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: Identify the changes.
A long solenoid carries current $I_1$ and gives field $B_1$. Then the current is cut to $20\%$ and the turns per cm are made five times larger. Find the new field $B_2$.
Step 2: Field of a long solenoid.
Inside a long solenoid, $B = \mu_0 n I$, where $n$ is turns per unit length and $I$ the current. So $B_1 = \mu_0 n_1 I_1$.
Step 3: New current.
Reduced to $20\%$ means $I_2 = 0.20\, I_1 = \dfrac{1}{5} I_1$.
Step 4: New turns density.
Five times the turns per cm gives $n_2 = 5 n_1$.
Step 5: Compute $B_2$.
$B_2 = \mu_0 n_2 I_2 = \mu_0 (5 n_1)\left(\dfrac{1}{5} I_1\right) = \mu_0 n_1 I_1$.
Step 6: Compare.
The factors $5$ and $\dfrac{1}{5}$ cancel, so $B_2 = \mu_0 n_1 I_1 = B_1$. The field is unchanged, option (1).
\[ \boxed{B_2 = B_1} \]