The magnetic field inside a long solenoid is 4 × 10−3 T when it carries a current of 2 A. If the current is increased to 6 A, the magnetic field becomes:
1.2 × 10−2 T
The magnetic field inside a long solenoid is \(B = \mu_0 n I\), where n is the number of turns per unit length.
For the same solenoid, \(\mu_0\) and n are fixed, so \(B \propto I\).
\(\frac{B_2}{B_1} = \frac{I_2}{I_1} = \frac{6}{2} = 3\).
\(B_2 = 3 \times 4 \times 10^{-3} = 12 \times 10^{-3} = 1.2 \times 10^{-2}\ \text{T}\).
Hence, the answer is 1.2 × 10−2 T.
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