A current-carrying long solenoid is formed by winding 200 turns per cm. If the number of turns per cm is increased to 201, keeping the current constant, then the magnetic field inside the solenoid will change by:
A current-carrying long solenoid is formed by winding 200 turns per cm. If the number of turns per cm is increased to 201, keeping the current constant, then the magnetic field inside the solenoid will change by:
- 4.0 A
- 1.0 A
- 0.5 A
- 2.0 A
- 9.0 A
The Correct Option is C
Approach Solution - 1
Given:
- Initial turns per unit length, \( n_1 = 200 \, \text{turns/cm} \)
- Final turns per unit length, \( n_2 = 201 \, \text{turns/cm} \)
- Current (\( I \)) is constant
Step 1: Magnetic Field Formula
The magnetic field inside a long solenoid is given by:
\[ B = \mu_0 n I \]
where \( \mu_0 \) is the permeability of free space, \( n \) is the number of turns per unit length, and \( I \) is the current.
Step 2: Calculate Initial and Final Fields
Initial magnetic field (\( B_1 \)):
\[ B_1 = \mu_0 n_1 I \]
Final magnetic field (\( B_2 \)):
\[ B_2 = \mu_0 n_2 I \]
Step 3: Determine Change in Field
The change in magnetic field (\( \Delta B \)) is:
\[ \Delta B = B_2 - B_1 = \mu_0 I (n_2 - n_1) \]
Substituting the values:
\[ \Delta B = \mu_0 I (201 - 200) = \mu_0 I \times 1 \, \text{turn/cm} \]
Step 4: Calculate Percentage Change
Percentage change in magnetic field:
\[ \text{Percentage Change} = \left( \frac{\Delta B}{B_1} \right) \times 100 = \left( \frac{\mu_0 I \times 1}{\mu_0 I \times 200} \right) \times 100 \]
Simplifying:
\[ \text{Percentage Change} = \left( \frac{1}{200} \right) \times 100 = 0.5\% \]
Conclusion:
The magnetic field inside the solenoid changes by 0.5%.
Answer: \(\boxed{C}\)
Approach Solution -2
Step 1: Recall the formula for the magnetic field inside a long solenoid.
The magnetic field \( B \) inside a long solenoid is given by:
\[ B = \mu_0 n I, \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( n \) is the number of turns per unit length, and
- \( I \) is the current through the solenoid.
Step 2: Analyze the change in the magnetic field.
The magnetic field \( B \) is directly proportional to the number of turns per unit length \( n \), as the current \( I \) is constant. Initially, \( n_1 = 200 \, \text{turns/cm} \), and after the change, \( n_2 = 201 \, \text{turns/cm} \).
The percentage change in the magnetic field is equal to the percentage change in \( n \), since \( B \propto n \). The percentage change in \( n \) is given by:
\[ \text{Percentage change in } n = \frac{n_2 - n_1}{n_1} \times 100. \]
Step 3: Substitute the values of \( n_1 \) and \( n_2 \).
Substitute \( n_1 = 200 \) and \( n_2 = 201 \):
\[ \text{Percentage change in } n = \frac{201 - 200}{200} \times 100 = \frac{1}{200} \times 100 = 0.5\%. \]
Final Answer: The magnetic field inside the solenoid will change by \( \mathbf{0.5\%} \), which corresponds to option \( \mathbf{(C)} \).
Top Questions on Magnetism and matter
- A magnetic dipole experiences a torque of \( 80\sqrt{3} \) N m when placed in a uniform magnetic field in such a way that the dipole moment makes an angle of \( 60^\circ \) with the magnetic field. The potential energy of the dipole is:
- JEE Main - 2025
- Physics
- Magnetism and matter
- A particle of charge 1.6 $\mu$C and mass 16 $\mu$g is present in a strong magnetic field of 6.28 T. The particle is then fired perpendicular to magnetic field. The time required for the particle to return to original location for the first time is _______ s. (Take $ \pi = 3.14 $)
- JEE Main - 2025
- Physics
- Magnetism and matter
Two long parallel wires X and Y, separated by a distance of 6 cm, carry currents of 5 A and 4 A, respectively, in opposite directions as shown in the figure. Magnitude of the resultant magnetic field at point P at a distance of 4 cm from wire Y is \( 3 \times 10^{-5} \) T. The value of \( x \), which represents the distance of point P from wire X, is ______ cm. (Take permeability of free space as \( \mu_0 = 4\pi \times 10^{-7} \) SI units.)
- JEE Main - 2025
- Physics
- Magnetism and matter
A particle of charge $ q $, mass $ m $, and kinetic energy $ E $ enters in a magnetic field perpendicular to its velocity and undergoes a circular arc of radius $ r $. Which of the following curves represents the variation of $ r $ with $ E $?
- JEE Main - 2025
- Physics
- Magnetism and matter
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : If oxygen ion (O\(^{-2}\)) and Hydrogen ion (H\(^{+}\)) enter normal to the magnetic field with equal momentum, then the path of O\(^{-2}\) ion has a smaller curvature than that of H\(^{+}\).
Reason R : A proton with same linear momentum as an electron will form a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly.
In the light of the above statements, choose the correct answer from the options given below- JEE Main - 2025
- Physics
- Magnetism and matter
Questions Asked in KEAM exam
- If $ f(x) = \log 3 - \sin x $, $ y = f(f(x)) $, find $ y(0) $.
- KEAM - 2025
- Functions
- The inward electric flux through a closed surface is \( 6 \times 10^{-5} \) and the outward flux is \( 3 \times 10^{-5} \). Then the total charge enclosed is?
- KEAM - 2025
- Electrostatics
- The element that has the highest melting point in the 3d series is:
- KEAM - 2025
- d -and f -Block Elements
- A planet revolves around the sun with a time period 27 times that of planet B. Planet A is at \( x \) times the distance of planet B from the sun. Find the value of \( x \).
- KEAM - 2025
- Astronomy
- Find the limit: \[ \lim_{x \to 0} \frac{\sin[x]}{[x]}, \text{ where } [x] \text{ represents greatest integer function} \]
- KEAM - 2025
- Continuity
Concepts Used:
Magnetic Field
The magnetic field is a field created by moving electric charges. It is a force field that exerts a force on materials such as iron when they are placed in its vicinity. Magnetic fields do not require a medium to propagate; they can even propagate in a vacuum. Magnetic field also referred to as a vector field, describes the magnetic influence on moving electric charges, magnetic materials, and electric currents.
A magnetic field can be presented in two ways.
- Magnetic Field Vector: The magnetic field is described mathematically as a vector field. This vector field can be plotted directly as a set of many vectors drawn on a grid. Each vector points in the direction that a compass would point and has length dependent on the strength of the magnetic force.
- Magnetic Field Lines: An alternative way to represent the information contained within a vector field is with the use of field lines. Here we dispense with the grid pattern and connect the vectors with smooth lines.
Properties of Magnetic Field Lines
- Magnetic field lines never cross each other
- The density of the field lines indicates the strength of the field
- Magnetic field lines always make closed-loops
- Magnetic field lines always emerge or start from the north pole and terminate at the south pole.