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KATRIN_1 [288]
3 years ago
15

1. Two wires - A and B - with circular cross-sections have identical lengths and are made of the same material. Yet, wire A has

four times the resistance of wire B. How many times greater is the diameter of wire B than wire A?
Physics
1 answer:
PolarNik [594]3 years ago
6 0

Answer:

Diameter of wire B is 2 times the diameter of wire A

Explanation:

We have given two wire A and B

They are made up of same material and length of both the wire is sane

So l_A=l_B

Let the resistivity of both the wire is \rho

It is given that wire A has 4 times the resistance as wire B

So R_A=4R_B

So \frac{\rho l_A}{a_A}=4\frac{\rho l_B}{a_B} ( As l_A=l_B )

\frac{a_A}{a_B}=\frac{1}{4}

\frac{d_A^2}{d_B^2}=\frac{1}{4}

\frac{d_A}{d_B}=\frac{1}{2}

d_B=2d_A

So diameter of wire B is 2 times the diameter of wire A

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When the current in a toroidal solenoid is changing at a rate of 0.0240 A/s , the magnitude of the induced emf is 12.4 mV . When
Gemiola [76]

Answer:

The number of turns in the solenoid is 230.

Explanation:

Given that,

Rate of change of current, \dfrac{dI}{dt}=0.0240\ A/s

Induced emf, \epsilon=12.4\ mV=12.4\times 10^{-3}\ V

Current, I = 1.5 A

Magnetic flux, \phi=0.00338\ Wb

The induced emf through the solenoid is given by :

\epsilon=L\dfrac{dI}{dt}

or

L=\dfrac{\epsilon}{(di/dt)}........(1)

The self inductance of the solenoid is given by :

L=\dfrac{N\phi}{I}.........(2)

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

\dfrac{\epsilon}{(di/dt)}=\dfrac{N\phi}{I}

N is the number of turns in the solenoid

N=\dfrac{\epsilon I}{\phi (dI/dt)}

N=\dfrac{12.4\times 10^{-3}\times 1.5}{0.00338 \times 0.024}

N = 229.28 turns

or

N = 230 turns

So, the number of turns in the solenoid is 230. Hence, this is the required solution.

3 0
3 years ago
A 0.55-μF capacitor is connected to a 3.5-V battery. How much charge is on each plate of the capacitor?
yan [13]

Answer:

1.925 μC

Explanation:

Charge: This can be defined as the product of the capacitance of a capacitor and the voltage. The S.I unit of charge is Coulombs (C)

The formula for the charge stored in a capacitor is given as,

Q = CV ................... Equation 1

Where Q = charge, C = Capacitor, V = Voltage.

Note: 1 μF  = 10⁻⁶  F

Given: C = 0.55 μF = 0.55×10⁻⁶ F, V = 3.5 V.

Substitute into equation 1

Q = 0.55×10⁻⁶×3.5

Q = 1.925×10⁻⁶ C.

Q = 1.925 μC

Hence the charge on the plate = 1.925 μC

6 0
3 years ago
Read 2 more answers
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