Question

What current would it need to carry in order to experience a very modest 1.0×10−3n⋅m torque?

Answer 1

The current required in the loop to experience the given torque is $\boxed{6.366\times{{10}^2}\,{\text{A}}}$  or $\boxed{636.6\,{\text{A}}}$ .

Further Explanation:

Given:

The diameter of the circular loop is $20\,{\text{cm}}$ .

The torque experienced by the circular loop is $1.0\times{10^{-3}}\,{\text{N}}\cdot{\text{m}}$ .

Concept:

Since the circular loop is kept in the effect of the Earth’s Magnetic field, it will experience a magnetic torque due to the magnetic lines of force passing through the area of cross-section of the loop.

The torque experienced by the loop is expressed as:

$\boxed{\tau =BIA}$

Here, $\tau$  is the torque experienced, $B$  is the magnetic field, $I$  is the current in the loop and $A$  is the area of cross-section of the loop.

The strength of the Earth’s magnetic field is $5\times{10^{-5}}\,{\text{T}}$ .

Substitute the values in the above expression.

$\begin{aligned}1.0\times{10^{-3}}&=\left({5\times{{10}^{-5}}}\right)\timesI\times\left({\pi \times{{\left({\frac{d}{2}}\right)}^2}}\right)\\I&=\frac{{1.0\times{{10}^{-3}}}}{{5\times{{10}^{-5}}\left({\pi {{\left({\frac{{0.20}}{2}}\right)}^2}}\right)}}\\&=6.366\times{10^2}\,{\text{A}}\\\end{aligned}$

The current required in the loop to experience the given torque is $\boxed{6.366\times{{10}^2}\,{\text{A}}}$  or $\boxed{636.6\,{\text{A}}}$ .

Learn More:

1. A capacitor with capacitance (c) = 4.50 microfarad is connected to a 12.0 v battery. What is the magnitude of the charge on each of the plates brainly.com/question/8892837

2. A very long, uniformly charged cylinder has radius r and linear charge density brainly.com/question/1834208

3. A small bulb has a resistance of 2 ohm when cold brainly.com/question/10421964

Answer Details:

Grade: College

Subject: Physics

Chapter: Electromagnetism

Keywords:

Earth’s magnetic field, torque, maximum torque, maximum current, through the loop, experience a modest torque, T=BIA, 636 A, wire is oriented.

Answer 2

By using the formula:

Tau=(I)(A)(B)Sinθ 

Get the area first of the circle:

A= πr^2 

A= π(0.08m)^2 

A= 0.02011 m^2 


Then solve, 

B(earth)= 5x10^-5 T 

θ = 0, since the angle of the wire is oriented for maximum torque in the earth's field. Which means the angle is 0. 

Reorder the formula to solve for the charge (I): 

I = (tau)/(A)(B)Sinθ 

I = (tau)/(A)(B) 

I = (1.0x10^-3) / (0.02011)(5x10^-5) 

I = 994 A