Numerical Problems: Moving Charges and Magnetism | Physics Class 12 - NEET PDF Download

Q6. A solenoid has an inductance of 10 henry and a resistance of 2Ω. It is connected to a 10V battery. The time taken by the magnetic energy to reach (1/4) of its maximum value is

Sol: The magnetic energy (μ) stored in the solenoid is given by the formula:

μ = 1/2 L i²

Where:

• L is the inductance (10 henry)
• i is the current

The magnetic energy as a function of time is:

μ = 1/2 L i₀² (1 - e^-Rt/L)²

Where:

• i₀ is the maximum current
• R is the resistance (2Ω)
• t is the time

We need to find the time when μ reaches (1/4) of its maximum value. The maximum value of μ is 1/2 L i₀², so (1/4) of the maximum value is:

μ' = 1/4

Therefore:

1 - e^-Rt/L = 1/2

e^-Rt/L = 1/2

Taking the natural logarithm on both sides:

-Rt/L = log 2

Since log 2 = 0.7 (given):

t = (L/R) log 2

Substituting the values L = 10 and R = 2:

t = (10/2) × 0.7

t = 5 × 0.7

t = 3.5 s

Q7. An electron with kinetic energy 5 keV enters a magnetic field of 0.01 T perpendicular to it. Find the radius of its path.

Sol: To find the radius of the electron's path, we use the relationship between the magnetic force and the centripetal force acting on the electron. The magnetic force provides the necessary centripetal force for circular motion.

The magnetic force on a charged particle is given by:

F = q v B

Where:

• q is the charge of the electron (1.6 × 10^-19 C)
• v is the velocity of the electron
• B is the magnetic field strength (0.01 T)

The centripetal force required for circular motion is:

F = m v² / r

Where:

• m is the mass of the electron (9.11 × 10^-31 kg)
• r is the radius of the path

Equating the magnetic force to the centripetal force:

q v B = m v² / r

Solving for the radius r:

r = m v / (q B)

First, we need to find the velocity v of the electron using its kinetic energy. The kinetic energy (KE) is given as 5 keV. Converting keV to joules:

1 keV = 1.6 × 10^-16 J

KE = 5 × 1.6 × 10^-16 = 8 × 10^-16 J

The kinetic energy is also given by:

KE = 1/2 m v²

Solving for v²:

v² = 2 KE / m

v² = 2 × (8 × 10^-16) / (9.11 × 10^-31)

v² ≈ 1.76 × 10¹⁵

v ≈ √(1.76 × 10¹⁵) ≈ 4.19 × 10⁷ m/s

Now substitute the values into the radius formula:

r = (m v) / (q B)

r = (9.11 × 10^-31 × 4.19 × 10⁷) / (1.6 × 10^-19 × 0.01)

r = (3.82 × 10^-23) / (1.6 × 10^-21)

r ≈ 2.39 × 10^-2 m

r ≈ 0.0239 m or 2.39 cm

Q8. Find the work done in rotating a magnetic dipole of $0.1\backslash ,\; \backslash text\{Am\}^2$0.1Am² from 0° to 90° in a field of $0.5\hspace{0.17em}T0.5\backslash ,\; \backslash text\{T\}$T.