Answer:
a) F = 2.2275 * 10^-4 N , upwards
b) F = 2.2275 * 10^-4 N , east to west
c) F = 0
Explanation:
Given:
- The straight wire of length L = 2.7 m
- The current I in the wire I = 1.50 A
- The magnetic Field B = 0.55 * 10^-4 T
Find:
Find the magnitude and direction of the force that our planet’s magnetic field exerts on this wire if it is oriented so that the current in it is running
a) from west to east
b) vertically upward
c) from north to south
d) Is the magnetic force ever large enough to cause significant effects under normal household conditions?
Solution:
- If current runs from west to east the angle between the magnetic field B is θ = 90°, so the magnitude of force due to magnetic field is given by Lorentz force.
F = B*I*L*sin(θ)
F = 0.55*1.5*2.7*sin(90)*10^-4
F = 2.2275 * 10^-4 N
- From Figure B points north and current I points east. From right hand rule, the direction of force is out of page, so its upward.
- If current runs upward the angle between the magnetic field B is θ = 90°, so the magnitude of force due to magnetic field is given by Lorentz force.
F = B*I*L*sin(θ)
F = 0.55*1.5*2.7*sin(90)*10^-4
F = 2.2275 * 10^-4 N
- From Figure B points north and current I points out of page . From right hand rule, the direction of force is out of page, so its east to west.
- If current runs north to south the angle between the magnetic field B is θ = 0°, so the magnitude of force due to magnetic field is given by Lorentz force.
F = B*I*L*sin(θ)
F = 0.55*1.5*2.7*sin(0)*10^-4
F = 0 N
The magnitude of the force that our planet's magnetic field exerts is mathematically given as
F=1.82* 10^{-4}N
What is the magnitude of the force that our planet's magnetic field exerts on this cord?
Question Parameter(s):
A straight 2.70 m wire carries a typical household current of 1.50
the earth's magnetic field is 0.550 gauss
Generally, the equation for the force exerted by the magnetic field is mathematically given as
F=ILB
Where
B=0.550 G
B= 0.55 *10^{-4}T
In conclusion,force exerted by the magnetic field is
F=(1.50 A)(2.20 m)(0.55* 10^{-4} T)
F=1.82* 10^{-4}N
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Complete Question
A straight 2.70 m wire carries a typical household current of 1.50 an (in one direction) at a location where the earth's magnetic field is 0.550 gauss from south to north.
Find the magnitude of the force that our planet's magnetic field exerts on this cord if is oriented so that the current in it is running from west to east
two pulleys one with adius 2 inches and the other with radius 8 inches are connected by a belt. If the 2 inch pulley is caused to rotate at 3 revolutions per minuite detirmine te revolutions per minute of the 8 inch pulley
Explanation:
we know that,
linear speed = circumference × revolution per minute
linear speed of belt = 2πr × revolution per minute
now we will compute the linear speed of a belt for 2 inch pulley that is,
linear speed for 2 inch pulley = (2π×2)×( 3 revolutions per minute) ∵ r =2
= 4π × 3 revolution per minute (1)
again we will compute the linear speed of a belt for 8 inch pulley,
linear speed of 8 inch pulley = (2π×8) × (x revolution per minute) ∵ r =8
= 16π×x revolutions per minute (2)
As the linear speed is same for both pulleys. by comparing equations (1) and (2).
4π×3 = 16π×x
x = 3/4
Thus, the revolutions per minute for the 8 inch pulley is 3/4.
Let two objects of equal mass m collide. Object 1 has initial velocity v, directed to the right, and object 2 is initially stationary.
A. If the collision is perfectly elastic, what are the final velocities v1 and v2 of objects 1 and 2?
Give the velocity v1 of object 1 followed by the velocity v2 of object 2, separated by a comma. Express each velocity in terms of v.
Answer:
(v, 0) or (0, v)
Explanation:
The law of conservation of linear momentum states total initial momentum equal total final momentum.
Total initial momentum = [tex]mv + m\times 0 = mv[/tex]
Total final momentum = [tex]mv_1 + mv_2 = m(v_1+v_2)[/tex]
Equating both,
[tex]mv = m(v_1+v_2)[/tex]
[tex]v = v_1+v_2[/tex]
For an elastic collision, kinetic energy is conserved i.e. total initial kinetic energy = total final kinetic energy
[tex]\frac{1}{2}mv^2 + \frac{1}{2}m\times0^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2[/tex]
[tex]v^2 = v_1^2+v_2^2[/tex]
From the first equation, [tex]v_1 = v-v_2[/tex].
Substituting this in the second equation,
[tex]v^2 = (v-v_2)^2+v_2^2[/tex]
[tex]v^2 = v^2-2vv_2 +v_2^2+v_2^2[/tex]
[tex]0 = 2v_2^2-2vv_2[/tex]
[tex]v_2(v_2-v) = 0[/tex]
[tex]v_2 = 0[/tex] or [tex]v_2 = v[/tex]
From [tex]v_1 = v-v_2[/tex],
[tex]v_1 = v-v = 0[/tex]
OR
[tex]v_1 = v-0 = v[/tex]