Problem: A square loop of side 10 cm is placed in a uniform magnetic field of 0.5 T. Calculate the magnetic flux through the loop when (a) the plane of the loop is perpendicular to the field, and (b) the plane makes an angle of 30° with the field direction.

Solution 2 Marks

Given:
Side of square loop, $a = 10\, \text{cm} = 0.1\, \text{m}$
Area, $A = a^2 = (0.1)^2 = 0.01\, \text{m}^2$
Magnetic field, $B = 0.5\, \text{T}$

Part (a): Plane perpendicular to field
Angle between normal to plane and field, $\theta = 0^\circ$
$\Phi = BA \cos \theta = (0.5)(0.01) \cos 0^\circ = 0.005 \times 1 = 0.005\, \text{Wb}$

Part (b): Plane at 30° to field
Angle between normal to plane and field, $\theta = 60^\circ$
$\Phi = BA \cos \theta = (0.5)(0.01) \cos 60^\circ = 0.005 \times 0.5 = 0.0025\, \text{Wb}$

$\boxed{\text{(a) } 0.005\,\text{Wb} \quad \text{(b) } 0.0025\,\text{Wb}}$

2. Induced EMF in a Changing Field

Problem: A circular coil of radius 8 cm has 50 turns. The magnetic field perpendicular to the coil changes from 0.2 T to 0.8 T in 0.5 seconds. Calculate the induced EMF in the coil.

Solution 2 Marks

Given:
Radius, $r = 8\, \text{cm} = 0.08\, \text{m}$
Number of turns, $N = 50$
Initial field, $B_1 = 0.2\, \text{T}$
Final field, $B_2 = 0.8\, \text{T}$
Time interval, $\Delta t = 0.5\, \text{s}$

Step 1: Calculate area
$A = \pi r^2 = \pi (0.08)^2 = 0.0201\, \text{m}^2$

Step 2: Calculate change in flux
Initial flux for one turn: $\Phi_1 = B_1 A = (0.2)(0.0201) = 0.00402\, \text{Wb}$
Final flux for one turn: $\Phi_2 = B_2 A = (0.8)(0.0201) = 0.01608\, \text{Wb}$
Change in flux per turn: $\Delta \Phi = \Phi_2 - \Phi_1 = 0.01608 - 0.00402 = 0.01206\, \text{Wb}$

Step 3: Apply Faraday’s Law
$\varepsilon = -N \frac{\Delta \Phi}{\Delta t} = -50 \times \frac{0.01206}{0.5} = -50 \times 0.02412 = -1.206\, \text{V}$
Magnitude of induced EMF = $1.206\, \text{V}$

$\boxed{1.206\,\text{V}}$

3. Direction of Induced Current

Problem: A bar magnet is moved towards a coil along its axis with the north pole facing the coil. What is the direction of the induced current in the coil when viewed from the magnet side?

Concept: This problem tests understanding of Lenz’s Law. The induced current will oppose the change in flux that produced it.

Solution 1 Mark

Analysis:
1. As the north pole approaches the coil, the magnetic flux through the coil (in the direction from magnet to coil) increases.
2. By Lenz’s Law, the induced current will create a magnetic field that opposes this increase.
3. Therefore, the face of the coil towards the magnet must become a north pole to repel the approaching north pole.
4. For the coil to have a north pole facing the magnet, the current must flow counterclockwise when viewed from the magnet side (using right-hand thumb rule).

$\boxed{\text{Counterclockwise}}$

4. Motional EMF in a Rotating Rod

Problem: A metallic rod of length 0.5 m rotates with a constant angular velocity of 10 rad/s about one end in a plane perpendicular to a uniform magnetic field of 0.4 T. Calculate the induced EMF between the ends of the rod.

Metallic rod rotating in magnetic field showing induced EMF calculation. — A rod rotating about one end in a perpendicular magnetic field.

Solution 3 Marks

Given:
Length of rod, $l = 0.5\, \text{m}$
Angular velocity, $\omega = 10\, \text{rad/s}$
Magnetic field, $B = 0.4\, \text{T}$

Step 1: Consider a small element
Consider a small element of length $dx$ at distance $x$ from the axis of rotation.
Linear velocity of this element: $v = \omega x$

Step 2: EMF in small element
The motional EMF in this element: $d\varepsilon = B v dx = B (\omega x) dx$

Step 3: Total EMF
Integrate from 0 to $l$ :
$\varepsilon = \int_0^l B \omega x \, dx = B \omega \left[\frac{x^2}{2}\right]_0^l = B \omega \frac{l^2}{2}$
$\varepsilon = (0.4)(10)\frac{(0.5)^2}{2} = 4 \times \frac{0.25}{2} = 4 \times 0.125 = 0.5\, \text{V}$

$\boxed{0.5\,\text{V}}$

5. Eddy Current Power Loss

Problem: A circular metal disc of radius 10 cm and thickness 2 mm rotates at 1200 rpm in a magnetic field of 0.25 T perpendicular to its plane. If the resistivity of the metal is $1.7 \times 10^{-8}\, \Omega \cdot \text{m}$ , estimate the power loss due to eddy currents.

Solution 3 Marks

Given:
Radius, $r = 10\, \text{cm} = 0.1\, \text{m}$
Thickness, $t = 2\, \text{mm} = 0.002\, \text{m}$
Angular speed, $N = 1200\, \text{rpm} = 20\, \text{rps}$
$\omega = 2\pi N = 40\pi\, \text{rad/s}$
Magnetic field, $B = 0.25\, \text{T}$
Resistivity, $\rho = 1.7 \times 10^{-8}\, \Omega \cdot \text{m}$

Step 1: Approximate power loss formula
For a rotating disc in perpendicular field, power loss due to eddy currents is approximately:
$P \approx \frac{\pi B^2 \omega^2 r^4 t}{8\rho}$

Step 2: Substitute values
$P = \frac{\pi (0.25)^2 (40\pi)^2 (0.1)^4 (0.002)}{8 \times 1.7 \times 10^{-8}} \approx 0.453\, \text{W}$

$\boxed{0.453\,\text{W}}$

6. Self Inductance of a Solenoid

Problem: A long solenoid has 2000 turns per meter and a cross-sectional area of 4 cm². Calculate its self-inductance per meter length. If the current changes from 2 A to 5 A in 0.1 s, what is the induced EMF?

Solution 3 Marks

Given:
Turns per meter, $n = 2000\, \text{turns/m}$
Area, $A = 4 \times 10^{-4}\, \text{m}^2$
$\Delta I = 3\, \text{A}, \Delta t = 0.1\, \text{s}$

Step 1: Calculate self-inductance
$L = \mu_0 n^2 A = (4\pi \times 10^{-7}) (2000)^2 (4 \times 10^{-4}) = 2.011 \times 10^{-3}\, \text{H}$

Step 2: Calculate induced EMF
$\varepsilon = -L \frac{\Delta I}{\Delta t} = -(0.002011) \times 30 = -0.06033\, \text{V}$
Magnitude = $0.06033\, \text{V}$

$\boxed{L = 2.011\,\text{mH}, \quad \varepsilon = 0.0603\,\text{V}}$

7. Mutual Inductance Between Coils

Problem: Two coaxial solenoids have lengths of 0.5 m. The inner solenoid has 500 turns and radius 2 cm, while the outer solenoid has 1000 turns. Calculate their mutual inductance.

Solution 2 Marks

Given:
$l = 0.5\, \text{m}, N_1 = 500, N_2 = 1000, r_1 = 0.02\, \text{m}$

Step 1: Calculate mutual inductance
$M = \mu_0 \frac{N_1 N_2 A_1}{l} = (4\pi \times 10^{-7}) \frac{(500)(1000)(\pi \times 0.02^2)}{0.5} = 1.58 \times 10^{-3}\, \text{H}$

$\boxed{1.58\,\text{mH}}$

8. Energy Stored in an Inductor

Problem: A coil of inductance 0.5 H carries a current of 4 A. How much energy is stored in the magnetic field? If the current is reduced to zero in 10 ms, what is the average induced EMF?

Solution 2 Marks

Given:
$L = 0.5\, \text{H}, I = 4\, \text{A}, \Delta t = 0.01\, \text{s}$

Step 1: Energy stored
$U = \frac{1}{2} L I^2 = \frac{1}{2} (0.5) (16) = 4\, \text{J}$

Step 2: Average induced EMF
$\varepsilon = -L \frac{\Delta I}{\Delta t} = -(0.5) \frac{-4}{0.01} = 200\, \text{V}$

$\boxed{U = 4\,\text{J}, \quad \varepsilon = 200\,\text{V}}$

9. AC Generator Output

Problem: An AC generator has a coil of 100 turns, each of area 0.03 m², rotating in a magnetic field of 0.2 T at 50 revolutions per second. Calculate (a) the peak voltage and (b) the rms voltage generated.

Solution 2 Marks

Given:
$N = 100, A = 0.03\, \text{m}^2, B = 0.2\, \text{T}, f = 50\, \text{Hz}$

Step 1: Peak voltage
$\varepsilon_0 = NBA\omega = (100)(0.2)(0.03)(100\pi) = 188.5\, \text{V}$

Step 2: RMS voltage
$\varepsilon_{rms} = \frac{188.5}{\sqrt{2}} = 133.3\, \text{V}$

$\boxed{\text{(a) } 188.5\,\text{V} \quad \text{(b) } 133.3\,\text{V}}$

10. RL Circuit Analysis

Problem: A coil of inductance 2 H and resistance 10 Ω is connected to a 20 V DC source. Calculate (a) the time constant of the circuit, (b) the current after 0.2 seconds, and (c) the energy stored in the inductor when the current reaches steady state.

Solution 3 Marks

Given:
$L = 2\, \text{H}, R = 10\, \Omega, V = 20\, \text{V}$

Part (a): Time constant
$\tau = \frac{L}{R} = 0.2\, \text{s}$

Part (b): Current after 0.2 s
$I = I_0 (1 - e^{-t/\tau}) = 2 (1 - e^{-1}) = 1.264\, \text{A}$

Part (c): Energy at steady state
$U = \frac{1}{2} L I_0^2 = \frac{1}{2} (2) (2)^2 = 4\, \text{J}$

$\boxed{\text{(a) } 0.2\,\text{s} \quad \text{(b) } 1.264\,\text{A} \quad \text{(c) } 4\,\text{J}}$

Exam Tip: For numerical problems involving electromagnetic induction:

• Always draw a diagram to visualize the situation
• Clearly identify what’s changing (flux, area, field, orientation)
• Apply Faraday’s Law with correct sign convention based on Lenz’s Law
• For motional EMF, use $\varepsilon = Blv$ or integrate for non-uniform motion
• For RL circuits, remember the exponential growth/decay formulas

← Back to EMI Theory Next: AC Numericals →

10 Solved Numericals on Electromagnetic Induction

10 Solved Numericals on Electromagnetic Induction

1. Magnetic Flux Through a Square Loop

2. Induced EMF in a Changing Field

3. Direction of Induced Current

4. Motional EMF in a Rotating Rod

5. Eddy Current Power Loss

6. Self Inductance of a Solenoid

7. Mutual Inductance Between Coils

8. Energy Stored in an Inductor

9. AC Generator Output

10. RL Circuit Analysis

Editorial Team

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1. Magnetic Flux Through a Square Loop

2. Induced EMF in a Changing Field

3. Direction of Induced Current

4. Motional EMF in a Rotating Rod

5. Eddy Current Power Loss

6. Self Inductance of a Solenoid

7. Mutual Inductance Between Coils

8. Energy Stored in an Inductor

9. AC Generator Output

10. RL Circuit Analysis

Editorial Team

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