Question

The magnetic field inside a current carrying toroidal solenoid is B. What will the magnetic field be if the radius and current are doubled.

Answer 1

2B

Answer 2

The magnetic field inside a current-carrying toroidal solenoid is given by the formula:

$B = \mu_0 n I$

where:

• $B$ is the magnetic field strength
• $\mu_0$ is the permeability of free space
• $n$ is the number of turns per unit length along the circumference of the toroid
• $I$ is the current flowing through the solenoid

The initial magnetic field is given as $B$. So, $B = \mu_0 n I$.

The question states that the current ($I$) is doubled, meaning the new current $I' = 2I$. The radius ($R$) is also doubled, meaning the new radius $R' = 2R$.

In problems of this nature, unless specified otherwise, it is generally assumed that the number of turns per unit length ($n$) remains constant. If $n$ is constant, then the magnetic field is directly proportional to the current $I$.

Let's calculate the new magnetic field $B'$:

$B' = \mu_0 n I'$

Substitute $I' = 2I$:

$B' = \mu_0 n (2I)$

$B' = 2 (\mu_0 n I)$

Since the initial magnetic field $B = \mu_0 n I$, we can substitute this into the equation for $B'$:

$B' = 2B$

Therefore, if the current is doubled and the number of turns per unit length remains constant, the magnetic field inside the toroidal solenoid will also double. The change in radius does not affect the magnetic field if $n$ is constant.