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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Answer:
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
[tex]I = \frac{emf}{R} = \frac{VBL}{R}[/tex]
magnitude of the force that the magnetic field exerts on the loop:
F = BIL
Substitute in the value of I
[tex]F = \frac{VB^2L^2}{R}[/tex]
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
[tex]F = \frac{3*2^2*(2*10^{-2})^2}{1} \\\\F = 4.8 *10^{-3} \ N[/tex]
Therefore, the magnitude of the force that the magnetic field exerts on the loop 4.8 x 10⁻³ N
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?
Answer:
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
