A long solenoid carrying a current produces a magnetic field \(B\) along its axis.
If the current is doubled and the number of turns per cm is halved, what will be the new value of the magnetic field?
1. \(B/2\)
2. \(B\)
3. \(2B\)
4. \(4B\)
In a current-carrying long solenoid, the field produced does not depend upon:
1. | Number of turns per unit length | 2. | Current flowing |
3. | Radius of the solenoid | 4. | All of the above |
1. | \(B \over 2\) | 2. | \(2B\) |
3. | \(B \over 4\) | 4. | \(2B \over 3\) |
1. | infinite | 2. | zero |
3. | \( \frac{\mu_0 2 i}{4 \pi} ~\text{T} \) | 4. | \( \frac{\mu_0 i}{2 r} ~\text{T} \) |
If a long hollow copper pipe carries a direct current along its length, then the magnetic field associated with the current will be:
1. | Only inside the pipe | 2. | Only outside the pipe |
3. | Both inside and outside the pipe | 4. | Zero everywhere |
1. | 2. | ||
3. | 4. |
1. | \(16 \times 10^{-4}~\text{T}\) | 2. | \(8 \times 10^{-4}~\text{T}\) |
3. | \(32 \times 10^{-4}~\text{T}\) | 4. | \(4 \times 10^{-4}~\text{T}\) |
1. | \(\frac{1}{2}\) | 2. | \(1\) |
3. | \(4\) | 4. | \(\frac{1}{4}\) |
Consider six wires with the same current flowing through them as they enter or exit the page. Rank the magnetic field's line integral counterclockwise around each loop, going from most positive to most negative.
1. \(B>C>D>A\)
2. \(B>C=D>A\)
3. \(B>A>C=D\)
4. \(C>B=D>A\)
If the value of integral \(\oint \vec {B} \cdot \vec {dl}\) for the loops \(C_1, C_2,~\text{and}~C_3\) \(2\mu_0, 4\mu_0~\text{and}~\mu_0\) in the units of N/A, respectively, then:
1. | \(I_1=3 A\) into the paper | 2. | \(I_2=3 A\) out of the paper |
3. | \(I_3=0\) | 4. | \(I_3=1 A\) out of the paper |