The solenoid's changing magnetic field is represented as: onI = B1(t) (t) The second coil's magnetic flux is 2 = onI(t).b2 As a result of the solenoid's changing magnetic field, the induced emf in...
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.
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]
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.
It = /R is the current induced in the loop. BA = 1/R.d/dt It = 1/R.d/dt It = 1/R.d/d vBd/R = it The angle formed by B and A is equal to zero. At t = o, the switch S is closed. Q(t) = Cv is the...
A rod of mass m and resistance R slides smoothly over two parallel perfectly conducting wires kept sloping at an angle θ 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.
The angle formed by B and PQ is 90 dϕ = B.dA dϕ = B v d cos θ -ε = B v d cos θ I = -Bvd/R cos θ Using Newton's second law to solve the preceding problem, we get v as v = α g sin θ [1 – e...
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 ω of the ring after the field vanishes.
When the magnetic field is lowered in t, the magnetic flux across the conducting ring drops to zero from its maximum. E2b = induced emf According to Faraday's law of emf, The induced emf is equal to...
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.
If the instantaneous current is t, then I(t) = 1/R d/dt I(t) If q is the charge that passes during time t, dQ/dt = I(t) 1/R d/dt = dQ/dt When we integrate the equation, we obtain Q = 0L1L2/2R log...
Consider an infinitely long wire carrying a current I (t ), with dI dt =λ= constant. Find the current produced in the rectangular loop of wire ABCD if its resistance is R.
The strip has a width of dr and a length of l, and it is located inside the rectangular box at a distance of r from the current-carrying conductor's surface. The magnetic field along the length of...
i) When the position of the rotating conductor is assumed to be at the time interval t = 0 to t = π/4ꞷ We get current I = 1/2 Bl2ꞷ/λl sec2 ꞷt cos ꞷt = Blꞷ/2λ cos ꞷt ii) When the position of the rotating conductor is at time interval π/4ꞷ < t < 3π/4ꞷ We get current I = 1/2 Blꞷ/λ sin ꞷt iii) When the position of the rotating conductor is at time interval 3π/4ꞷ < t < π/ꞷ We get current I = 1/2 Blꞷ/ λ sin ꞷt
I If the spinning conductor's location is considered to be in the time period t = 0 to t = 4 We obtain current I = 1/2 Bl2/l sec2 t cos t = Bl/2 cos t = Bl/2 cos t = Bl/2 cos t = Bl/2 cos t = Bl/2...
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.
m = B.A = BA cos m The area vector is A, while the magnetic field vector is B. e1 = -dB(t)/dt lx e1 e2 = B(t) lv (t) The total emf in the circuit is equal to the emf owing to field change plus the...
A magnetic field B = Bo sin ωt 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?
Allow wire AB to travel with velocity v at time t = 0. x(t) = vt at time t AB = e1 = Blv Motional emf across (Bo sin t)vd = e1 (-j) d(B)/dt = e2 e2 = -B0 cos tx (t)d e2 = -B0 cos tx (t)d e2 = -B0...
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?
The current flowing through the coil is denoted by Ia. Mutual induction between A and B is known as Mab. The number of turns in coil A is Na. The number of turns in coil B is Nb. an is the flux...
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.
We may deduce from the graph that when the rate of change of magnetic flux reaches its highest, the electromagnetic force, which is proportional to the rate of change of current, reaches its...
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 acting on PQ's free charge particle. The motional emf is calculated by multiplying E along the PQ by the effective length of the PQ. As a result, the induced current will be vBd/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 that pass through are identical to those that flow through the surface. The magnetic field lines in an area A with magnetic flux B are represented by = B1dA1 + B2dA2. As a...
A 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 magnetic field is directed along the z-axis. B.A = BA cos = B.A cos = B.A cos = B.A cos = B.A cos Using the electromagnetic induction law of Faraday, R sin t = Boa2/R sin t = I = Boa2/R sin t =...
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.
The magnetic flux across the pipe changes when a cylindrical bar magnet with a radius of 0.8 cm is dropped through it, causing eddy currents to form. The existence of eddy current causes the magnet...
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 already know that current was flowing through the solenoid, causing it to act like a magnet with the S pole on the upper side. As a result, the ring has no induced current. When the current is...
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.
The metal ring leaps up when the current is quickly turned on because the magnetic flux across the ring is enhanced.
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.
The magnetic flux rises when an iron core is introduced into the solenoid. According to Lenz's law, as the flux increases, the current flow through the coil decreases.
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 spiral coil is extended and there are gaps between consecutive components, the current rises. Reactance must drop in order for current to rise.
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
There will be no current induced since there is no change in the magnet or the circuit area. In addition, there has been no change in the angle
The mutual inductance M12 of coil 1 with respect to coil 2 (a) increases when they are brought nearer. (b) depends on the current passing through the coils. (c) increases when one of them is rotated about an axis. (d) is the same as M21 of coil 2 with respect to coil 1.
(a) increases when they are brought nearer. (d) is the same as M21 of coil 2 with respect to coil 1.
An e .m.f is produced in a coil, which is not connected to anexternal voltage source. This can be due to (a) the coil is in a time-varying magnetic field. (b) the coil moving in a time-varying magnetic field. (c) the coil moving in a constant magnetic field. (d) the coil is stationary in an external spatially varying magnetic field, which does not change with time.
(a) the coil is in a time-varying magnetic field. (b) the coil moving in a time-varying magnetic field. (c) the coil moving in a constant magnetic field.
A metal plate is getting heated. It can be because (a) a direct current is passing through the plate. (b) it is placed in a time-varying magnetic field. (c) it is placed in a space varying magnetic field, but does not vary with time. (d) a current (either direct or alternating) is passing through the plate.
(a) The plate is receiving a direct current. (b) it is exposed to a magnetic field that changes over time. (c) it is put in a magnetic field that varies in space but not in time.
The self-inductance L of a solenoid of length l and area of cross-section A, with a fixed number of turns N increases as (a) l and A increase. (b) l decreases and A increases. (c) l increases and A decreases. (d) both l and A decrease.
(b) l decreases and A increases.
Same as problem 4 except the coil A is made to rotate about a vertical axis. No current flows in B if A is at rest. The current in coil A, when the current in B (at t = 0) is counterclockwise and the coil A is as shown at this instant, t = 0, is (a) constant current clockwise. (b) varying current clockwise. (c) varying current counterclockwise. (d) constant current counterclockwise.
(a) clockwise steady current
A loop, made of straight edges has six corners at A(0,0,0), B(L,O,0) C(L,L,0), D(0,L,0) E(0,L,L) and F(0,0,L). A magnetic field B = Bo (i + k)T is present in the region. The flux passing through the loop ABCDEFA (in that order) is (a) Bo L2 Wb. (b) 2 Bo L2 Wb. (c) 2 Bo L2 Wb. (d) 4 Bo L2 Wb.
(b) 2 Bo L2 Wb. is the correct answer
A square of side L meters lies in the x-y plane in a region, where the magnetic field is given by B = Bo (2i + 3j + 4k) T where Bo is constant. The magnitude of flux passing through the square is (a) 2 Bo L2 Wb. (b) 3 Bo L2 Wb. (c) 4 Bo L2 Wb. (d) √29 B Lo2 Wb
(c) 4 Bo L2 Wb. is the correct answer
A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis. A uniform magnetic field extends over a circular region within the rim. It is given by,
\[B=-{{B}_{0}}k(r\le a;a<R)\] = 0 (otherwise) What is the angular velocity of the wheel after the field is suddenly switched off? Answer – Line charge per unit length is given by the expression –...
(i) We are given a long straight wire and a square loop of given size (refer to figure). Find out an expression for the mutual inductance between both.
(ii) Now, consider that we passed an electric current through the straight wire of 50 A, and the loop is then moved to the right with constant velocity, v = 10 m/s.Find the emf induced in the loop...
We have an air-cored solenoid having a length of 30 cm, whose area is 25cm2 and a number of turns are 500. And the solenoid has carried a current of 2.5 A. Suddenly the current is turned off and the time is taken for it is 10−3s. What would be the average value of the induced back -emf by the ends of the open switch in the circuit? (Neglect the variation in the magnetic fields near the ends of the solenoid.)
Answer – Given, Length of the solenoid is given by l = 30 cm = 0.3 m Area of the solenoid, then becomes A =25 cm2 = 25 × 10−4m2 Number of turns on the solenoid is given by N = 500 Current in...
In the given figure we have a metal rod PQ which is put on the smooth rails AB and these are kept in between the two poles of permanent magnets. All these three (rod, rails and the magnetic field ) are in mutually perpendicular direction. There is a galvanometer ‘G’ connected through the rails by using a switch ‘K’. Given, Rod’s length = 15 cm , Magnetic field strength, B = 0.50 T, Resistance produced by the closed-loop = 9.0mΩ. Let’s consider the field is uniform.
(i) Determine the polarity and the magnitude of the induced emf if we will keep the K open and the rod will be moved with the speed of 12 cm/s in the direction shown in the figure. (ii) When...
We have a powerful loudspeaker magnet and have to measure the magnitude of the field between the poles of the speaker. And a small search coil is placed normal to the field direction and then quickly removed out of the field region, the coil is of 2cm2 area and has 25 closely wound turns. Similarly, we can give the coil a quick 90 degree turn to bring its plane parallel to the field direction. We have measured the total charge flown in the coil by using a ballistic galvanometer and it comes to 7.5 mC. Total resistance after combining the coil and the galvanometer is 0.50Ω. Estimate the field strength of the magnet.
Ans: Given, Coil’s Area is A = 2cm2 = 2×10-4m2 Number of turns on the coil is given by N = 25 Total Charge in the coil is given by Q = 7.5 mC = 7.5×10−3C The combo of coil and galvanometer...
We have a square loop having side as 12 cm and its sides are parallel to x and y-axis is moved with a velocity of 8 cm /s in the positive x-direction in a region which have a magnetic field in the direction of positive z-axis. The field is not uniform whether in case of its space or in the case of time. It has a gradient of 10−3 T cm−1 along the negative x-direction(i.e its value increases by 10−3 T cm−1 as we move from positive to negative direction ), and it is reducing in the case of time with the rate of 10−3 T s−1 . Calculate the magnitude and direction of induced current in the loop (Given: Resistance = 4.50mΩ).
Answer – We have, Side of the Square loop is given by s = 12cm = 0.12m Area of the loop becomes – A = s × s = 0.12 × 0.12 A = 0.0144 m2 Velocity of the loop is given by v = 8 cm-1 =...
Let us assume that the loop in question number 4 is stationary or constant but the current source which is feeding the electromagnet which is producing the magnetic field is slowly decreased. It was having an initial value of 0.3 T and the rate of reducing the field is 0.02 T / sec. If the cut is joined to form the loop having a resistance of 1.6 \Omega1.6Ω. Calculate how much power is lost in the form of heat? What is the source of this power?
Answer – A rectangular loop is given having sides as 8 cm and 2 cm. Therefore, the area of the loop will be will be given by A = L × B A = 8 cm × 2 cm = 16 cm2 A = 16×10−4m2 Value of...
A jet plane is travelling towards the west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth’s magnetic field at the location
has a magnitude of 5 × 10–4 T and the dip angle is 30°.
Answer – We have been provided the following data – Speed of the plane with which it is moving is given by v = 1800 km/h = 500 m/s Wing span of the jet is given by l = 25 m Magnetic field...
A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?
Ans: Given, Mutual inductance µ = 1.5H Current at initial point is given by I1= 0 A Current at final point is given by I2 = 20 A Therefore, change in current becomes dI = I1- I2 = =...
Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.
Ans: Current at initial point is given by I1 = 5.0 A Current at final point is given by I2 = 0.0 A Therefore the change in current is given by dI = I1- I2 = 5 A Total time taken is...
A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s–1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 × 10–4 Wb m–2.
(a) What is the instantaneous value of the emf induced in the wire?
(b) What is the direction of the emf?
(c) Which end of the wire is at the higher electrical potential?
Answer...
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s–1 in a uniform horizontal magnetic field of magnitude 3.0 × 10–2 T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10 Ω, calculate the maximum value of current in the coil. Calculate the average power loss due to Joule heating. Where does this power come from?
Answer – Maximum emf induced is given = 0.603 V Average emf induced is given by= 0 V Maximum current in the coil = 0.0603 A Power loss (average) is given =...
A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of the uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s–1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?
Answer – Length of the wired loop is given by l = 8 cm = 0.08 m Width of the wired loop is given by b = 2 cm = 0.02 m Since the loop is a rectangle, the area of the wired loop is given by – A = lb =...
A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?
Answer – We are given - Number of turns – 15 turns / cm = 1500 turns / m Number of turns per unit length is given by n = 1500 turns The solenoid has a small loop of area which is given by A = 2.0...
We are rotating a 1 m long metallic rod with an angular frequency of 400 red s^{-1}s−1 with an axis normal to the rod passing through its one end. And on to the other end of the rod, it is connected with a circular metallic ring. There exist a uniform magnetic field of 0.5 T which is parallel to the axis everywhere. Find out the emf induced between the centre and the ring.
Answer – Length of the rod is given = 1m Angular frequency is given by = ω = 400 rad/sec Magnetic field strength is given by B = 0.5 T The rod has zero linear velocity at one end and a linear...
Predict the direction of induced current in the situations described by the following figures
Answer – Lenz’s law predicts the direction of the induced current in a closed loop. When the North pole of a bar magnet is moved towards a closed loop, a current is induced in the loop which flows...