Set of sure shot questions of Magnetic effect of Current for class XII CBSE Board Exam Reg.

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Deborah O’Brien’
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1 Set of sure shot questions of Magnetic effect of Current for class XII CBSE Board Exam Reg. 1 Two Parallel Conducting wires carrying current in the same direction attract each other.why? I 1 A B I 2 2. X The direction of magnetic field at the conductor B due to the current I 1 through the conductor A is inward. By flahming Left hand Rule the direction of force( )acting on the conductor B to its length l due to the current I 2 through it is towards the conductor A(Left) Similarly, the direction of magnetic field at the conductor A due to the current I 2 through the conductor B is outward. By flahming Left hand Rule the direction of force( )acting on the conductor A of its length l due to the current I 1 through it is towards the conductor B(Right) 2 Determine the trajectory of a charged particle which enters a region of magnetic field B with velocity v making an angle α with the field B. Consider a charged particle q enters in a region of magnetic field B with velocity v making an angle α with the field B. The perpendicular component v sinα of the initial velocity makes the charge move along a circular path of radius r = 2 The period of revolution is T = The Parallel component vcosα makes it move along the direction of the magnetic field.hence the resultant path of the charged particle will be a helix, with its axis along the direction of B. The distance moved along the magnetic field in one rotation is called the pitch of the helical path. Pitch= xt=vcosαx

2 3 To increase the current sensitivity of a moving coil galvanometer by 50%, its resistance is increased so that the new resistance becomes twice its initial resistance.by what factor does its voltage sensitivity change. Current sensitivity, = 2 Voltage sensitivity = = = New Current sensitivity = + = New Voltage Sensitivity = = = =0.75 The new voltage sensitivity becomes 75% of its initial value i.e decreases by 25% 4 A charge q moving in a straight line is accelerated by a potential difference V. It enters a uniform magnetic field B perpendicular to its path.deduce in terms of V an expression for the radius of the circular path in which it travels. Gain in kinetic energy of the q when accelerated through V volts is 3 m =qv Hence = As the force due to the perpendicular magnetic field provides the required centripetal force. Hence =qbv or r= = 5 A galvanometer with a coil of resistance 12.0 Ohm shows full scale defection for a current of 2.5 ma.how will you convert the meter into: (a) An ammeter of range 0 to 7.5 A (b) A voltmeter of range 0 to 10.0V? 3 (i) For Conversion into Ammeter: x12 = =4.0x Ω So by connecting a shunt resistance of 4.0x Ω in parallel with the galvanometer,we get an ammeter of range 0 to 7.5 A (ii)

3 For conversion into Voltmeter: R= - R= -12= =3988Ω So by connecting a resistance of 3988Ω in series with the galvanometer,we get a voltmeter of range 0 to 10 V 6 As shown in figure, a cell is connected across two points A and B of a uniform circular conductor.prove that the magnetic field at its centre O will be zero. C O B I 22 C I 11 A 3 Suppose I 1 and I 2 be the current through the segments ACB and ACD respectively. The P.D across the points A and B will remain the same. I 1 R 1 = I 2 R 2 but R α l (length) Hence I 1 l 1 = I 2 l (1) Therefore the magnetic field at the centre O due to the currents I 1 and I 2 respectively = = And = = But I 1 l 1 = I 2 l 2 Hence As the currents I 1 and I 2 are oppositely directed, their magnetic field will be opposite to each other. Hence the resultant field at the centre O is zero. 7 State Biot Savart law. Use it to obtain the magnetic field at the axial point at a distance x from the centre of a circular coil of radius a carrying a current I. Hence,Compare the magnitudes of the magnetic field of this coil at the centre and at an axial point for which x= Statement: Biot Savart s Law: 5

dl, sin θ and inversely proportional to the square of the distance (r 2 ) where θ

4 The strength of magnetic field db due to a small current element dl carrying a current I at a point P distant r from the element is directly proportional to I, dl, sin θ and inversely proportional to the square of the distance (r 2 ) where θ is the angle between dl and r. ) db α I ii) db α dl iii) db α sin θ iv) db α 1 / r 2

how is the magnetic field inside a given solenoid made stronger?

5 Derivation: = = = = =8 8 Using Ampere s circuital law, Obtain the expression for the magnetic field at a point inside the air-cored solenoid on its axis.how is the magnetic field inside a given solenoid made stronger? Sketch the magnetic field lines for a finite solenoid The magnetic field inside a closely wound long solenoid is uniform everywhere and zero outside it. 5

= X Total current through the loop PQRS B= ni It can be easily shown that the magnetic field at the end of the solenoid is just one

6 Figure shows the sectional view of a long solenoid. To determine the magnetic field B at any inside point, consider a rectangular closed path PQRS. According to the Ampere s circuital law. = X Total current through the loop PQRS B= ni It can be easily shown that the magnetic field at the end of the solenoid is just one half of that at its middle. Thus = ni The field inside a solenoid can be increased by (a) Inserting an iron core inside it (b) Increasing number of turns per unit length, and (c) Increasing the current through the solenoid.

9 Draw a schematic diagram of a cyclotron.explain its underlying principle and working stating clearly the function of the electric and magnetic field applied on a charged particle.

7 9 Draw a schematic diagram of a cyclotron.explain its underlying principle and working stating clearly the function of the electric and magnetic field applied on a charged particle. Deduce an expression for the period of revolution and show that it does not depend on the speed of the charged particle. Principal: A charged particle can be accelerated to very high energy by making it pass through a moderate electric field a number of times. This can be done with the help of a perpendicular magnetic field which throws the charged particle in a circular motion, the frequency of which does not depend on the speed of the particle and the radius of the circular orbit. Working: Suppose a positive ion, says a proton, enters the gap between the two dees and finds dee D1 to be negative. It gets accelerated towards dee D1. As it enters the dee D1, it does not experience any electrical field due to shielding effect of the metallic dee. The perpendicular magnetic fields throw it in to a circular path. At the instant the proton comes out of the dee D1, it finds dee D1 describing a larger semi-circle than before. Thus the frequency the applied voltage is kept exactly the same as the frequency of the revolution of the proton, the every time proton reach the gap between the two Dees, the electric field reversed and the proton receive a push and finally it acquire very high-energy. This is called the cyclotron s resonance condition. The proton follows spiral path. The accelerated proton is ejected through a window by deflecting voltage and hits the target. 5 Construction: Diagram

8 Period of revolution of the charged particle is given by

10 (a) With the help of a diagram, explain the principle and working of a moving coil galvanometer.

9 T= =. = Hence frequency of the revolution of the particle will be f c = =. The frequency of the ions does not depend on the speed of the particle and the radius of the circular orbit. 10 (a) With the help of a diagram, explain the principle and working of a moving coil galvanometer. (b) What is the importance of a radial magnetic field and how is it produced? (c) Increasing the current sensitivity of a galvanometer may not necessarily increase its voltage sensitivity Justify this statement. 5 Principle: A current carrying coil placed in a magnetic field experiences a torque, the magnitude of which depends on the strength of the current.

Voltage sensitivity, = = If the current sensitivity is increased by increasing the number turns N,the

10 Current sensitivity of a galvanometer is the deflection produced per unit flow of current while voltage sensitivity is the deflection produced per unit applied potential difference Current sensitivity, = = Voltage sensitivity, = = If the current sensitivity is increased by increasing the number turns N,the resistance R will also increase. So the voltage sensitivity might not increase on increasing the current sensitivity.

Tridib s Physics Tutorials. NCERT-XII / Unit- 4 Moving charge and magnetic field

Tridib s Physics Tutorials. NCERT-XII / Unit- 4 Moving charge and magnetic field MAGNETIC FIELD DUE TO A CURRENT ELEMENT The relation between current and the magnetic field, produced by it is magnetic effect of currents. The magnetic fields that we know are due to currents or moving