Magnetic Field - Free MCQ Practice Test with solutions, NEET Physics

An electric current as well as the permanent magnet produces a magnetic field whereas a temporary magnet fails to do so.

Gauss’s law for magnetism states that no magnetic monopoles exist and that the total flux through a closed surface must be zero.

• The discovery that one particular pole of a magnet orients northward, whereas the other pole orients southward allowed people to identify the north and south poles of any magnet.
• It was then noticed that the north poles of two different magnets repel each other, and likewise for the south poles. Conversely, the north pole of one magnet attracts the south pole of other magnets.
• This situation is analogous to that of electric charge, where like charges repel and unlike charges attract. In magnets, we simply replace the charge with a pole: Like poles repel and unlike poles attract.

The magnetic field at any axial point is given by, B =  2μ_o/ 4πx³
Similarly, the field at any equatorial point is given by, B = μ_o/ 4πx³
Thus, the field at any equatorial point is half of what it is at an axial point.
 

The magnetic field at the center of a square loop of current is given by the formula

where: μ₀ = permeability of free space = 4π × 10⁻⁷ T·m/A, I = current in the loop, and, a = side length of the square loop
Calculation:

Let B be the magnetic field due to single side
then 
= 
∴ B_net at centre O = 4B
= 8 × 10^–6 
P = 8 

• The apparatus used by Faraday to demonstrate that magnetic fields can create currents is illustrated in Figure B.
• When the switch is closed, a magnetic field is produced in the coil on the top part of the iron ring and transmitted to the coil on the bottom part of the ring.
  So, B is the correct option.

The magnetic field due to a current element can be calculated using the Biot-Savart law.

The Biot-Savart law states that the magnetic field (dB) at a point due to a current element (Idl) is given by:

dB = (μ₀/4π) * (Idl × r̂) / r²

Where:

• μ₀ = 4π × 10⁻⁷ Tm/A (permeability of free space)
• I = current (in Amperes)
• dl = length of the current element (in meters)
• r̂ = unit vector in the direction from the current element to the point where the field is being calculated
• r = distance from the current element to the point (in meters)

Calculation:

Given values:

I = 10 A

dl = 0.05 m

r = 1 m

Using the Biot-Savart law:

⇒ dB = (μ₀/4π) * (I * dl × r̂) / r²

Since the angle between dl and r̂ is 90 degrees:

⇒ dB = (4π × 10⁻⁷ / 4π) * (10 * 0.05) / 1²

⇒ dB = 10⁻⁷ * 0.5

⇒ dB = 5 × 10⁻⁸ T

∴ The value of the magnetic field is 5.0 × 10⁻⁸ T.

Magnetic Moment and Torque on a Coil:
The torque (moment of force) acting on a rectangular coil placed in a magnetic field is given by the formula:
τ = n × B × A × I × sin(θ)

τ = torque (moment of force),
n = number of turns of the coil,
B = magnetic field strength,
A = area of the coil,
I = current flowing through the coil,
θ = angle between the normal to the coil and the magnetic field.

In this case, the angle θ is given as 60°.

The torque formula uses the angle between normal to coil and magnetic field.
If plane makes 60^∘, then normal makes:

θ = 90^∘ − 60^∘ = 30^∘

Calculation:

Given,

Number of turns, n = 400

Magnetic field strength, B = 1 T

Area of the coil, A = 10^-2 m²

Current, I = 0.5 A

Angle, θ = 30°

Using the formula for torque:

τ = n × B × A × I × sin(θ)

Substitute the values:

τ = 400 × 1 × 10^-2 × 0.5 × sin(30°)

τ = 400 × 10^-2 × 0.5 × (1/ 2)

τ = 400 × 10^-2 × 0.5 × 0.5

τ = 1 N·m

∴ The initial moment of force acting on the coil is 1 N·m.
Hence, the correct option is C) 1.

• A compass has a small magnet inside it with its north pole towards the tip of the needle.
• This magnet aligns itself with the earth's magnetic field thus points towards the magnetic south and geographical north.

For a current flowing through an infinitely long, straight, thin-walled pipe, we can use Ampère's Law to determine the magnetic field inside and around the pipe. According to Ampère's Law, the magnetic field inside a conductor is related to the current flowing through it. However, for an infinitely long pipe with a thin wall, we are concerned with the magnetic field at points inside the pipe.

• When a current flows along a thin-walled pipe, the magnetic field at any point inside the pipe is determined by the symmetry of the system. For a thin-walled pipe, the magnetic field at any point inside the pipe is zero due to the nature of the setup.
• This is because the magnetic fields produced by the current in each segment of the pipe cancel each other out at all points inside the pipe. Essentially, there is no net magnetic field within the pipe itself.

The correct answer is: The magnetic field at any point inside the pipe is zero (option 4).