A wire in a uniform magnetic field of 0.350 T carries a current of 3.50 A. If the magnitude of the magnetic force per unit lengt
1 answer:
Answer:
13.12°
Explanation:
Parameters given:
Magnetic field strength, B = 0.350 T
Current in the wire, I = 3.5 A
Force per unit length, F/L = 0.278 N/m
The force on a wire due to the presence of a magnetic field is given as:
F = I*L*B*sinФ
where Ф = angle between the magnetic field and current
The force per unit length, F/L, will be:
F/L = I*B*sinФ
0.278 = 3.5 * 0.35 * sinФ
=> sinФ =
sinФ = 0.227
Ф =
Ф = 13.12°
The angle between the magnetic field and the current is 13.12°.
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Refer to the diagram shown below.
When an athlete is in motion, he/she exerts a vertical force (the person's weight, W) on the ground. The ground exerts an equal and opposite force, N, the normal reaction on the athlete, so that W = N.
At the same time, the ground exerts a horizontal force, F, o n the athlete so that he/she does not slip.
The magnitude of the horizontal force is
F = μN = μW
where μ = the dynamic coefficient of friction.
Answer:
The horizontal force is μW,
where
W = the weight of the athlete and,
μ = the dynamic coefficient of friction.
When an athlete is in motion, he/she exerts a vertical force (the person's weight, W) on the ground. The ground exerts an equal and opposite force, N, the normal reaction on the athlete, so that W = N.
At the same time, the ground exerts a horizontal force, F, o n the athlete so that he/she does not slip.
The magnitude of the horizontal force is
F = μN = μW
where μ = the dynamic coefficient of friction.
Answer:
The horizontal force is μW,
where
W = the weight of the athlete and,
μ = the dynamic coefficient of friction.