is wound over an aluminium frame to provide electromagnetic damping because aluminium is dimagnetic and it does not interact with external magnetic field. Here, G = 12 Ω, I1 = 2.5 mA = 2.5 x 10^-3 A, I = 7.5 A S = (I1 * G) / (I - I1) S = (2.5 x 10^-3 * 12) / (7.5 - 2.5 x 10^-3) S = 0.004 Ω ### Q. 4. Obtain formula for the potential energy of a magnetic dipole placed in an external magnetic field. [2017, 20, 24] **Ans.** Magnetic dipole moment: Magnetic dipole moment is the torque acting on a dipole held perpendicular to a uniform magnetic field of unit strength. In figure, a uniform magnetic field B is represented by equidistant parallel lines. NS is a bar magnet of length 2l and strength of each pole m. The magnet is held at an angle θ with the direction of B. Force on N-pole = *mB, along B* Force on S-pole = *mB, opposite to B* where, m = strength of each pole and B = strength of magnetic field. When a magnetic dipole of moment *M* is held at an angle θ with the direction of a uniform magnetic field... ### 56 Where N is the number of turns and A is area of loop. Here, N = 1, i = 8 x 10^-4 A and A = πr^2 A = 3.14 (0.5 x 10^-10)^2 M = N i A M = 1 x 8 x 10^-4 x 3.14 (0.5 x 10^-10)^2 M = 6.28 x 10^-24 A-m^2 ### Q. 9. An electron of energy 45 eV is revolving in a circular path in magnetic field of intensity 9 x 10^5 Wb/m^2. Find the radius of circular path. [2023] **Ans.** Given, K = 45 eV = 45 x 1.6 x 10^-19 J = 72 x 10^-19 J, m = 9.0 x 10^-31 kg, e = 1.6 x 10^-19 C, B = 9 x 10^5 Wb/m^2 We know that, r = (√2mK)/(eB) So, r = √(2 x 9 x 10^-31 x 72 x 10^-19) / (1.6 x 10^-19 x 9 x 10^5) r = √(36 x 36 x 10^-50) / (14.4 x 10^-24) r = √(36 * 10^-25) / (14.4 x 10^-24) r = 6 x 10^-13 / (14.4 x 10^-24) r = 0.25 meter ### Q. 10. What is Fleming’s left hand rule? [2023] **Ans.** If thumb, forefinger and middle finger of left hand are adjusted or arranged such that thumb is perpendicular to both fingers and forefinger represents direction of B, and middle finger represents the direction of velocity v of the particle (see fig.), then thumb represents the direction of magnetic force Facting on positively charged particle. This law is known as Fleming’s left hand rule. ### Q. 1. A proton, deutron and an alpha particle with same spacing entered parallel in magnetic field. (i) Comparing these kinetic energies. (ii) If radius of circular path of proton is 10 cm, then what will be the radius of path of deutron and a-particle? (iii) Find the ratio of periodic time of proton and a-particle. [2023] **Ans.** (i) Kinetic energy of charge Q and potential V volt: K = QV joule Kinetic energy of proton, Kp = eV (charge q = e) Kinetic energy of deutron, Kd = eV Kinetic energy of a-particle, Ka = 2eV Kp: Kd: Ka = 1: 1: 2 (ii) In magnetic field B, for circular path of radius r, charged particle is moving with velocity v (mass m, charge q) r^2 = mv^2 / (q^2B^2) r = (mv^2 / (q^2B^2))^1/2 = (mv^2/qB)^1/2 For proton, mass m, charge e and kinetic energy 2mKp rp = (2mKp/e^2B^2)^1/2 For deutron mass 2m, charge e and kinetic energy 4mKd rd = (4mKd/e^2B^2)^1/2 For a-particle, mass 4m, charge 2e and kinetic energy 2Ka ra = (2Ka / (2e)^2B^2)^1/2 = (2mKa/e^2B^2)^1/2 From equations (1) and (2), 2Kd = 2Kp Kp: Kd = 1: 1 rp = (2m Kp/e^2B^2)^1/2 = √2 x 10 = 1.414 x 10 = 14.14 cm From eqns. (1) and (3) 2Ka = 4Kp Kp: Ka = 1: 2 ra = (2m Ka/e^2B^2)^1/2 = √4 x 10 = 1.414 x 10 = 14.14 cm (iii) Periodic time, T = 2πm / (Bq) Tp = 2π mp / (Bqp) Ta = 2π ma / (Bqa) Divide eq (1) by eq (2), Ta / Tp = (2π ma / (Bqa)) / (2π mp / (Bqp)) Ta / Tp = (ma / qa) / (mp / qp) Ta / Tp = (2m / q) / (m / q) Ta / Tp = 2 / 1 Ta : Tp = 2: 1 ### Q. 2. Deduce an expression for magnetic field at centre of current carrying coll. [2017, 18, 22, 23] ### 55 Total resistance of ammeter, G’ = (G * S) / (G + S) G = (20 x 0.012) / (20 + 0.012) G = 12 x 10^-3 Ω ### Q. 5. A galvanometer is of 500 resistance and current for full scale deflection is 0.05 A. What length of the shunt will be required so as to convert the galvanometer into an ammeter of 5A range? Area of cross-section of wire 2.97 x 10^-2 cm². Specific resistivity of the wire material = 5 x 10^-7 Ω-m. [2019, 24] **Ans.** Given, resistance of galvanometer, R1 = 50 Ω, Current, I1 = 0.05 A, Ammeter current, I = 5 A, Cross-section of wire = 2.97 x 10^-2 cm^2 = 2.97 x 10^-2 x 10^-4 m^2 = 2.97 x 10^-6 m^2 , And specific resistivity of the wire = 5 x 10^-7 Ω-m S = (R1 * I1) / (I - I1) S = (50 x 0.05) / (5 - 0.05) S = 2.5/4.95 = 50/99 Ω Also, S = ρl/A l = (S * A) / ρ l = (50 x 2.97 x 10^-6) / (5 x 10^-7) l = (50 x 2.97) / (5 x 99) l = ≈ 3 m Hence, length of the shunt required so as to convert the galvanometer into an ammeter of 5A range will be ≈3m. ### Q. 3. Define magnetic dipole moment. Determine the expression for torque on magnetic dipole held in a uniform magnetic field. [2023, 24] **Ans.** Suppose a bar magnet of magnetic length 2l is placed in a uniform magnetic field B and the pole strength of the magnet is m. See given fig. Let the angle between the magnet and the magnetic field is θ. τ = *mB x 2l sinθ* = MB sin θ Here, M = *m(2l)*, is the magnetic moment of the bar magnet. In vector form: The torque in vector form is τ = *M x B* The direction of the torque given in equation (1) is perpendicular to the plane containing M and B, given by right handed screw rule. Some times, the magnetic dipole moment is defined AS: Let B=1 and θ=90°, then using equation (2), we have τ = *MB* [ sin 90° = 1 ] Therefore, the magnetic dipole moment is defined as the torque acting on the dipole, placed in unit magnetic field perpendicular to it. Its unit is ampere-metre and dimensional is [AL²]. ### Q. 7. When a coil, magnetic torque of 3 x 10^-8 A-m² is placed in a magnetic field like that its surface is parallel to the area then torque of force applied on it is 75 x 10^-9 N-m. Find the value of magnetic field. If area of the coil is 1.5 cm² and number of turns in it is 15, then what is the value of current flowing in it? [2018, 23] **Ans.** Given, M = 3 x 10^-8 A-m^2, N = 15, A = 1.5 cm^2 = 1.5 x 10^-4 m^2, τmax = 7.5 x 10^-9 N-m, B = ?, i = ? When surface of coil is parallel, then torque of force is maximum. From τmax = MB Magnetic field, B = τmax/M = (7.5 x 10^-9) / (3 x 10^-8) B = 0.25 N/A-m Now, from M = NIA