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: What changes.
A long solenoid makes a magnetic field $B$ along its axis. We double the number of turns per centimetre and cut the current to one third. We want the new field.

Step 2: The field formula.
Inside a long solenoid the field is $B = \mu_0 n I$, where $n$ is turns per unit length and $I$ is the current.

Step 3: Write the new $n$ and $I$.
New turns per length: $n' = 2n$. New current: $I' = \dfrac{I}{3}$.

Step 4: Build the new field.
$B' = \mu_0 n' I' = \mu_0 (2n)\left(\dfrac{I}{3}\right)$.

Step 5: Collect the numbers.
Pull the factors out: $B' = \dfrac{2}{3}\,\mu_0 n I = \dfrac{2}{3} B$.

Step 6: State the result.
The new field is $\frac{2}{3}B$, which is option (C).
\[ \boxed{B' = \frac{2}{3}B} \]