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Practice each question, then open the answer or worked solution.
Consider a magnet surrounded by a wire with an on/off switch S. If the switch is thrown from the off position (open circuit) to the on position (closed circuit), will a current flow in the circuit? Explain.
No current will be induced as there is no change in any of the magnet or in the area of the circuit. Also, there is no change in the angle.
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.
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.
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.
When an iron core is inserted into the solenoid, the magnetic flux increases. According to Lenz’s law, when the flux increases, there is a decrease in the current flow through the coil.
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.
When the current is suddenly switched on, the metal ring jumps up because the magnetic flux is increased across the ring.
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?
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 decrease in the magnetic flux such that the N pole is on the lower side and is attracted by the solenoid.
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.
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 the magnet.
Short AnswersA magnetic field in a certain region is given by B = Bo cos ωt 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 =π ω /2, t =π /2ω and t =3π/ω
The direction of the magnetic field is along the z-axis.
Φ = B.A = BA cos θ
Using Faraday’s law of electromagnetic induction,
I = Boπa2ꞷ/R sin ꞷt
Therefore, current at t = π ω /2 = I = Boπa2ꞷ/R sin ꞷt = 1
Current at t = π /2ω = I = Boπa2ꞷ/R sin ꞷt = -Boπa2ꞷ/R
Current t = 3π/ω = I = Boπa2ꞷ/R sin ꞷt = I = -Boπa2ꞷ/R
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 φ= 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.
The magnetic flux lines passing through is the same as the magnetic flux lines passing through the surface. φ= B1dA1 + B2dA2 represents the magnetic field lines in an area A with magnetic flux B. Therefore, the number of lines passing through S1 is the same as the lines passing through S2.
Find the current in the wire for the configuration. Wire PQ has negligible resistance. B, the magnetic field is coming out of the paper. θ is a fixed angle made by PQ travelling smoothly over two conducting parallel wires separated by a distance d.
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.
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.
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 be maximum when the time axis makes a maximum angle at AB.
Therefore, when t = 3 seconds, s = slope of OA
Therefore, the rate of change of current at t = 3 = ¼ A/s
Therefore, the electromotive force is L/5.
When the emf is between 5 to 10 seconds, it is -3e
When the emf is between 10 to 30 seconds, it is +1/2e
When the emf is at 40 seconds, dI/dt = 0