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S_A_V [24]
3 years ago
12

An electric charge in a magnetic field will experience no force if the charge is moving _______ to the field.

Physics
2 answers:
andreev551 [17]3 years ago
7 0

No force is exerted on electric charge when it moves parallel to magnetic field.

Answer: Parallel

<u>Explanation: </u>

When the electric charge moves parallel in a magnetic field, the component of velocity and field becomes zero which results in net force as zero.

This due to the sine component which makes zero angle and thus component as zero too. The force on the charge moving in magnetic field is perpendicular to velocity and field.

BlackZzzverrR [31]3 years ago
6 0

Answer:

Parallel

Explanation:

An electric charge in a magnetic field will experience no force if the charge is moving parallel to the field.

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When you moved the compass near a bar magnet, the needle pointed toward the magnet's magnetic field and not toward the north.

Explanation:

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Which of these is most likely to lead directly to a black market?
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What would be the kinetic energy k2q of charge 2q at a very large distance from the other charges? express your answer in terms
Pavlova-9 [17]

Answer:

K_{2q}=\frac{7.76kq^2}{d}

Explanation:

At the corner of the square, the potential energy of interaction of other charges with the charge 2q  is given by U_{2q}

So

U_{2q,i}=k\frac{(2q)(q)}{d}+k\frac{(2q)(5q)}{d}+k\frac{(2q)(-3q)}{\sqrt{2}d}=\frac{7.76 kq^2}{d}

Also, since K_{2q,i}=0

The initial energy of the system is given by;

E_i=U_{2q ,i}+K_{2q,i}=\frac{7.76kq^2}{d}+0=\frac{7.76kq^2}{d}

Since U_{2q,f}=0

, the final energy of the system is obtained by

E_f=U_{2q ,f}+K_{2q,f}=0+K_{2q,f}

From the law of conservation of energy, E_i=E_f

Therefore, K_{2q}=\frac{7.76kq^2}{d}

7 0
3 years ago
A particle of mass, m = 3.0kg movers with a uniform velocity, v = (5m/s)I +(3m/s)j. At the time, t, the particle passes through
quester [9]

Answer:

The magnitude is 63 kg m²/s, and the direction is -k.

Explanation:

Plug the values into the equation:

L = m v×r

L = (3.0 kg) (<5 i + 3j> m/s × <2i − 3j> m)

Take the cross product.  The cross product of two dimensional vectors is:

<v₁ i + v₂ j> × <r₁ i + r₂ j> = <(v₁ r₂ − v₂ r₁) k>

Therefore:

L = (3.0 kg) <((5)(-3) − (3)(2)) k m²/s>

L = (3.0 kg) <-21 k m²/s>

Multiply:

L = <-63 k kg m²/s>

The magnitude is 63 kg m²/s, and the direction is -k.

6 0
3 years ago
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