Solved GATE Paper

GATE 2016 Electromagnetic Fields Questions and Solutions

Instructor: Prof. Mithun Mondal Institution: BITS Pilani Subject: Electromagnetic Fields
Question 01

Question 1

A soft-iron toroid is concentric with a long straight conductor carrying a direct current \(I\). If the relative permeability \(\mu_r\) of soft-iron is 100, the ratio of the magnetic flux densities at two adjacent points located just inside and just outside the toroid, is ________ .

Solution

The magnetic field is generated by the long straight conductor. Let the two points be at approximately the same radial distance \(r\).

The flux density inside the toroid (soft-iron, \(\mu = \mu_0 \mu_r\)) is:

Equation
\[B_{\text{in}} = \frac{\mu_0 \mu_r I}{2\pi r}\]

The flux density just outside the toroid (air, \(\mu = \mu_0\)) is:

Equation
\[B_{\text{out}} = \frac{\mu_0 I}{2\pi r}\]

Taking the ratio:

Equation
\[\text{Ratio} = \frac{B_{\text{in}}}{B_{\text{out}}} = \frac{\frac{\mu_0 \mu_r I}{2\pi r}}{\frac{\mu_0 I}{2\pi r}} = \mu_r\]
Equation
\[\therefore \text{Ratio} = 100\]
Final Answer
Answer: 100.
Question 02

Question 2

A rotating conductor of 1m length is placed in a radially outward (about the z-axis) magnetic flux density (\(B\)) of 1 Tesla as shown in figure below. Conductor is parallel to and at 1m distance from the z-axis. The speed of the conductor in r.p.m. required to induce a voltage of 1V across it, should be ________ .

Rotating conductor configuration
Rotating conductor configuration

Solution

The induced EMF (Voltage) for a conductor cutting magnetic field lines is given by:

Equation
\[\text{Voltage} = v B \ell\]

Rearranging for velocity \(v\):

Equation
\[v = \frac{\text{Voltage}}{B \ell} = \frac{1}{1 \times 1} = 1 \text{ m/s}\]

Converting linear velocity to rotational speed: The distance covered in one full rotation is the circumference:

Equation
\[d = 2\pi r = 2\pi (1) = 2\pi \text{ m}\]

The time taken for one rotation is \(T = \frac{d}{v} = 2\pi\) seconds.

Speed in RPM (Rotations Per Minute):

Equation
\[N_{\text{rpm}} = \frac{60 \text{ seconds}}{T} = \frac{60}{2\pi}\]
Equation
\[N_{\text{rpm}} \approx \frac{60}{6.283} \approx 9.55 \text{ rpm}\]
Final Answer
Answer: 9.55.
Question 03

Question 3

A parallel plate capacitor field with two dielectrics is shown in the figure below. If the electric field in the region A is 4 kV/cm, the electric field in the region B, in kV/cm, is:

Capacitor with two dielectrics
Capacitor with two dielectrics
  1. 1
  2. 2
  3. 4
  4. 16

Solution

The figure shows two dielectrics placed side-by-side (parallel arrangement) between the two parallel plates.

  • The potential difference (\(V\)) across both regions is the same (connected to the same plates).
  • The distance (\(d\)) between the plates is the same for both regions.

The Electric Field is given by:

Equation
\[E = \frac{V}{d}\]

Since both \(V\) and \(d\) are constant for region A and region B:

Equation
\[E_A = E_B\]
Equation
\[E_B = 4 \text{ kV/cm}\]
C
Final Answer
Correct answer: C.
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GATE Electromagnetic Fields