Answer:
time constant will decrease and steady state current will decrease on increasing the resistance
Explanation:
As we know that the EMF of cell is E which is used to connected across a resistor and an inductor.
So we will have

here we know that

now here we have

so if we increase the value of resistance of the wire then the time constant will decrease
and hence it will take less time to reach near the steady state value
also the steady state current will be smaller in that case
Answer:
he polarity of the electromagnet is determined by the direction the current. The north pole of the electromagnet is determined by using your right hand. Wrap your fingers around the coil in the same direction as the current is flowing (conventional current flows from + to -).
Explanation:
Answer: option 4: A wire that is 2-mm thick and coiled.
Explanation:
The current in each wire is same. The magnetic field due to a current carrying wire increases if the wire is coiled with the more number of turns. A thick wire would cause low resistance to the current. Hence, a 2-mm thick wire which is coiled would produce the strongest magnetic field.
Answer:
8F_i = 3F_f
Explanation:
When two identical spheres are touched to each other, they equally share the total charge. Therefore, When neutral C is first touch to A, they share the initial charge of A equally.
Let us denote that the initial charge of A and B are Q. Then after C is touched to A, their respective charges are Q/2.
Then, C is touched to B, and they share the total charge of Q + Q/2 = 3Q/2. Their respective charges afterwards is 3Q/4 each.
The electrostatic force, Fi, in the initial configuration can be calculated as follows.
![F_i = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{Q^2}{r^2}[/tex}The electrostatic force, Ff, in the final configuration is [tex]F_f = \frac{1}{4\pi\epsilon_0}\frac{q_Aq_B}{r^2} = \frac{1}{4\pi\epsilon_0}\frac{3Q^2/8}{r^2}[/tex}Therefore, the relation between Fi and Ff is as follows[tex]F_i = F_f\frac{3}{8}\\8F_i = 3F_f](https://tex.z-dn.net/?f=F_i%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7Bq_Aq_B%7D%7Br%5E2%7D%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7BQ%5E2%7D%7Br%5E2%7D%5B%2Ftex%7D%3C%2Fp%3E%3Cp%3EThe%20electrostatic%20force%2C%20Ff%2C%20in%20the%20final%20configuration%20is%20%3C%2Fp%3E%3Cp%3E%5Btex%5DF_f%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7Bq_Aq_B%7D%7Br%5E2%7D%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%5Cepsilon_0%7D%5Cfrac%7B3Q%5E2%2F8%7D%7Br%5E2%7D%5B%2Ftex%7D%3C%2Fp%3E%3Cp%3ETherefore%2C%20the%20relation%20between%20Fi%20and%20Ff%20is%20as%20follows%3C%2Fp%3E%3Cp%3E%5Btex%5DF_i%20%3D%20F_f%5Cfrac%7B3%7D%7B8%7D%5C%5C8F_i%20%3D%203F_f)
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