A long solenoid carrying a current produces a magnetic field B along its axis. If the current is doubled and the number of turns per cm is halved the new value of the magnetic field is

Question

A long straight current-carrying wire is placed in a uniform magnetic field of strength \( B = 0.5 \, \text{T} \). If the current in the wire is \( I = 2 \, \text{A} \) and the wire makes an angle of \( 30^\circ \) with the magnetic field, find the force per unit length on the wire.

Answer 1

To solve this problem, we need to understand the relationship between the magnetic field inside a solenoid, the current, and the number of turns per unit length.

The magnetic field inside an ideal long solenoid is given by the formula:

B = \mu_0 n I

where \mu_0 is the permeability of free space, n is the number of turns per unit length (turns/cm), and I is the current through the solenoid.

Let's look at the changes made to the solenoid:

The current I is doubled. Initially, if the current is I, now the current is 2I.
The number of turns per cm n is halved. If initially the number of turns is n, it becomes \frac{n}{2} now.

Let's substitute these changes into the formula for the magnetic field:

B' = \mu_0 \left(\frac{n}{2}\right)(2I)

On simplifying, we get:

B' = \mu_0 \cdot n \cdot I = B

This implies that the new magnetic field B' remains the same as the initial magnetic field B.

Hence, the correct answer is: B.

This shows that when the current is doubled and the number of turns per cm is halved, the net effect is that the magnetic field remains unchanged.

Answer 2