Question

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. 4.0 A
• B. 1.0 A
• C. 0.5 A
• D. 2.0 A
• E. 9.0 A

Answer 1

The Correct Option is C

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: \(\boxed{C}\)

Answer 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 \( \mathbf{(C)} \).