Question

A curent-carrying 2.0-cm long segment of wire is inside a long solenoid adius - 40 cm - 800 turns/m, current - 50 mA). The wire segment is oriented pependicalrly to the axis of the solenoid. If the curreat in the wire segment is 12 A, what is temagnitade of the magnetic ferce on this segment?

Answer 1

The magnitude of the magnetic force on a current-carrying wire segment inside a solenoid can be calculated using the formulas for force and magnetic field strength. Substituting the given values, the magnetic force can be calculated as 0.0192 N.

Step-by-step explanation:

The magnitude of the magnetic force on a current-carrying wire segment inside a solenoid can be calculated using the formula:

Force = current × length of the wire segment × magnetic field strength

In this case, the current is 12 A, the length of the wire segment is 2.0 cm (0.02 m), and the magnetic field strength inside the solenoid can be determined using the formula:

Magnetic field strength = permeability of free space × number of turns per meter × current

Substituting the given values, the magnetic field strength can be calculated as:

Magnetic field strength = (4π×10-7 T×m/A) × (800 turns/m) × (50×10-3 A) = 0.08 T

Now, substituting the calculated magnetic field strength and the other given values into the formula for force:

Force = 12 A × 0.02 m × 0.08 T = 0.0192 N

Therefore, the magnitude of the magnetic force on the wire segment is 0.0192 N.

Answer 2

The value is
`F_(max)  = 12 \muN`

Step-by-step explanation:

From the question we are told that

The length is
`l = 2.0 \  cm  =  0.02 \  m`

The radius of the the solenoid is
`r =40 \ cm`

The number of turns per meter is
`N  =  800 \  turns / m`

The current through the solenoid is
`I  =  50 \  mA  =  50*10^(-3) \  A`

The current through the segment is
`I_s  =  12 \  A`

Generally the magnetic force is mathematically represented as

`F  = B  *  I_s l sin(\theta)`

At maximum
`\theta =  90^o`

So

`F_(max)  = B  *  I_s l`

Here B is the magnetic field is mathematically represented as

`B  =  \mu_o  * N  * I`

Here
`\mu_o` is the permeability of free space with value

`\mu_o  =   4\pi * 10^(-7) N/A^2`

So

`B  =    4\pi * 10^(-7)  * 800  *50*10^(-3)`

=>
`B  =   5.0272 *10^(-5)   \  T`

So

`F_(max)  =  5.0272 *10^(-5)  * 12 *  0.02`

`F_(max)  = 1.2 *10^(-5)`

`F_(max)  = 12 \muN`