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A proton has a magnetic field due to its spin on its axis. the field is similar to that created by a circular current loop 0.650×10−15 m in radius with a current of 1.05×104 a (no kidding). find the maximum torque on a proton in a 2.50-t field

Respuesta :

Answer:

[tex]\tau=3.48*10^{-26}Nm[/tex]  

Explanation:

The torque equation is given by:

[tex]\tau=IABsin(\theta)[/tex]

Where:

  • I is the current
  • A is the area of the loop (A=πR²)
  • θ is the angle between the normal vector of A and B, in this case it is 90 degrees

So:

[tex]\tau=1.05*10^{4}*\pi *(0.65*10^{-15})^{2}*2.5[/tex]  

Therefore the torque is:

[tex]\tau=3.48*10^{-26}Nm[/tex]    

I hope it helps you!

Answer:

The maximum torque on a proton in a 2.50 T field is  1.39 × 10⁻²⁵  N.m

Explanation:

The equation for torque in circular current loop as a result of the magnetic force is given by:

T = IABsin(Θ)

Where:

T is the torque

I is the current in the loop

A is the surface area of the circular current loop

B is the magnetic field intensity

Θ is the angle between the loop plane and the direction of the magnetic field.

Given that:

I = 1.05 × 10⁴ A

the radius r = 0.65 × 10⁻¹⁵ m

A = 4πr² = 4 × π × (0.65 × 10⁻¹⁵)² = 5.308 × 10⁻³⁰ m²

B = 2.50 T

The torque is maximum at Θ = 90°

Since T = IABsin(Θ)

substituting values

T =  1.05 × 10⁴ × 5.308 × 10⁻³⁰ × 2.5 × sin(90)

T = 1.39 × 10⁻²⁵  N.m

The maximum torque on a proton in a 2.50 T field is  1.39 × 10⁻²⁵  N.m