Question:

A long solenoid with 20 turns per cm has a small loop of area 2.0 cm^2 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 sec, calculate the induced emf in the loop while the current is changing.

Answer:

1. Calculate the number of turns in the solenoid: Number of turns = 20 turns/cm x 2 cm = 40 turns

2. Calculate the rate of change of current: Rate of change of current = (4.0 A - 2.0 A) / 0.1 sec = 20 A/sec

3. Calculate the induced emf in the loop: Induced emf = N x (Rate of change of current) = 40 x 20 A/sec = 800 V

Question:

Current in a circuit falls from 5.0A to 0.0A in 0.1 s. If an average emf of 200 V induced, find an estimate of the self-inductance of the circuit.

Answer:

1. Use the equation E = L(di/dt) to calculate the self-inductance of the circuit.

2. E = 200 V, di/dt = -50 A/s

3. L = E/(di/dt)

4. L = 200 V/(-50 A/s)

5. L = -4H

Question:

A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8cms^−1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10^−3Tcm^−1 along the negative x-direction (that is it increases by 10^−3Tcm^−1 as one moves in the negative x-direction), and it is decreasing in time at the rate of 10^−3Ts^−1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.50 mΩ.

Answer:

1. Calculate the area of the loop. A = (12 cm)^2 = 144 cm^2

2. Calculate the magnetic flux through the loop. Φ = B * A = (10^−3Tcm^−1) * (144 cm^2) = 0.144 T cm^2

3. Calculate the rate of change of magnetic flux. dΦ/dt = (10^−3Tcm^−1) * (12 cm) + (10^−3Ts^−1) * (144 cm^2) = 0.156 T cm^2/s

4. Calculate the induced emf in the loop. E = -dΦ/dt = -0.156 T cm^2/s

5. Calculate the induced current in the loop. I = E/R = -0.156 T cm^2/s / (4.50 mΩ) = -34.67 mA

6. Determine the direction and magnitude of the induced current. The induced current is in the negative x-direction, with a magnitude of 34.67 mA.

Question:

A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0ms−1, at right angles to the horizontal component of the earth’s magnetic field 0.30×10^−4WBm^−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) The instantaneous value of the emf induced in the wire is 0.50×10^−3 V.

(b) The direction of the emf is perpendicular to the direction of the wire and the magnetic field.

(c) The end of the wire at the west is at the higher electrical potential.

Question:

Suppose the loop in Exercise is stationary but the current feeding the electromagnet that produces the magnetic field is gradually reduced so that the field decreases from its initial value of 0.3 T at the rate of 0.02 T s^−1. If the cut is joined and the loop has a resistance of 1.6 ohm , how much power is dissipated by the loop as heat? What is the source of this power? Exercise : [ A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of 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:

Answer:

The power dissipated by the loop as heat is given by P = I^2R, where I is the current in the loop and R is the resistance of the loop. Since the current is decreasing at a rate of 0.02 T s^−1, the current at any time t is given by I = 0.3 - 0.02t. Substituting this value of I and the resistance of 1.6 ohms into the equation, we get P = (0.3 - 0.02t)^2 * 1.6 = 0.36 - 0.12t + 0.0004t^2.

The source of this power is the decreasing magnetic field which is causing the current to decrease. As the magnetic field decreases, the current in the loop decreases, resulting in a decrease in power dissipated as heat.

Question:

An air-cored solenoid with length 30 cm, area of cross-section 25cm2 and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10^−3s. How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.

Answer:

1. Calculate the magnetic field strength, B, of the solenoid: B = (μNI)/L = (4π10^-7500*2.5)/0.3 = 0.167 Tesla

2. Calculate the back emf, E, induced across the switch: E = -LdB/dt = -0.30.167/10^-3 = 0.0501 volts

Question:

A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of 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:

a) The emf developed across the cut when the velocity of the loop is 1 cm s^−1 in a direction normal to the longer side is:

E = B x v x l

where B is the magnetic field, v is the velocity, and l is the length of the side.

E = 0.3 T x 1 cm s^−1 x 8 cm

E = 2.4 V

b) The emf developed across the cut when the velocity of the loop is 1 cm s^−1 in a direction normal to the shorter side is:

E = B x v x l

where B is the magnetic field, v is the velocity, and l is the length of the side.

E = 0.3 T x 1 cm s^−1 x 2 cm

E = 0.6 V

For how long does the induced voltage last in each case?

The induced voltage lasts as long as the loop is moving through the region of uniform magnetic field.

Question:

It is desired to measure the magnitude of field between the poles of a powerful loud speaker magnet. A small flat search coil of area 2cm^2 with 25 closely wound turns, is positioned normal to the field direction, and then quickly snatched out of the field region. Equivalently, one can give it a quick 90∘ turn to bring its plane parallel to the field direction). The total charge flown in the coil (measured by a ballistic galvanometer connected to coil) is 7.5 mC. The combined resistance of the coil and the galvanometer is 0.50 Ω. Estimate the field strength of magnet.

Answer:

1. Calculate the total charge flown in coil: Q = 7.5 mC

2. Calculate the combined resistance of the coil and the galvanometer: R = 0.50 Ω

3. Calculate the induced emf in the coil: E = Q/R = 7.5 mC/0.50 Ω = 15 V

4. Calculate the number of turns in the coil: N = 25

5. Calculate the area of the coil: A = 2 cm^2

6. Calculate the magnetic flux through the coil: Φ = E/N = 15 V/25 = 0.6 Vm

7. Calculate the field strength of the magnet: B = Φ/A = 0.6 Vm/2 cm^2 = 0.3 T

Question:

A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rads^−1 in a uniform horizontal magnetic field of magnitude 3.0×10^−2T. Obtain the maximum and average emf induced in the coil. If the coil forms a closed loop of resistance 10Ω , 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:

1. Maximum EMF induced in the coil: EMF = BANω = (3.0×10^−2T)(20)(π)(8.0 cm)(50 rads^−1) = 7.4 V

2. Average EMF induced in the coil: Average EMF = (maximum EMF)/2 = 3.7 V

3. Maximum value of current in the coil: I = EMF/R = (7.4 V)/(10Ω) = 0.74 A

4. Average power loss due to Joule heating: P = I^2R = (0.74 A)^2(10Ω) = 5.52 W

5. The power comes from the energy stored in the magnetic field of the coil. When the coil is rotated, the magnetic field lines are cut by the coil and this generates an EMF which causes a current to flow in the coil. This current then dissipates energy in the form of heat due to Joule heating.

Question:

A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s^−1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

Answer:

1. Calculate the angular velocity (ω) of the rod: ω = 400 rads^−1

2. Calculate the linear velocity (v) of the rod: v = ω x 1.0 m = 400 m/s

3. Calculate the flux of the magnetic field through the area of the rod: Flux = B x A = 0.5 T x (π x 1.0 m^2) = 1.57 Tm^2

4. Calculate the rate of change of flux: Rate of change of flux = dF/dt = ω x Flux = 400 rads^−1 x 1.57 Tm^2 = 628 Tm^2/s

5. Calculate the induced emf: emf = -Rate of change of flux x L = -628 Tm^2/s x 1.0 m = -628 V

Question:

A pair of adjacent coil has a mutual inductance of 1.5H. If the current in one coil changes from 0 to 20A in 0.5 sec, what is the change of flux linkage with the other coil?

Answer:

Step 1: Calculate the rate of change of current in the first coil: Rate of change of current = (20A - 0A) / 0.5 sec = 40 A/sec

Step 2: Calculate the rate of change of flux linkage with the other coil: Rate of change of flux linkage = Mutual inductance (M) x Rate of change of current = 1.5H x 40 A/sec = 60 Wb/sec

Question:

A jet plane is travelling towards west at a speed of 1600 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earths magnetic field at the location has a magnitude of 5 ×10^−4 T and the dip angle is 30o. A 4.1 V B 2.2 V C 3.2 V D 3.8 V

Answer:

Step 1: Calculate the magnetic flux crossing the wing. Magnetic flux = (Magnetic field)(Wing span)(cosine of dip angle) = (5 × 10^−4 T)(25 m)(cos 30o) = 0.00625Tm

Step 2: Calculate the voltage difference. Voltage difference = (Magnetic flux) × (Speed of jet plane) = (0.00625Tm)(1600 km/h) = 10 V

Step 3: Compare the answer with the given options. A 4.1 V B 2.2 V C 3.2 V D 3.8 V

Answer: A 4.1 V