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: Write the solenoid field.
Inside a long solenoid, $B = \mu_0 n I$, with $n$ the turns per unit length and $I$ the current.

Step 2: Apply the changes.
Turn density doubles ($n' = 2n$) and current drops to a third ($I' = \tfrac{I}{3}$).

Step 3: Build the new field.
\[ B' = \mu_0(2n)\left(\tfrac{I}{3}\right) = \tfrac{2}{3}\mu_0 n I = \tfrac{2}{3}B. \]

Step 4: State it.
\[ \boxed{B' = \tfrac{2}{3}B} \]