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saveliy_v [14]
2 years ago
9

QUICK: A circular loop of radius r is rotated through a magnetic field B, which of the following would increase the magnetic flu

x through the loop by a factor of 16?
SELECT ALL THAT APPLY
A. Doubling B and quadrupling r.
B. Quadrupling B and doubling r.
C. B remains constant and quadrupling r.
D. Doubling B and doubling r.
Physics
2 answers:
WITCHER [35]2 years ago
6 0
Doubling B and quadrupling r
xeze [42]2 years ago
3 0
Ok ty you back to my yyyyyy
It’s B
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Answer:

\vec{F}= -3.52\times 10^{-13}\hat{i}\ N

Explanation:

given,

charge = -5.0 μC

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force would a proton experience = ?

we know

\vec{F} = q \vec{E}

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\vec{E} =-2.2 \times 10^{6}\hat{i}

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using formula

\vec{F} = q \vec{E}

\vec{F}= 1.6 \times 10^{-19}\times -2.2 \times 10^{6}\hat{i}

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Force experienced by the photon in the same field is equal to \vec{F}= -3.52\times 10^{-13}\hat{i}\ N

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I dropped an apple (mass 0.1kg) from the window because i'm weird. (15m above the ground). How fast was it going when it hit the
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Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge
Taya2010 [7]

Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃=  2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So,  the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs.  F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

\frac{k*q_{1}*q_3 }{(d_{13})^{2}  } = \frac{k*q_{2}*q_3 }{(d_{23})^{2}  }

We divide by (k * q3) on both sides of the equation

\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }

q_{1} = \frac{q_{2}*(d_{13})^{2}   }{(d_{23} )^{2}  }

q_{1} = \frac{5*(2)^{2} }{(4 )^{2}  }

q₁ = + 1.25 nC

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