Ans: 1) A is paramagnetic material. 2) B is diamagnetic material. THREE MARKS QUESTIONS 1. State law for magnetic field produced at a point due to a small current element. How will you find the direction of magnetic field? Ans. Consider a current element dl of a conductor carrying current I. let P be a point having . Let and position vector r with respect to current element dl be the angle between dl r According to Biot-Savart law, the magnetic field dB at point P due to current element is (i) dB I (ii) dB dl (iii) dB sin (iv) dB 1 r 2 Combining all the four factors, we get dB I dl r 2 dB = K I dl r 2 The proportionality constant K depends on the medium between point P and the current element and the system of units chosen. For free space and in SI units, K 0 10 T dB I dl r 2 and , so from the above As the direction of dB is perpendicular to the plane of dl r equation, we get the vector form of the Biot-Savart law as sin sin 4 7
(vd If n is the number of free electrons per unit volume, then total number of electrons in the conductor is N = n x volume = nAl Total force on the conductor is F vd If represents a current element vector in the direction of current, then vectors and I l l vdwill have opposite directions and we can take l vd= l F B) [ ] F I (l B) The magnitude of the force is F = I l B sin (i) If =0 or 180 , then F = I l B(0) = 0 Thus a current carrying conductor placed parallel to the direction of the magnetic field does not experience any force. (ii) If =90 , then F = I l B sin90o or Fmax = I l B Thus a current carrying conductor placed perpendicular to the direction of a magnetic field experiences a maximum force. 4. Define current sensitivity and voltage sensitivity of a galvanometer. State the factors on which the sensitivity of a moving coil galvanometer depends. Ans. Current sensitivity: It is the deflection produced in the galvanometer when a unitcurrent flows through it. Current sensitivity, Is NBA I k Voltage sensitivity: It is the deflection produced in the galvanometer when a unitpotential difference is applied across its ends. Voltage sensitivity, Vs=α/V =NBA/k RG
Let RG be the resistance of galvanometer and Ig be the current with which galvanometer gives full scale deflection. To measure a maximum potential difference V, a high resistance R is connected in series with it. Total resistance of the device = R + RG or R V RG I g Therefore by Ohm’s law V I g(R R I g
Reflection on the basis of Wave Theory Let t be the time in which the incident wave front reach from Q to P’ in the same time the reflected wave front starting from P reaches Q’ QP’=PQ’=v t In triangle PQP’ and triangle PQ’P’ Angle PQP’= Angle PQ’P’= 90 PP’ is the common side QP’=PQ’=v t Triangle PQP’ Triangle PQ’P’ AngleQPP’ = AngleQ’PP’ i = r Refraction on the basis of wave theory Let t be the time in which the incident wave front reach from Q to P’ in the same time the refracted wave front starting from P reaches Q’ such that QP’=C1t and PQ’=C2t In triangle PQP’ In triangle Hence from eq. (1) and (2) INTERFERENCE OF LIGHT CONDITION FORCONSTRUCTIVE AND DESTRUCTIVE INTERFERENCE OF LIGHT Let the waves from two coherent source of light be represented by __________________(1) ______________(2) According to principal of superposition, resultant displacement is given by y = y1 + y2 = a = 2 a cos cos (t + 2) The amplitude of the resultant displacement is 2a cos (/2) and therefore the intensity at that point will be
where A and B are constants. Plot a graph showing the dependence of µon λ. Ans: How does power of lens vary when incident red light is replaced by blue light? Ans: Wavelength decreases hence power increases An object is held at the principal focus of a concave lens of focal length f. Where is the image formed? Ans: That is image will be formed between optical centre and focus of lens; towards the side of the object. What is the geometrical shape of the wavefront when a plane wave passes through a convex lens? Ans: The wavefront is spherical of decreasing radius. What is the angle between the plane of polariser and analyser, in order that the intensity of transmitted through analyser reduces to half? Ans: 450 A diverging lens of focal length ‘F’ is cut into two identical parts each forming a plano-concave lens. What is the focal length of each part? Ans: Focal length of each half part will be twice the focal length of initial diverging lens. How the angular separation of interference fringes in Young’s double slit experiment change when the distance between the slits and screen is doubled? Ans: Angular separation between fringes, θ = λ/d where λ = wavelength, d = separation between coherent sources. So, θ is independent of distance between the slits and screen. So angular separation (θ ) will remain unchanged. Two thin lenses of power +6 D and – 2 D are in contact. What is the focal length of the combination? Ans: Net power of lens combination P = P1 + P2 = + 6 D - 2 D = + D ∴ Focal length, f = 1/P = ¼ m = 25 cm How does the angular separation between fringes in single-slit diffraction experiment change when the distance of separation between the slit and screen is doubled? Ans: Angular separation is θ = β / D = λ / d Since θ is independent of D, angular separation would remain same. 12. How does the fringe width, in Young’s double-slit experiment, change when the distance of separation between the slits and screen is doubled? Ans: The fringe width is, β = D λ / d