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kkurt [141]
3 years ago
8

An electron moving with a velocity v⃗ = 5.0 × 107 m/s i^ enters a region of space where perpendicular electric and a magnetic fi

elds are present. The electric field is E⃗ = 104 V/m j^. What magnetic field will allow the electron to go through the region without being deflected?
Physics
1 answer:
stiv31 [10]3 years ago
3 0

Answer:

Magnetic field, B=2\times 10^{-4}\ T

Explanation:

Given that,

Velocity of electron, v=5\times 10^7\ m/s

It enters  a region of space where perpendicular electric and a magnetic fields are present.

Magnitude of electric field, E=10^4\ V/m

We need to find the magnetic field will allow the electron to go through the region without being deflected.

Magnetic force on the electron, F_m=qvB\ sin\theta.......(1)

Electric force on the electron, F = q E........(2)

From equation (1) and (2) we get:

qvB\ sin\theta=qE

B=\dfrac{E}{v}

B=\dfrac{10^4\ V/m}{5\times 10^7\ m/s}

B = 0.0002 T

or

B=2\times 10^{-4}\ T

Hence, this is the required solution.

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Answer:

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Explanation:

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Every one is the multiplication of a scalar by a vector.

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Thus, in a graph the resulting vector is represented with a parallel arrow and pointing in the same direction as the original vector but with a length equal to the original length multiplied by the magnitude of the scalar.

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Tanya [424]

Answer:

(a) K_{small}=4839.3J

(b) K_{larger}=17406.4J

Explanation:

Given data

The angular velocity of two cylinders ω=257 rad/s

The mass of the two cylinders m=2.88 kg

The radius of small cylinder r₁=0.319 m

The radius of larger cylinder r₂=0.605 m

For Part (a)

The rotational kinetic energy of the cylinder is given by:

K=\frac{1}{2}Iw^2

Where I is rotational of inertia of solid cylinder about its central axis.

So

K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr^2)w^2

Substitute the given values

So

K_{small}=\frac{1}{4}(2.88kg)(0.319)^2(257rad/s)^2 \\K_{small}=4839.3J

For Part (b)

K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr_{2}^2)w^2

Substitute the given values

K_{larger}=\frac{1}{4}mr_{2}^2w^2\\ K_{larger}=\frac{1}{4}(2.88kg)(0.605m)^2(257rad/s)^2\\ K_{larger}=17406.4J

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Answer:

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Vector is a quantity having direction as well as magnitude, especially as determining the position of one point in space relative to another.

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Explanation:

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Mandarinka [93]

Answer:

Number of Significant Figures: 2

The Significant Figures are 3 6

Explanation:

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