the magnetic field inside the solenoid and then use it to determine the torque on the coil.

The solenoid has 500 turns and a length of 0.4 meters. The current it carries is 3 A. The magnetic field $ B $ inside the solenoid is given by:

$ B = \mu_0 \frac{N}{l} \times I $

Where:

$\mu_0 = 4 \pi \times 10^{-7} \, \mathrm{T \cdot m/A}$ is the permeability of free space,

$N = 500$ is the number of turns,

$l = 0.4 \, \mathrm{m}$ is the length of the solenoid,

$I = 3 \, \mathrm{A}$ is the current through the solenoid.

Substituting the values, we get:

$ B = 4 \pi \times 10^{-7} \times \frac{500}{0.4} \times 3 $

The coil has 10 turns, a radius of 0.01 m, and carries a current of 0.4 A. The torque $ \tau $ on the coil can be calculated using:

$ \tau = N \cdot I \cdot A \cdot B $

Where:

$N = 10$ is the number of turns in the coil,

$I = 0.4 \, \mathrm{A}$ is the current in the coil,

$A = \pi r^2$ is the area of the coil ($r = 0.01 \, \mathrm{m}$ is the radius).

Thus, the torque is:

$ \tau = 10 \cdot 0.4 \cdot (\pi \times (0.01)^2) \cdot B $

Substituting the expression for $B$ into the torque formula and simplifying it step by step:

$ \tau = 10 \cdot 0.4 \cdot \pi \cdot (0.01)^2 \cdot \left( \frac{4 \pi \times 10^{-7} \times 500 \times 3}{0.4} \right) $

Finally, calculating this expression results in:

$ \tau = 5.92 \times 10^{-6} \, \mathrm{N \cdot m} $

Therefore, the torque required is $5.92 \times 10^{-6} \, \mathrm{N \cdot m}$.

Moving Charges And Magnetism

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