This set of Class 12 Physics Chapter 4 Multiple Choice Questions & Answers (MCQs) focuses on “Ampere and Forces between Two Parallel Currents”.
1. What is the expression for the force experienced by two parallel current carrying conductors of unequal length?
a) F=\(\frac {\mu_o}{4\pi } \times \frac {2I_1 I_2}{r}\) × l
b) F=\(\frac {\mu_o}{4\pi } \times \frac {I_1 I_2}{r}\) × l
c) F=\(\frac {\mu_o}{4\pi } \times \frac {\frac {2I_1 I_2}{r}}{l}\)
d) F=\(\frac {\mu_o}{4\pi } \times \frac {2I_1 I_2}{4r}\) × l
View Answer
Explanation: If two linear current carrying conductors of unequal length are held parallel to each other, then the force on a long conductor is due to magnetic field interaction due to currents of short conductor and long conductor. The force on long conductor is force to that on the short conductor given by:
F=\(\frac {\mu_o}{4\pi } \times \frac {2I_1 I_2}{r}\) × l
2. Two parallel, long wires carry currents I1 and I2 where I1 > I2. When the currents are in the same direction, the magnetic field at a point midway between the wires is 50 μT. If the direction of I2 is revered, the field becomes 100 μT. What is the value of \(\frac {I_1}{I_2}\)?
a) 1
b) 2
c) 3
d) 4
View Answer
Explanation: At the midpoint ➔ CASE 1:
B1=\(\frac {\mu_o I_1}{2\pi d} – \frac {\mu_o I_2}{2\pi d}\)
B1=\(\frac {\mu_o (I_1 – I_2)}{2\pi d}\)=50 μT
At the midpoint ➔ CASE 2:
B2=\(\frac {\mu_o I_1}{2\pi d} + \frac {\mu_o I_2}{2\pi d}\)
B1=\(\frac {\mu_o (I_1 + I_2)}{2\pi d}\)=100 μT
\(\frac {B1}{B2}=\frac {I_1-I_2}{I_1+I_2}=\frac {50 \mu T}{100 \mu T}\)
\(\frac {I_1-I_2}{I_1+I_2}=\frac {1}{2}\)
Using componendo and Dividendo rule:
\(\frac {(I_1+I_2) \, + \, (I_1-I_2)}{(I_1-I_2) \, – \, (I_1+I_2)}=\frac {1+2}{1-2}\)
\(\frac {I_1}{-I_2}=\frac {3}{-1}\)
Therefore, \(\frac {I_1}{I_2}\)=3
3. There are two conductors X and Y carrying a current I and moving in the same direction. p and q are two electron beams also moving in the same direction. Will there be attraction or repulsion between the 2 conductors and between the two electron beams separately?
a) The electron beams will repel each other and conductors attract each other
b) The electron beams will attract each other and the conductors also attract each other
c) The electron beams will attract each other and the conductors repel each other
d) The electron beams will repel each other and the conductors also repel each other
View Answer
Explanation: Since, both the current carrying conductors are moving in the same direction, they will attract each other. Moving electron beams is equivalent to an electric current in opposite direction. Therefore, there will attraction between the conductors X and Y as well as between the electron beams p and q.
4. When infinitely long parallel wires carry equal currents in the same direction, the magnetic field at the midpoint in between the two wires is zero.
a) True
b) False
View Answer
Explanation: Yes, when infinitely long parallel wires carry equal currents in the same direction, the magnetic field at the midpoint in between the two wires is zero. This is because, when the wires carry equal currents in the same direction, the magnetic field created at the midpoint will cancel out each other and thus, the field is zero.
5. Two long conductors, separated by a distance r carry current I1 and I2 in the same direction. They exert a force F on each other. Now, the current in one them is increased to 3 times and the direction is reversed. The distance is also increased to 5r. What is the new value of the force between them?
a) \(\frac {3F}{5}\)
b) \(\frac {-3F}{5}\)
c) \(\frac {5F}{3}\)
d) \(\frac {-5F}{3}\)
View Answer
Explanation: First case:
F=\(\frac {\mu_o}{2\pi } \big [ \frac {I_1 I_2}{r} \big ] \)l
Second case:
F‘=\(\frac {\mu_o}{2\pi } \big [ \frac {(-3I_1)(I_2)}{5r} \big ] \)l
\(\frac {F^{‘}}{F} = \frac {\frac {\mu_o}{2\pi } \big [ \frac {(-3I_1)(I_2)}{5r} \big ] l}{\frac {\mu_o}{2\pi } \big [ \frac {I_1 I_2}{r} \big ] l}\)
\(\frac {F^{‘}}{F}= -\frac {3}{5}\)
Therefore, F‘=\((-\frac {3}{5})\)F=\(\underline{\frac {-3F}{5}}\)
6. There are 2 long parallel conductors AB and CD. AB carries 4A current and CD carries 3A current. The magnetic field at the midpoint of these 2 conductors is B. If 4A current is switched off, then what is the magnetic field at the midpoint now?
a) \(\frac {B}{3}\)
b) \(\frac {2B}{3}\)
c) B
d) 3B
View Answer
Explanation: Firstly, the magnetic field at the midpoint due to current in AB:
BAB=μo×\(\frac {2 \times 4}{4\pi d}=\frac {8\mu_o}{4\pi d}\) ………………..1
Magnetic field at midpoint due to current in CD:
BCD=μo×\(\frac {2 \times 3}{4\pi d}=\frac {6\mu_o}{4\pi d}\) …………………..2
The net magnetic field = BAB – BCD ➔ 1 – 2
B = \(\frac {2 \mu_o}{4\pi d}\) ……………………….3
Now, when 4A is switched off, then the magnetic field at the midpoint will be due to the current in CD, i.e. due to 3A current
B‘ = μo×\(\frac {2 \times 3}{4\pi d}=\frac {6\mu_o}{4\pi d}\) ……………..4
Comparing 3 and 4:
Therefore, B‘ = 3B
7. Pick out the expression for force per unit length between two parallel carrying conductors if the current passing through both of them are the same?
a) \(\frac {\mu_o 2I^2}{4πb}\)
b) \(\frac {\mu_o 2I^3}{4πb}\)
c) \(\frac {\mu_o I^2}{4πb}\)
d) \(\frac {\mu_o 4I^2}{4πb}\)
View Answer
Explanation: The expression for force between 2 parallel current carrying conductors is given by:
f=\(\frac {\mu_o}{4\pi } \big [ \frac {I_1 I_2}{b} \big ] \)
In this case, the current passing through both of them are the same, i.e. I1=I2
Therefore, the force per unit length is given as:
f=\(\frac {\mu_o}{4\pi } \big [ \frac {I^2}{b} \big ] \)
Sanfoundry Global Education & Learning Series – Physics – Class 12.
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