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katrin2010 [14]
3 years ago
8

A square current-carrying loop is placed in a uniform magnetic field B with the plane of the loop parallel to the magnetic field

. (See drawing). The dashed line is the axis of rotation. The magnetic field exerts:
a) a net torque, but not a net force, on the loop

b) a net force and a net torque on the loop

c) a net force, but not a net torque, on the loop

d) neither a net force nor a net torque on the loop
Physics
1 answer:
Radda [10]3 years ago
5 0

Answer:

(A) a net torque but no net force on the loop.

Explanation:

The total force on the loop is zero because the forces on the opposite sides of the loop are equal but act in opposite directions and as a result they cancel each other out. The two forces on opposite sides to the axis of rotation each give rise to a torque about the axis of rotation. This torque is directed along the axis of rotation.

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Answer:

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

Explanation:

Energy conservation law: In isolated system the amount of total energy remains constant.

The types of energy are

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Kinetic energy =\frac{1}{2} mv^2

Potential energy =\frac{Kq_1q_2}{d}

Here, q₁= +5.00×10⁻⁴C

q₂=-3.00×10⁻⁴C

d= distance = 4.00 m

V = velocity = 800 m/s

Total energy(E) =Kinetic energy+Potential energy

                      =\frac{1}{2} mv^2+ \frac{Kq_1q_2}{d}

                     =\frac{1}{2} \times 4.00\times 10^{-3}\times(800)^2 +\frac{9\times10^9\times 5\times10^{-4}\times(-3\times10^{-4})}{4}

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                    =942.5 J

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Therefore,

E =\frac{1}{2} mv^2 + \frac{Kq_1q_2}{d}

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\Rightarrow 942.5 = 2\times10^{-3}v^2 -6750

\Rightarrow 2 \times10^{-3}\times v^2= 942.5+6750

\Rightarrow v^2 = \frac{7692.5}{2\times 10^{-3}}

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