Question

A long solenoid carrying current $I_1$ produces magnetic field $B_1$ along its axis. If the current is reduced to $20\%$ and number of turns per cm are increased five times then new magnetic field $B_2$ is equal to

• A. $B_1$
• B. $\frac{B_1}{5}$
• C. $5B_1$
• D. $0.25B_1$

Hint:

Since $B \propto nI$, any simultaneous changes can be treated as scaling factors. If one variable scales by $k$ and the other by $\frac{1}{k}$, their product factor is $k \times \frac{1}{k} = 1$, meaning the net physical property stays constant.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem asks for the relationship between the final magnetic field ($B_2$) and the initial magnetic field ($B_1$) of a long solenoid after modifying its operating current and the winding turns density.

Step 2: Key Formula or Approach:
The magnetic field $B$ deep inside an ideal long solenoid along its central axis 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 (turns per cm), and $I$ is the electric current.

Step 3: Detailed Explanation:
Let the initial parameters of the solenoid be turns density $n_1$ and current $I_1$. The initial magnetic field is:
$$B_1 = \mu_0 n_1 I_1$$ According to the given updates:
1. The current is reduced to $20\%$ of its original value:
$$I_2 = 20\% \text{ of } I_1 = 0.20 I_1 = \frac{1}{5} I_1$$ 2. The number of turns per cm is increased five times:
$$n_2 = 5n_1$$ Now, substitute these new values into the core solenoid equation to calculate $B_2$:
$$B_2 = \mu_0 n_2 I_2$$ $$B_2 = \mu_0 (5n_1) \left(\frac{1}{5}I_1\right)$$ The factors of $5$ and $\frac{1}{5}$ cancel out perfectly:
$$B_2 = \mu_0 n_1 I_1 = B_1$$ Therefore, the magnetic field remains completely unchanged.

Step 4: Final Answer:
The new magnetic field $B_2$ is equal to $B_1$, which corresponds to option (A).