Question

There is a metallic ring of radius \(20\,\text{cm}\) placed in a uniform unsteady magnetic field \( B = (2t^2 + 2t + 3)\,\text{T} \) perpendicular to the plane of the ring. If the resistance of the ring is \(2\,\Omega\), find the induced current at \(t = 2\,\text{s}\). \([\pi = 22/7]\)

• A. \(0.63\,\text{A}\)
• B. \(0.063\,\text{A}\)
• C. \(0.3\,\text{A}\)
• D. \(6.3\,\text{A}\)

Hint:

When magnetic field varies with time but the loop is stationary: \[ \varepsilon = A\frac{dB}{dt} \] Always differentiate the magnetic field first and then substitute the time.

Answer 1

The Correct Option is A

Concept: Induced emf in a loop due to changing magnetic field is given by Faraday's law: \[ \varepsilon = \frac{d\phi}{dt} \] Magnetic flux: \[ \phi = BA \] Thus, \[ \varepsilon = A\frac{dB}{dt} \] Current induced in the ring: \[ I = \frac{\varepsilon}{R} \]
Step 1: Find rate of change of magnetic field. \[ B = 2t^2 + 2t + 3 \] \[ \frac{dB}{dt} = 4t + 2 \]
Step 2: Find area of the ring. Radius \[ r = 20\,\text{cm} = 20\times10^{-2}\,\text{m} \] Area \[ A = \pi r^2 \] \[ A = \pi (20\times10^{-2})^2 \] \[ A = 4\pi \times10^{-2} \]
Step 3: Calculate induced emf. \[ \varepsilon = A\frac{dB}{dt} \] \[ \varepsilon = 4\pi\times10^{-2}(4t+2) \] At \(t=2\), \[ \varepsilon = 4\pi\times10^{-2}(8+2) \] \[ \varepsilon = 0.4\pi \]
Step 4: Calculate induced current. \[ I=\frac{\varepsilon}{R} \] \[ I=\frac{0.4\pi}{2} \] \[ I=0.2\pi \] Using \(\pi=\frac{22}{7}\), \[ I=\frac{4.4}{7} \] \[ I\approx0.628\approx0.63\,\text{A} \] \[ \boxed{I=0.63\,\text{A}} \]