Typical numericals on Self-Inductance and Energy in Magnetic Field (Electromagnetic Induction) for JEE and CBSE board exams:

JEE Advanced

1. A coil of 200 turns has a self-inductance of 1 H. What is the energy stored in the magnetic field when a current of 2 A flows through the coil? Answer: The energy stored in the magnetic field of a coil is given by the formula:

   E = 1/2 * L * I^2
   

   where:

   • E is the energy stored in joules (J)
   • L is the self-inductance of the coil in henries (H)
   • I is the current flowing through the coil in amperes (A)

   In this case, L = 1 H and I = 2 A, so the energy stored in the magnetic field is:

   E = 1/2 * 1 H * (2 A)^2 = 2 J
   

2. A solenoid of length 50 cm has 1000 turns. When a current of 5 A flows through the solenoid, the magnetic field inside it is found to be 0.2 T. Calculate the energy stored in the magnetic field of the solenoid. Answer: The energy stored in the magnetic field of a solenoid is given by the formula:

   E = (1/2) * (μ₀ * n^2 * A * l) * B^2
   

   where:

   • E is the energy stored in joules (J)
   • μ₀ is the permeability of free space (4π × 10^-7 T·m/A)
   • n is the number of turns in the solenoid
   • A is the cross-sectional area of the solenoid in square meters (m²)
   • l is the length of the solenoid in meters (m)
   • B is the magnetic field inside the solenoid in teslas (T)

   In this case, n = 1000 turns, A = π * (0.5/100)^2 m² = 1.96 × 10^-4 m², l = 50/100 m = 0.5 m, and B = 0.2 T. Plugging these values into the formula, we get:

   E = (1/2) * (4π × 10^-7 T·m/A) * (1000 turns)^2 * (1.96 × 10^-4 m²) * (0.5 m) * (0.2 T)^2 = 3.89 × 10^-3 J
   

3. A conducting loop of area 0.1 m2 is placed perpendicular to a uniform magnetic field of strength 0.5 T. If the magnetic field is increased at a constant rate of 0.1 T/s, calculate the magnitude of the induced emf in the loop. Answer: The magnitude of the induced emf in a conducting loop is given by the formula:

   ε = -N * dΦ/dt
   

   where:

   • ε is the induced emf in volts (V)
   • N is the number of turns in the loop
   • Φ is the magnetic flux through the loop in webers (Wb)
   • t is the time in seconds (s)

   In this case, N = 1 turn, Φ = B * A = 0.5 T * 0.1 m² = 0.05 Wb, and dΦ/dt = 0.1 T/s. Plugging these values into the formula, we get:

   ε = -1 turn * (0.05 Wb / 1 s) = -0.05 V
   

4. A metal rod of length 1 m is moved at a constant velocity of 2 m/s perpendicular to a uniform magnetic field of strength 1 T. Calculate the magnitude of the induced emf in the rod. Answer: The magnitude of the induced emf in a moving conductor is given by the formula:

   ε = B * l * v * sinθ
   

   where:

   • ε is the induced emf in volts (V)
   • B is the magnetic field strength in teslas (T)
   • l is the length of the conductor in meters (m)
   • v is the velocity of the conductor in meters per second (m/s)
   • θ is the angle between the conductor and the magnetic field

   In this case, B = 1 T, l = 1 m, v = 2 m/s, and θ = 90°. Plugging these values into the formula, we get:

   ε = 1 T * 1 m * 2 m/s * sin 90° = 2 V
   

CBSE Board Exams

1. A coil of 100 turns has a self-inductance of 0.5 H. What is the energy stored in the magnetic field when a current of 1 A flows through the coil? Answer: The energy stored in the magnetic field of a coil is given by the formula:

   E = 1/2 * L * I^2
   

   where:

   • E is the energy stored in joules (J)
   • L is the self-inductance of the coil in henries (H)
   • I is the current flowing through the coil in amperes (A)

   In this case, L = 0.5 H and I = 1 A, so the energy stored in the magnetic field is:

   E = 1/2 * 0.5 H * (1 A)^2 = 0.25 J
   

2. A solenoid of length 20 cm has 500