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Pavel [41]
3 years ago
6

The Hall effect can be used to calculate the charge-carrier number density in a conductor. A conductor carrying current of 2.0 A

is 0.50 mm thick, and the Hall voltage is 4.5 x 10-6 V when it is in a uniform magnetic field of 1.2 T. What is the density of charge carriers in the conductor? a) n = 4.6 x 1027 charges/m b) n = 1.7 x 1027 charges/m c) n = 1.0 x 1028 charges/m d) n = 6.7 x 1027 charges/m²
Physics
1 answer:
marysya [2.9K]3 years ago
7 0

Answer:

option D

Explanation:

given,

A conductor is carrying current = 2.0 A is 0.5 mm thick

Hall voltage = 4.5 x 10-6 V

uniform magnetic field  =  1.2 T

density of the charge = n =?

hall voltage =V_h =\dfrac{i\ B}{n\ e\ L}

n = \dfrac{i\ B}{V\ e\ L}

n = \dfrac{2 \times 1.2 }{4.5 \times 10^{-6}\times 1.6 \times 10^{-19} \times 0.5 \times 10^{-3}}

n = 6.67 × 10²⁷ charges/m

hence the correct answer is option D

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The inner conductor of a coaxial cable has a radius of 0.800 mm, and the outer conductor’s inside radius is 3.00 mm. The space b
ZanzabumX [31]

Answer:

The maximum potential difference is 186.02 x 10¹⁵ V

Explanation:

formula for calculating maximum potential difference

V = \frac{2K_e \lambda}{k}ln(\frac{b}{a})

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k is the dielectric constant = 2.3

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Substitute in these values in the above equation;

V = \frac{2K_e \lambda}{k}ln(\frac{b}{a}) =  \frac{2*8.99*10^9*18*10^6 }{2.3}ln(\frac{3}{0.8}) =140.71 *10^{15} *1.322 \\\\V= 186.02 *10^{15} \ V

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what happens to a circuits resistance, voltage, and current when you increase the diameter of the wire in the circuit?
zhannawk [14.2K]

Answer:

Option D.

Resistance (R) decreases

Voltage (V) is constant

Current (I) increases

Explanation:

We'll begin by writing an equation relating resistance and diameter of a wire. This is given below:

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A = πr² (since the wire is circular in shape)

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A = πd²/4

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R = ρL/A

R = ρL ÷ A

R = ρL ÷ πd²/4

R = ρL × 4/πd²

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R is the resistance of the wire.

ρ is the resistivity of the wire.

L is the length of the wire.

A is the cross sectional area of the wire.

r is the radius.

d is the diameter of the wire

From equation (1) above, we can say that the resistance (R) is inversely proportional to the square of the diameter of the wire. This implies that an increase in the diameter of the wire will result in a decrease of the resistance. Also, a decrease in the diameter of the wire will result in an increase in the resistance of the wire.

1. Since the diameter of the wire is increase, therefore, the resistance of the wire will decrease.

2. From ohm's law,

V = IR

Divide both side by I

R = V/I

Where:

R is the resistance

V is the voltage

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From the above equation, the resistance (R) is directly proportional to the voltage (V) and inversely proportional to the current (I).

If we keep the voltage constant, this means that an increase in the resistance will lead to a decrease in the current. Also, a decrease in the resistance will lead to an increase in the current.

Since the resistance of the wire decrease, the current will increase.

From the illustrations made above, an increase in the diameter of the wire will lead to:

1. Decrease in resistance.

2. Voltage is constant.

3. Increase in current.

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