When a voltage measuring device is connected to AC mains,
Question: When a voltage measuring device is connected to AC mains, the meter shows the steady input voltage of 220V. This means (a) input voltage cannot be AC voltage, but a DC voltage. (b) maximum input voltage is 220V. (c) the meter reads not v but v2 and is calibrated to read v2 . (d) the pointer of the meter is stuck by some mechanical defect. Solution: (c) the meter reads not v but v2 and is calibrated to read v2...
Read More →An alternating current generator has an internal
Question: An alternating current generator has an internal resistance Rg and an internal reactance Xg. It is used to supply power to a passive load consisting of a resistance Rg and a reactance XL. For maximum power to be delivered from the generator to the load, the value of XL is equal to (a) zero. (b) Xg. (c) Xg. (d) Rg . Solution: (c) Xg....
Read More →If the rms current in a 50 Hz ac circuit is 5 A,
Question: If the rms current in a 50 Hz ac circuit is 5 A, the value of the current 1/300 seconds after its value becomes zero is (a) 5 2 A (b) 5 3/2 A (c) 5/6 A (d) 5/ 2 A Solution: (b) 5 3/2 A...
Read More →A long solenoid ‘S’ has ‘n’ turns per meter,
Question: A long solenoid S has n turns per meter, with diameter a. At the centre of this coil, we place a smaller coil of N turns and diameter b (where b a). If the current in the solenoid increases linearly, with time, what is the induced emf appearing in the smaller coil. Plot graph showing nature of variation in emf, if current varies as a function of mt2 + C. Solution: The varying magnetic field in the solenoid is given as: B1(t) = onI(t) Magnetic flux in the second coil is Φ2 = onI(t).b2 T...
Read More →A metallic ring of mass m and radius l (ring being horizontal)
Question: A metallic ring of mass m and radius l (ring being horizontal) is falling under gravity in a region having a magnetic field. If z is the vertical direction, the z- component of the magnetic field is Bz = Bo (1+ z). If R is the resistance of the ring and if the ring falls with a velocity v, find the energy lost in the resistance. If the ring has reached a constant velocity, use the conservation of energy to determine v in terms of m, B, and acceleration due to gravity g. Solution: The r...
Read More →Find the current in the sliding rod AB (resistance = R).
Question: Find the current in the sliding rod AB (resistance = R). B is constant and is out of the paper. Parallel wires have no resistance. v is constant. Switch S is closed at time t = 0. Solution: The angle between A and B = 0o Therefore, the emf is Bvd. -LdI(t)/dt + Bvd = IR LdI(t)/dt + RI (t) = Bvd Solving the equation, we get I = Bvd/R [1-e-Rt/2]...
Read More →Find the current in the sliding rod AB (resistance = R).
Question: Find the current in the sliding rod AB (resistance = R). B is constant and is out of the paper. Parallel wires have no resistance. v is constant. Switch S is closed at time t = 0. Solution: The current induced in the loop is It = /R It = 1/R.d/dt BA It = vBd/R The angle between B and A is zero. As the switch S is closed at t = o, Charge through the capacitor, is Q(t) = Cv Current through the capacitor, I(c) = Q(t)/RC Using the above information, we can calculate the current through the...
Read More →A rod of mass m and resistance R slides smoothly over two parallel
Question: A rod of mass m and resistance R slides smoothly over two parallel perfectly conducting wires kept sloping at an angle with respect to the horizontal. The circuit is closed through a perfect conductor at the top. There is a constant magnetic field B along the vertical direction. If the rod is initially at rest, find the velocity of the rod as a function of time. Solution: The angle between B and PQ = 90o dϕ = B.dA dϕ = B v d cos - = B v d cos I = -Bvd/R cos Solving the above equation u...
Read More →A magnetic field B is confined to a region r a ≤ and
Question: A magnetic field B is confined to a region r a and points out of the paper (the z-axis), r = 0 being the centre of the circular region. A charged ring (charge = Q) of radius b, b a and mass m lies in the x-y plane with its centre at the origin. The ring is free to rotate and is at rest. The magnetic field is brought to zero in time ∆t. Find the angular velocity of the ring after the field vanishes. Solution: The magnetic flux across the conducting ring reduces to zero from maximum when...
Read More →A rectangular loop of wire ABCD is kept
Question: A rectangular loop of wire ABCD is kept close to an infinitely long wire carrying a current II ( ) t = o (1 /t T ) for 0 t T and I (0) = 0 for t T. Find the total charge passing through a given point in the loop, in time T. The resistance of the loop is R. Solution: If t is the instantaneous current then, I(t) = 1/R dϕ/dt If q is the charge passing in time t I(t) = dQ/dt dQ/dt = 1/R dϕ/dt Integrating the equation we get, Q = 0L1L2/2R log (L2 + x/x)...
Read More →Consider an infinitely long wire carrying a current I (t ),
Question: Consider an infinitely long wire carrying a current I (t ), with dI dt == constant. Find the current produced in the rectangular loop of wire ABCD if its resistance is R. Solution: The width of the strip is dr and the length is l which is inside the rectangular box at a distance r from the surface of the current-carrying conductor. The magnetic field across the strip length is given as l = B(r) = oI/2r B(r) is perpendicular to the paper upward. Flux in the strip = ϕ = oIl/2 [loger] =-d...
Read More →ODBAC is a fixed rectangular conductor of negligible resistance (CO is not connected)
Question: ODBAC is a fixed rectangular conductor of negligible resistance (CO is not connected) and OP is a conductor which rotates clockwise with an angular velocity . The entire system is in a uniform magnetic field B whose direction is along the normal to the surface of the rectangular conductor ABDC. The conductor OP is in electric contact with ABDC. The rotating conductor has a resistance of per unit length. Find the current in the rotating conductor, as it rotates by 180. Solution: (i) Whe...
Read More →A conducting wire XY of mass m and negligible resistance
Question: A conducting wire XY of mass m and negligible resistance slides smoothly on two parallel conducting wires. The closed-circuit has a resistance R due to AC. AB and CD are perfect conductors. There is a magnetic field B = B t(k ). (i) Write down the equation for the acceleration of the wire XY. (ii) If B is independent of time, obtain v(t) , assuming v (0) = u0. (iii) For (b), show that the decrease in kinetic energy of XY equals the heat lost in R. Solution: The magnetic flux linked wit...
Read More →A magnetic field B = Bo sin ωt k covers a large region
Question: A magnetic field B = Bo sin t k covers a large region where a wire AB slides smoothly over two parallel conductors separated by a distance d. The wires are in the x-y plane. The wire AB (of length d) has resistance R and the parallel wires have negligible resistance. If AB is moving with velocity v, what is the current in the circuit? What is the force needed to keep the wire moving at constant velocity? Solution: Let wire AB at t = 0 move with velocity v. At t, x(t) = vt Motional emf ...
Read More →Evaluate the integral:
Question: Evaluate the integral: $\int \frac{x}{x^{2}+3 x+2} d x$ Solution: $I=\int \frac{x}{x^{2}+3 x+2} d x$ As we can see that there is a term of $x$ in numerator and derivative of $x^{2}$ is also $2 x$. So there is a chance that we can make substitution for $x^{2}+3 x+2$ and I can be reduced to a fundamental integration. As, $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}+3 \mathrm{x}+2\right)=2 \mathrm{x}+3$ $\therefore$ Let, $x=A(2 x+3)+B$ $\Rightarrow x=2 A x+3 A+B$ On comparing both ...
Read More →There are two coils A and B separated
Question: There are two coils A and B separated by some distance. If a current of 2 A flows through A, a magnetic flux of 10-2 Wb passes through B (no current through B). If no current passes through A and a current of 1 A passes through B, what is the flux through A? Solution: Ia is the current passing through the coil Mab is the mutual induction between A and B Na is the number of turns in coil A Nb is the number of turns in coil B ϕa is the flux linked to coil A due to coil B ϕb is the flux l...
Read More →A (current vs time) graph of the current passing
Question: A (current vs time) graph of the current passing through a solenoid is shown in Fig 6.9. For which time is the back electromotive force (u) a maximum. If the back emf at t = 3s is e, find the back emf at t = 7 s, 15s and 40s. OA, AB and BC are straight line segments. Solution: From the graph, we can say that when there is a maximum rate of change of magnetic flux, the electromagnetic force will be maximum which is proportional to the rate of change of current. The rate of current will ...
Read More →Find the current in the wire for the configuration.
Question: Find the current in the wire for the configuration. Wire PQ has negligible resistance. B, the magnetic field is coming out of the paper. is a fixed angle made by PQ travelling smoothly over two conducting parallel wires separated by a distance d. Solution: F is the force on the free charge particle of PQ. The motional emf is given as the product of E along the PQ and the effective length PQ. Therefore, the induced current will be vBd/R which is independent of q....
Read More →Consider a closed loop C in a magnetic field such
Question: Consider a closed loop C in a magnetic field such that the flux passing through the loop is defined by choosing a surface whose edge coincides with the loop and using the formula = B1dA1 + B2dA2 + . Now if we chose two different surfaces S1 and S2 having C as their edge, would we get the same answer for flux. Justify your answer. Solution: The magnetic flux lines passing through is the same as the magnetic flux lines passing through the surface. = B1dA1 + B2dA2 represents the magnetic ...
Read More →A magnetic field in a certain region
Question: A magnetic field in a certain region is given by B = Bo cos t k and a coil of radius a with resistance R is placed in the x-y plane with its centre at the origin in the magnetic field. Find the magnitude and the direction of the current at (a, 0, 0) at t = /2, t = /2 and t =3/ Solution: The direction of the magnetic field is along the z-axis. Φ = B.A = BA cos Using Faradays law of electromagnetic induction, I = Boa2ꞷ/R sin ꞷt Therefore, current at t = /2 = I = Boa2ꞷ/R sin ꞷt = 1 Curren...
Read More →Consider a metallic pipe with an inner radius of 1 cm.
Question: Consider a metallic pipe with an inner radius of 1 cm. If a cylindrical bar magnet of radius 0.8cm is dropped through the pipe, it takes more time to come down than it takes for a similar unmagnetised cylindrical iron bar dropped through the metallic pipe. Explain. Solution: When a cylindrical bar magnet with radius 0.8 cm is dropped through the pipe, the magnetic flux across the pipe changes and there is a production of eddy currents. The presence of eddy current opposes the motion of...
Read More →Consider a metal ring kept (supported by a cardboard)
Question: Consider a metal ring kept (supported by a cardboard) on top of a fixed solenoid carrying a current I. The centre of the ring coincides with the axis of the solenoid. If the current in the solenoid is switched off, what will happen to the ring? Solution: We know that the current was already flowing through the solenoid making it behave like a magnet such that the S pole is on the upper side. Therefore, there is no induced current in the ring. When the current is turned off, there is a ...
Read More →Consider a metal ring kept on top of a fixed
Question: Consider a metal ring kept on top of a fixed solenoid such that the centre of the ring coincides with the axis of the solenoid. If the current is suddenly switched on, the metal ring jumps up. Explain. Solution: When the current is suddenly switched on, the metal ring jumps up because the magnetic flux is increased across the ring....
Read More →A solenoid is connected to a battery so
Question: A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, will the current increase or decrease? Explain. Solution: When an iron core is inserted into the solenoid, the magnetic flux increases. According to Lenzs law, when the flux increases, there is a decrease in the current flow through the coil....
Read More →A wire in the form of a tightly wound solenoid
Question: A wire in the form of a tightly wound solenoid is connected to a DC source, and carries a current. If the coil is stretched so that there are gaps between successive elements of the spiral coil, will the current increase or decrease? Explain. Solution: When the coil is stretched, there are gaps between successive elements of the spiral coil then the current will increases. For current to increase, reactance must decrease....
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