Question

A long wire carr1es a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is \(\mathrm{B}\). It is then bent into a circular Loop of n turns. The magnetic field at the centre of the coil for same current will be. (a) \(\mathrm{nB}\) (b) \(\mathrm{n}^{2} \mathrm{~B}\) (c) \(2 \mathrm{nB}\) (d) \(2 \mathrm{n}^{2} \mathrm{~B}\)

Short answer

The magnetic field at the center of the coil for the same current with 'n' turns will be (a) nB.

Step-by-step solution

Recall relationship between magnetic field and coil attributes

The magnetic field at the center of a circular loop is given by the formula: \[B = \frac{\mu_0 * I * R}{2 R^2}\] Where: - \(B\) is the magnetic field at the center of the loop, - \(\mu_0\) is the permeability of free space, - \(I\) is the current through the wire, - \(R\) is the radius of the circular loop. In this case, when the coil has one turn, the magnetic field is given as \(B\).

Find the expression for magnetic field with n turns

Now we need to extend this expression for a coil with 'n' turns. The magnetic field produced by each turn is the same, given by the formula in step 1. Since the loops are wound on top of each other, the magnetic field at the center will add up. Thus, the total magnetic field at the center of the coil with 'n' turns would be: \[B_{total} = n * B\] Where: - \(B_{total}\) is the total magnetic field at the center, - \(n\) is the number of turns. Now let's find the correct answer from the given options.

Match the result with given options

Comparing our derived expression with the given options, we can see that: \[B_{total} = n * B\] Corresponds to option (a) nB. So, the magnetic field at the center of the coil for the same current with 'n' turns will be: (a) nB