Question

A long solenoid carrying a current produces a magnetic field B along its axis. If the number of turns per cm is doubled and the current is made 1/3 rd, then the new value of the magnetic field will be

• A. $B/3$
• B. $3B$
• C. $2/3\ B$
• D. $3/2\ B$

Hint:

For solenoid problems, the field is directly proportional to both the number of turns per unit length and the current. Simply multiply the factors of change: $2 \times (1/3) = 2/3$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to find the new magnetic field $B'$ of a long solenoid after modifying both the turn density ($n$) and the current ($I$).

Step 2: Key Formula or Approach:
The magnetic field $B$ inside a long solenoid is given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length and $I$ is the current.

Step 3: Detailed Explanation:
1. Initial magnetic field: $B = \mu_0 n I$.
2. Modified turn density: $n' = 2n$.
3. Modified current: $I' = \frac{I}{3}$.
4. New magnetic field: $B' = \mu_0 n' I' = \mu_0 (2n) (\frac{I}{3})$.
5. $B' = \frac{2}{3} (\mu_0 n I) = \frac{2}{3} B$.

Step 4: Final Answer:
The new magnetic field is $2/3\ B$, which is option (C).