Step 1: Understand the relationship between turns and magnetic induction. Magnetic induction \( B \) is directly proportional to the number of turns \( N \) and inversely proportional to the area \( A \), assuming the current \( I \) is the same. \[ B \propto \frac{N}{A} \] Step 2: Calculate the ratio. For the first coil with two turns: \[ B_1 \propto \frac{2}{A_1} \] For the second coil with four turns: \[ B_2 \propto \frac{4}{A_2} \] Given that the wire length and hence the total area is the same, \[ \frac{B_1}{B_2} = \frac{2}{4} = 1:4 \].
A wire of 60 cm length and mass 10 g is suspended by a pair of flexible leads in a magnetic field of 0.60 T as shown in the figure. The magnitude of the current required to remove the tension in the supporting leads is:

