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A long solenoid has 500 turns per cm and carries a current I. The magnetic field at its center is 7.54 × 10^-2 Wb m^-2. Another long solenoid has 300 turns per cm and it carries a current \(\frac {I}{3}\). What is the value of magnetic field at the center?(a) 1.50 × 10^-3 Wb/m^2(b) 1.50 × 10^-5 Wb/m^2(c) 1.05 × 10^-3 Wb/m^2(d) 1.50 × 10^-2 Wb/m^2I got this question during an online exam.My query is from Solenoid and Toroid topic in division Moving Charges and Magnetism of Physics – Class 12

Answer»

Correct answer is (d) 1.50 × 10^-2 Wb/m^2

To explain I would say: The magnetic field induction at the center of a long solenoid is B = μonI→B ∝ NIN = number of turns per unit length of solenoid. I = current PASSING through the solenoid.

\(\frac {B1}{B2} = \frac {n1}{n2} [ \frac {I1}{I2} ] \)

\(\frac {B1}{B2} = \frac {500}{300} \, \TIMES \, \frac {I}{(\frac {I}{3})}\)

\(\frac {B1}{B2}\) = 5

B2 = \(\frac {B1}{5}\)

B2 = \(\frac {7.54 \, \times \, 10^{-2}}{5}\)

B2 = 0.01508 = 1.50 × 10^-2Wb/m^2


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