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Page No 230: - Chapter 6 Electromagnetic Induction Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024]


Question 6.2:

Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.19:

(a) A wire of irregular shape turning into a circular shape;

(b) A circular loop being deformed into a narrow straight wire.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/4/250/14103/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_66096be4.jpg

Answer:

According to Lenz’s law, the direction of the induced emf is such that it tends to produce a current that opposes the change in the magnetic flux that produced it.

(a) When the shape of the wire changes, the flux piercing through the unit surface area increases. As a result, the induced current produces an opposing flux. Hence, the induced current flows along adcb.

(b) When the shape of a circular loop is deformed into a narrow straight wire, the flux piercing the surface decreases. Hence, the induced current flows along https://img-nm.mnimgs.com/img/study_content/curr/1/12/4/250/14103/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_70491e69.gif

Question 6.3:

A long solenoid with 15 turns per cm has a small loop of area 2.0 cmplaced 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:

Number of turns on the solenoid = 15 turns/cm = 1500 turns/m

Number of turns per unit length, = 1500 turns

The solenoid has a small loop of area, A = 2.0 cm2 = 2 × 10−4 m2

Current carried by the solenoid changes from 2 A to 4 A.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7052/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_4dd19828.gifChange in current in the solenoid, di = 4 − 2 = 2 A

Change in time, dt = 0.1 s

Induced emf in the solenoid is given by Faraday’s law as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7052/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_m529a038a.gif

Where,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7052/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_2fac22b2.gif = Induced flux through the small loop

BA … (ii)

= Magnetic field

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7052/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_55b7b537.gif

μ0 = Permeability of free space

4π×10−7 H/m

Hence, equation (i) reduces to:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7052/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_m6b6d0405.gif

Hence, the induced voltage in the loop is https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7052/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_4eefdc3d.gif

Question 6.4:

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:

Length of the rectangular wire, l = 8 cm = 0.08 m

Width of the rectangular wire, b = 2 cm = 0.02 m

Hence, area of the rectangular loop,

A = lb

= 0.08 × 0.02

= 16 × 10−4 m2

Magnetic field strength, B = 0.3 T

Velocity of the loop, v = 1 cm/s = 0.01 m/s

(a) Emf developed in the loop is given as:

e = Blv

= 0.3 × 0.08 × 0.01 = 2.4 × 10−4 V

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7053/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_m20310a2b.gif

Hence, the induced voltage is 2.4 × 10−4 V which lasts for 2 s.

(b) Emf developed, e = Bbv

= 0.3 × 0.02 × 0.01 = 0.6 × 10−4 V

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7053/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_5c35c0c4.gif

Hence, the induced voltage is 0.6 × 10−4 V which lasts for 8 s.

Question 6.5:

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:

Length of the rod, l = 1 m

Angular frequency,ω = 400 rad/s

Magnetic field strength, B = 0.5 T

One end of the rod has zero linear velocity, while the other end has a linear velocity of lω.

Average linear velocity of the rod,https://img-nm.mnimgs.com/img/study_content/curr/1/12/4/250/14104/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_76745b08.gif

Emf developed between the centre and the ring,

https://img-nm.mnimgs.com/img/study_content/curr/1/12/4/250/14104/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_m3ec7bf18.gif

Hence, the emf developed between the centre and the ring is 100 V.

Question 6.6:

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 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:

Max induced emf = 0.603 V

Average induced emf = 0 V

Max current in the coil = 0.0603 A

Average power loss = 0.018 W

(Power comes from the external rotor)

Radius of the circular coil, r = 8 cm = 0.08 m

Area of the coil, A = πr2 = π × (0.08)m2

Number of turns on the coil, N = 20

Angular speed, ω = 50 rad/s

Magnetic field strength, B = 3 × 10−2 T

Resistance of the loop, R = 10 Ω

Maximum induced emf is given as:

e = Nω AB

= 20 × 50 × π × (0.08)2 × 3 × 10−2

= 0.603 V

The maximum emf induced in the coil is 0.603 V.

Over a full cycle, the average emf induced in the coil is zero.

Maximum current is given as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7055/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_6d7d81e5.gif

Average power loss due to joule heating:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7055/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_m611ce8e6.gif

The current induced in the coil produces a torque opposing the rotation of the coil. The rotor is an external agent. It must supply a torque to counter this torque in order to keep the coil rotating uniformly. Hence, dissipated power comes from the external rotor.

Question 6.7:

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:

Length of the wire, l = 10 m

Falling speed of the wire, v = 5.0 m/s

Magnetic field strength, B = 0.3 × 10−4 Wb m−2

(a) Emf induced in the wire,

e = Blv

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7056/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_m604dd834.gif

(b) Using Fleming’s right hand rule, it can be inferred that the direction of the induced emf is from West to East.

(c) The eastern end of the wire is at a higher potential.

Question 6.8:

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.

Answer:

Initial current, I1 = 5.0 A

Final current, I2 = 0.0 A

Change in current,https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7057/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_m2ef75ded.gif

Time taken for the change, t = 0.1 s

Average emf, e = 200 V

For self-inductance (L) of the coil, we have the relation for average emf as:

e = L https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7057/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_5637e516.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7057/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_m626a791d.gif

Hence, the self induction of the coil is 4 H.

Question 6.9:

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?

Answer:

Mutual inductance of a pair of coils, µ = 1.5 H

Initial current, I1 = 0 A

Final current I2 = 20 A

Change in current, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7058/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_m48fa83bb.gif

Time taken for the change, t = 0.5 s

Induced emf, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7058/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_6f066621.gif

Where https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7058/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_4b1b6ff1.gif is the change in the flux linkage with the coil.

Emf is related with mutual inductance as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7058/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_m4aa3af85.gif

Equating equations (1) and (2), we get

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7058/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_m5a49709a.gif

Hence, the change in the flux linkage is 30 Wb.

Question 6.10:

A jet plane is travelling towards 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:

Speed of the jet plane, v = 1800 km/h = 500 m/s

Wing spanof jet plane, l = 25 m

Earth’s magnetic field strength, B = 5.0 × 10−4 T

Angle of dip, https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7059/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_m153b1308.gif

Vertical component of Earth’s magnetic field,

BV = B sinhttps://img-nm.mnimgs.com/img/study_content/curr/1/12/16/250/7059/NS_18-11-08_Sravana_12_Physics_6_17_NRJ_html_m14aad32c.gif

= 5 × 10−4 sin 30°

= 2.5 × 10−4 T

Voltage difference between the ends of the wing can be calculated as:

e = (BV) × l × v

= 2.5 × 10−4 × 25 × 500

= 3.125 V

Hence, the voltage difference developed between the ends of the wings is

3.125 V.

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