Physics, 04.04.2020 12:18 shanicar33500

At the instant when the speed of the loop is 3.00 m/sm/s and it is still partially in the field region, what is the magnitude of the force that the magnetic field exerts on the loop?

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Answer from: copelandgarret9972

Complete question:

A rectangular loop of wire with dimensions 2.0 cm by 10.0 cm and resistance 1.0 Ω is being pulled to the right out of a region of uniform magnetic field. The magnetic field has magnitude of 2.0 T and is directed into the plane. At the instant when the speed of the loop is 3.00 m/s and it is still partially in the field region, what is the magnitude of the force that the magnetic field exerts on the loop?

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The magnitude of the force that the magnetic field exerts on the loop 4.8 x 10⁻³ N

Explanation:

Given;

resistance of the wire; R = 1.0 Ω

magnitude of magnetic field strength, B = 2.0 T

speed of the loop, v = 3.00 m/s

Induced emf is given as;

ε = IR

$I = \frac{emf}{R} = \frac{VBL}{R}$

magnitude of the force that the magnetic field exerts on the loop:

F = BIL

Substitute in the value of I

$F = \frac{VB^2L^2}{R}$

where;

L is the displacement vector between the initial and final end of the portion of the wire inside the field = 2.0 cm

Substitute the given values and solve for F

$F = \frac{3*2^2*(2*10^{-2})^2}{1} \\\\F = 4.8 *10^{-3} \ N$

Therefore, the magnitude of the force that the magnetic field exerts on the loop 4.8 x 10⁻³ N

At the instant when the speed of the loop is 3.00 m/sm/s and it is still partially in the field regi

Answer from: daltondespain2

Complete Question:

A rectangular loop of wire with dimensions 1.50 cm by 8.00 cm and resistance 0.700 Ω is being pulled to the right out of a region of uniform magnetic field. The magnetic field has magnitude 2.20 T and is directed into the plane of.

At the instant when the speed of the loop is 3.00 m/s and it is still partially in the field region, what is the magnitude of the force that the magnetic field exerts on the loop?

F = 0.133 N

Explanation:

Magnitude of the magnetic field, B = 2.20 T

Length of the loop = 1.5 cm = 0.015 m

The speed of the loop, v = 3.00 m/s

The emf induced in the loop , e = Blv

e = 2.20 * 0.015 * 3

e = 0.099 V

Current induced in the loop, I = e/R

I = 0.099/0.7

I = 0.1414 A

The magnitude of the force is given by, F = I *l *B sin90

F = 0.1414 * 0.015 * 2.20

F = 0.00467 N

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