Magnetic Effects Of Current And Magnetism

Torque on a magnetic dipole in a magnetic field is given by: Where:
- , magnetic moment
- , magnetic field
-

Step 1: Identify the straight segments and their contribution.
The wire is bent into a semi-circular arc of radius 2 cm, with straight segments extending infinitely on either side of the arc. The figure indicates a semi-circle centered at point , with the straight segments lying along a line (the diameter of the circle). We need the magnetic field at due to the straight segments (not the arc). For a straight wire carrying current , the magnetic field at a perpendicular distance is given by the Biot-Savart law or Ampere’s law:
Step 2: Analyze the geometry and calculate.
The straight segments are along the diameter of the semi-circle, and the center is at a distance equal to the radius (2 cm = 0.02 m) from the straight wire. However, the key insight is the positioning: the straight segments are semi-infinite (extending from the ends of the arc to infinity). For a semi-infinite wire, the magnetic field at a point perpendicular to the wire at distance is: Here, A, m. There are two straight segments (one from each end of the arc), both contributing to the field at . Compute for one segment: The other segment produces a field of the same magnitude but in the opposite direction (using the right-hand rule, the currents in the two straight segments produce fields at that cancel out). Thus, the net field due to the straight segments is: Final Answer: The magnetic field due to the straight segments is .

From Lorentz force law: - : along
- : along
- (for electron)
Using the right-hand rule for : So the force is in the negative z-direction.

Current loop in magnetic field experiences inward magnetic force. A flexible wire attains minimum potential energy in circular shape.

Step 1: Relate potential difference to kinetic energy and velocity
When an electron (charge , mass ) is accelerated from rest through a potential difference , its potential energy is converted into kinetic energy. From this, we can express the velocity as: Step 2: Relate velocity to the maximum magnetic force
When a charged particle moves in a uniform magnetic field , the magnetic force on it is given by: For an electron ( ), the force is . The maximum force occurs when (i.e., the velocity is perpendicular to the magnetic field), so: Step 3: Combine the relationships for the first case
Substitute equation (1) into equation (2): Here, is the maximum force when the potential difference is . Step 4: Combine the relationships for the second case
When the potential difference is changed to , the maximum force becomes . Let the new force be . Step 5: Find the ratio
Divide equation (3) by equation (4): To remove the square root, square both sides: So, the ratio is . Step 6: Analyze Options
\begin{itemize} \item Option (1): 1 : 4. Incorrect. \item Option (2): 4 : 1. Incorrect. \item Option (3): 2 : 1. Incorrect. \item Option (4): 1 : 16. Correct, as it matches our calculated ratio. \end{itemize}

- A charged particle moving perpendicular to a magnetic field experiences a Lorentz force , which provides the centripetal force for circular motion.
- Centripetal force required: .
- Equate the forces:
- Solve for :
- This matches option (C).

Step 1: Recall the Formula for Magnetic Field at the Centre of a Circular Coil
The magnetic field ( ) at the centre of a circular coil with turns, radius , and carrying a current is given by the formula: Where is the permeability of free space (a constant). Step 2: Identify the Initial Conditions
Let the initial magnetic field be .
Initial number of turns =
Initial radius =
Initial current =
So, the initial magnetic field is: Step 3: Identify the New Conditions
The problem states that the radius of the circular coil is doubled and the number of turns is halved, while the current remains the same.
New number of turns =
New radius =
New current =
Step 4: Calculate the New Magnetic Field
Let the new magnetic field be . Substitute the new conditions into the formula for : Step 5: Compare the New Magnetic Field with the Initial Magnetic Field
Divide equation (2) by equation (1): Therefore, . The magnetic field becomes one fourth of its initial value. Step 6: Analyze Options
\begin{itemize} \item Option (1): becomes doubled. Incorrect. \item Option (2): becomes one fourth. Correct, as it matches our calculated result. Option (3): becomes quadrupled. Incorrect. \item Option (4): remains unchanged. Incorrect. \end{itemize}