Physics, 07.03.2020 02:49 Jasten

A single-turn square loop of wire, 2.00 cm on each edge, carries a clockwise current of 0.150 A. The loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30.0 turns/cm and carries a clockwise current of 15.0 A.

(a) Find the force on each side of the loop.
magnitude = µN.

(b) Find the magnitude of the torque acting on the loop.
= N · m

Answers: 2

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Answer from: xojade

The force acting on one of the sides the square loop is 169.6 N

The torque acting on the loop is ZERO

Explanation:

From the question we are given that

permeability of free space $\mu_0$ $= 4\pi*10^{-7} T\cdot m/A$

Number of turns of the solenoid per unit length , n = 30 turns per cm

= 3000 turns per m

The current flowing through the solenoid I_1 = 15A

Length of the side of the loop, L = 2 cm

= 2 cm [ $\frac{1*10^{-2}m}{1cm}$ ]

= $2*10^{-2} m$

Angle , $\theta =90$ °

The current flowing through the wire I_2 = 0.15A

Generally The formula for Magnetic field inside of a solenoid is

$B = \mu_0nI_1$

Where

$\mu_0$ is the permeability of free space

n is the number of turns of the solenoid per unit length

I_1 is the current through the solenoid

Also the formula for force acting on one of the sides of a square loop is

$F = BI_2Lsin \theta$

Where I_2 is the current through the wire ,

L is the length of the sides of the loop,and

$\theta$ is the angle between B and I_2

Substituting the given values into the first formula

$B = (4 \pi *10^{-7})(3000)(15)$

$= 5.65*10^{-2} T$

To obtain the force acting on the loop by substituting the appropriate value into the second formula

$F = (5.65*10^{-2}(0.15)(2*10^{-2}(sin90)))$

$169.6 \mu N$

The torque acting on the lop is zero because the

The forces on the loop are acting in the plane of the loop

These force are perpendicular to the sides

and their direction is away from the inner side of the loop

the only effect the orientation of these forces can cause is stretching of the loop

This way by which the forces are oriented cannot cause any rotation

Answer from: Quest

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more

Answer from: Quest

real, smaller and upside-down. so the answer should be b.

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