362702
\(A\) and \(B\) are two infinitely long straight parallel conductors. \(C\) is another straight conductor of length 1\(m\) kept parallel to A and B as shown in the figure. Then the force experienced by \(C\) is
1 Towards \(A\) equal to \(0.6 \times {10^{ - 5}}\;N\)
2 Towards \(B\) equal to \(5.4 \times {10^{ - 5}}\;N\)
3 Towards \(A\) equal to \(5.4 \times {10^{ - 5}}\;N\)
4 Towards \(B\) equal to \(0.6 \times {10^{ - 5}}\;N\)
Explanation:
Conductor A attracts conductor \(C\) towards \(A\) Force towards A \(\begin{aligned}& =\dfrac{\mu_{0}}{2 \pi} \dfrac{I_{1} I_{2} l}{r}=\dfrac{\mu_{0}}{2 \pi} \dfrac{2 \times 3 \times 1}{0.05}=\dfrac{3 \mu_{0}}{\pi} \times 20 \\& =\dfrac{60 \times 4 \pi \times 10^{-7}}{\pi}=240 \times 10^{-7} \mathrm{~N}\end{aligned}\) Again force towards \(\begin{aligned}& B=\dfrac{\mu_{0}}{2 \pi} \times \dfrac{3 \times 4 \times 1}{0.08}=\dfrac{3 \mu_{0}}{\pi} \times 25 \\& =\dfrac{3 \times 4 \pi \times 10^{-7}}{\pi} \times 25=300 \times 10^{-7} \mathrm{~N}\end{aligned}\) \(\therefore\) Resultant force towards B \(=(300-240)\) \(10^{-7} \mathrm{~N}\) \(=0.6 \times 10^{-5} \mathrm{~N}\).
KCET - 2007
PHXII04:MOVING CHARGES AND MAGNETISM
362703
Three long, straight parallel wires carrying current, are arranged as shown in figure. The force experienced by a 25\(cm\) length of wire \(C\) is
1 \({10^3}\;N\)
2 \(2.5 \times {10^3}\;N\)
3 zero
4 \(1.5 \times {10^{ - 3}}\;N\)
Explanation:
Force on wire \(C\) due to wire \(D\) \({F_A} = {10^{ - 7}} \times \frac{{2 \times 30 \times 10}}{{3 \times {{10}^{ - 2}}}} \times 25 \times {10^{ - 2}} = 5 \times {10^{ - 4}}\;N\) (towards right) Force on wire \(C\) due to wire \(B\) \({F_B} = {10^{ - 7}} \times \frac{{2 \times 20 \times 10}}{{2 \times {{10}^{ - 2}}}} \times 25 \times {10^{ - 2}} = 5 \times {10^{ - 4}}\;N\) (towards left) \(\Rightarrow\) Net force on wire \(C\) is \(F_{n e t}=F_{A}-F_{B}=0\)
PHXII04:MOVING CHARGES AND MAGNETISM
362704
Two parallel wires 1\(m\) apart carry current of 1\(A\) and 3\(A\) respectively in opposite directions. The force per unit length acting between these two wires is
The force per unit length acting between two parallel wires carrying currents \(I_{1}\) and \(I_{2}\) and placed a distance \(d\) apart is \(f = \frac{{{\mu _0}{I_1}{I_2}}}{{2\pi d}}\) Here, \({I_1} = 1A,{I_2} = 3,A,d = 1\;m\) \({\mu _0} = 4\pi \times {10^{ - 7}}Tm{A^{ - 1}}\) \(\therefore f = \frac{{\left( {4\pi \times {{10}^{ - 7}}Tm{A^{ - 1}}} \right)(1\;A)(3\;A)}}{{2\pi (1\;m)}} = 6 \times {10^{ - 7}}\;N\;{m^{ - 1}}\) As the currents are in opposite directions, so \(\mathrm{f}\) is repulsive
KCET - 2015
PHXII04:MOVING CHARGES AND MAGNETISM
362705
Assertion : Two electron beams moving parallel come close each other Reason : Wires carrying current in opposite direction attracy each other
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Two parallel conductors carrying current in same direction attract each other, conductors carrying current in opposite direction repel each other.
PHXII04:MOVING CHARGES AND MAGNETISM
362706
An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current \(I\) along the same direction as shown in fig. Magnitude of force per unit length on the middle wire \(B'\) is given by :-
362702
\(A\) and \(B\) are two infinitely long straight parallel conductors. \(C\) is another straight conductor of length 1\(m\) kept parallel to A and B as shown in the figure. Then the force experienced by \(C\) is
1 Towards \(A\) equal to \(0.6 \times {10^{ - 5}}\;N\)
2 Towards \(B\) equal to \(5.4 \times {10^{ - 5}}\;N\)
3 Towards \(A\) equal to \(5.4 \times {10^{ - 5}}\;N\)
4 Towards \(B\) equal to \(0.6 \times {10^{ - 5}}\;N\)
Explanation:
Conductor A attracts conductor \(C\) towards \(A\) Force towards A \(\begin{aligned}& =\dfrac{\mu_{0}}{2 \pi} \dfrac{I_{1} I_{2} l}{r}=\dfrac{\mu_{0}}{2 \pi} \dfrac{2 \times 3 \times 1}{0.05}=\dfrac{3 \mu_{0}}{\pi} \times 20 \\& =\dfrac{60 \times 4 \pi \times 10^{-7}}{\pi}=240 \times 10^{-7} \mathrm{~N}\end{aligned}\) Again force towards \(\begin{aligned}& B=\dfrac{\mu_{0}}{2 \pi} \times \dfrac{3 \times 4 \times 1}{0.08}=\dfrac{3 \mu_{0}}{\pi} \times 25 \\& =\dfrac{3 \times 4 \pi \times 10^{-7}}{\pi} \times 25=300 \times 10^{-7} \mathrm{~N}\end{aligned}\) \(\therefore\) Resultant force towards B \(=(300-240)\) \(10^{-7} \mathrm{~N}\) \(=0.6 \times 10^{-5} \mathrm{~N}\).
KCET - 2007
PHXII04:MOVING CHARGES AND MAGNETISM
362703
Three long, straight parallel wires carrying current, are arranged as shown in figure. The force experienced by a 25\(cm\) length of wire \(C\) is
1 \({10^3}\;N\)
2 \(2.5 \times {10^3}\;N\)
3 zero
4 \(1.5 \times {10^{ - 3}}\;N\)
Explanation:
Force on wire \(C\) due to wire \(D\) \({F_A} = {10^{ - 7}} \times \frac{{2 \times 30 \times 10}}{{3 \times {{10}^{ - 2}}}} \times 25 \times {10^{ - 2}} = 5 \times {10^{ - 4}}\;N\) (towards right) Force on wire \(C\) due to wire \(B\) \({F_B} = {10^{ - 7}} \times \frac{{2 \times 20 \times 10}}{{2 \times {{10}^{ - 2}}}} \times 25 \times {10^{ - 2}} = 5 \times {10^{ - 4}}\;N\) (towards left) \(\Rightarrow\) Net force on wire \(C\) is \(F_{n e t}=F_{A}-F_{B}=0\)
PHXII04:MOVING CHARGES AND MAGNETISM
362704
Two parallel wires 1\(m\) apart carry current of 1\(A\) and 3\(A\) respectively in opposite directions. The force per unit length acting between these two wires is
The force per unit length acting between two parallel wires carrying currents \(I_{1}\) and \(I_{2}\) and placed a distance \(d\) apart is \(f = \frac{{{\mu _0}{I_1}{I_2}}}{{2\pi d}}\) Here, \({I_1} = 1A,{I_2} = 3,A,d = 1\;m\) \({\mu _0} = 4\pi \times {10^{ - 7}}Tm{A^{ - 1}}\) \(\therefore f = \frac{{\left( {4\pi \times {{10}^{ - 7}}Tm{A^{ - 1}}} \right)(1\;A)(3\;A)}}{{2\pi (1\;m)}} = 6 \times {10^{ - 7}}\;N\;{m^{ - 1}}\) As the currents are in opposite directions, so \(\mathrm{f}\) is repulsive
KCET - 2015
PHXII04:MOVING CHARGES AND MAGNETISM
362705
Assertion : Two electron beams moving parallel come close each other Reason : Wires carrying current in opposite direction attracy each other
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Two parallel conductors carrying current in same direction attract each other, conductors carrying current in opposite direction repel each other.
PHXII04:MOVING CHARGES AND MAGNETISM
362706
An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current \(I\) along the same direction as shown in fig. Magnitude of force per unit length on the middle wire \(B'\) is given by :-
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII04:MOVING CHARGES AND MAGNETISM
362702
\(A\) and \(B\) are two infinitely long straight parallel conductors. \(C\) is another straight conductor of length 1\(m\) kept parallel to A and B as shown in the figure. Then the force experienced by \(C\) is
1 Towards \(A\) equal to \(0.6 \times {10^{ - 5}}\;N\)
2 Towards \(B\) equal to \(5.4 \times {10^{ - 5}}\;N\)
3 Towards \(A\) equal to \(5.4 \times {10^{ - 5}}\;N\)
4 Towards \(B\) equal to \(0.6 \times {10^{ - 5}}\;N\)
Explanation:
Conductor A attracts conductor \(C\) towards \(A\) Force towards A \(\begin{aligned}& =\dfrac{\mu_{0}}{2 \pi} \dfrac{I_{1} I_{2} l}{r}=\dfrac{\mu_{0}}{2 \pi} \dfrac{2 \times 3 \times 1}{0.05}=\dfrac{3 \mu_{0}}{\pi} \times 20 \\& =\dfrac{60 \times 4 \pi \times 10^{-7}}{\pi}=240 \times 10^{-7} \mathrm{~N}\end{aligned}\) Again force towards \(\begin{aligned}& B=\dfrac{\mu_{0}}{2 \pi} \times \dfrac{3 \times 4 \times 1}{0.08}=\dfrac{3 \mu_{0}}{\pi} \times 25 \\& =\dfrac{3 \times 4 \pi \times 10^{-7}}{\pi} \times 25=300 \times 10^{-7} \mathrm{~N}\end{aligned}\) \(\therefore\) Resultant force towards B \(=(300-240)\) \(10^{-7} \mathrm{~N}\) \(=0.6 \times 10^{-5} \mathrm{~N}\).
KCET - 2007
PHXII04:MOVING CHARGES AND MAGNETISM
362703
Three long, straight parallel wires carrying current, are arranged as shown in figure. The force experienced by a 25\(cm\) length of wire \(C\) is
1 \({10^3}\;N\)
2 \(2.5 \times {10^3}\;N\)
3 zero
4 \(1.5 \times {10^{ - 3}}\;N\)
Explanation:
Force on wire \(C\) due to wire \(D\) \({F_A} = {10^{ - 7}} \times \frac{{2 \times 30 \times 10}}{{3 \times {{10}^{ - 2}}}} \times 25 \times {10^{ - 2}} = 5 \times {10^{ - 4}}\;N\) (towards right) Force on wire \(C\) due to wire \(B\) \({F_B} = {10^{ - 7}} \times \frac{{2 \times 20 \times 10}}{{2 \times {{10}^{ - 2}}}} \times 25 \times {10^{ - 2}} = 5 \times {10^{ - 4}}\;N\) (towards left) \(\Rightarrow\) Net force on wire \(C\) is \(F_{n e t}=F_{A}-F_{B}=0\)
PHXII04:MOVING CHARGES AND MAGNETISM
362704
Two parallel wires 1\(m\) apart carry current of 1\(A\) and 3\(A\) respectively in opposite directions. The force per unit length acting between these two wires is
The force per unit length acting between two parallel wires carrying currents \(I_{1}\) and \(I_{2}\) and placed a distance \(d\) apart is \(f = \frac{{{\mu _0}{I_1}{I_2}}}{{2\pi d}}\) Here, \({I_1} = 1A,{I_2} = 3,A,d = 1\;m\) \({\mu _0} = 4\pi \times {10^{ - 7}}Tm{A^{ - 1}}\) \(\therefore f = \frac{{\left( {4\pi \times {{10}^{ - 7}}Tm{A^{ - 1}}} \right)(1\;A)(3\;A)}}{{2\pi (1\;m)}} = 6 \times {10^{ - 7}}\;N\;{m^{ - 1}}\) As the currents are in opposite directions, so \(\mathrm{f}\) is repulsive
KCET - 2015
PHXII04:MOVING CHARGES AND MAGNETISM
362705
Assertion : Two electron beams moving parallel come close each other Reason : Wires carrying current in opposite direction attracy each other
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Two parallel conductors carrying current in same direction attract each other, conductors carrying current in opposite direction repel each other.
PHXII04:MOVING CHARGES AND MAGNETISM
362706
An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current \(I\) along the same direction as shown in fig. Magnitude of force per unit length on the middle wire \(B'\) is given by :-
362702
\(A\) and \(B\) are two infinitely long straight parallel conductors. \(C\) is another straight conductor of length 1\(m\) kept parallel to A and B as shown in the figure. Then the force experienced by \(C\) is
1 Towards \(A\) equal to \(0.6 \times {10^{ - 5}}\;N\)
2 Towards \(B\) equal to \(5.4 \times {10^{ - 5}}\;N\)
3 Towards \(A\) equal to \(5.4 \times {10^{ - 5}}\;N\)
4 Towards \(B\) equal to \(0.6 \times {10^{ - 5}}\;N\)
Explanation:
Conductor A attracts conductor \(C\) towards \(A\) Force towards A \(\begin{aligned}& =\dfrac{\mu_{0}}{2 \pi} \dfrac{I_{1} I_{2} l}{r}=\dfrac{\mu_{0}}{2 \pi} \dfrac{2 \times 3 \times 1}{0.05}=\dfrac{3 \mu_{0}}{\pi} \times 20 \\& =\dfrac{60 \times 4 \pi \times 10^{-7}}{\pi}=240 \times 10^{-7} \mathrm{~N}\end{aligned}\) Again force towards \(\begin{aligned}& B=\dfrac{\mu_{0}}{2 \pi} \times \dfrac{3 \times 4 \times 1}{0.08}=\dfrac{3 \mu_{0}}{\pi} \times 25 \\& =\dfrac{3 \times 4 \pi \times 10^{-7}}{\pi} \times 25=300 \times 10^{-7} \mathrm{~N}\end{aligned}\) \(\therefore\) Resultant force towards B \(=(300-240)\) \(10^{-7} \mathrm{~N}\) \(=0.6 \times 10^{-5} \mathrm{~N}\).
KCET - 2007
PHXII04:MOVING CHARGES AND MAGNETISM
362703
Three long, straight parallel wires carrying current, are arranged as shown in figure. The force experienced by a 25\(cm\) length of wire \(C\) is
1 \({10^3}\;N\)
2 \(2.5 \times {10^3}\;N\)
3 zero
4 \(1.5 \times {10^{ - 3}}\;N\)
Explanation:
Force on wire \(C\) due to wire \(D\) \({F_A} = {10^{ - 7}} \times \frac{{2 \times 30 \times 10}}{{3 \times {{10}^{ - 2}}}} \times 25 \times {10^{ - 2}} = 5 \times {10^{ - 4}}\;N\) (towards right) Force on wire \(C\) due to wire \(B\) \({F_B} = {10^{ - 7}} \times \frac{{2 \times 20 \times 10}}{{2 \times {{10}^{ - 2}}}} \times 25 \times {10^{ - 2}} = 5 \times {10^{ - 4}}\;N\) (towards left) \(\Rightarrow\) Net force on wire \(C\) is \(F_{n e t}=F_{A}-F_{B}=0\)
PHXII04:MOVING CHARGES AND MAGNETISM
362704
Two parallel wires 1\(m\) apart carry current of 1\(A\) and 3\(A\) respectively in opposite directions. The force per unit length acting between these two wires is
The force per unit length acting between two parallel wires carrying currents \(I_{1}\) and \(I_{2}\) and placed a distance \(d\) apart is \(f = \frac{{{\mu _0}{I_1}{I_2}}}{{2\pi d}}\) Here, \({I_1} = 1A,{I_2} = 3,A,d = 1\;m\) \({\mu _0} = 4\pi \times {10^{ - 7}}Tm{A^{ - 1}}\) \(\therefore f = \frac{{\left( {4\pi \times {{10}^{ - 7}}Tm{A^{ - 1}}} \right)(1\;A)(3\;A)}}{{2\pi (1\;m)}} = 6 \times {10^{ - 7}}\;N\;{m^{ - 1}}\) As the currents are in opposite directions, so \(\mathrm{f}\) is repulsive
KCET - 2015
PHXII04:MOVING CHARGES AND MAGNETISM
362705
Assertion : Two electron beams moving parallel come close each other Reason : Wires carrying current in opposite direction attracy each other
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Two parallel conductors carrying current in same direction attract each other, conductors carrying current in opposite direction repel each other.
PHXII04:MOVING CHARGES AND MAGNETISM
362706
An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current \(I\) along the same direction as shown in fig. Magnitude of force per unit length on the middle wire \(B'\) is given by :-
362702
\(A\) and \(B\) are two infinitely long straight parallel conductors. \(C\) is another straight conductor of length 1\(m\) kept parallel to A and B as shown in the figure. Then the force experienced by \(C\) is
1 Towards \(A\) equal to \(0.6 \times {10^{ - 5}}\;N\)
2 Towards \(B\) equal to \(5.4 \times {10^{ - 5}}\;N\)
3 Towards \(A\) equal to \(5.4 \times {10^{ - 5}}\;N\)
4 Towards \(B\) equal to \(0.6 \times {10^{ - 5}}\;N\)
Explanation:
Conductor A attracts conductor \(C\) towards \(A\) Force towards A \(\begin{aligned}& =\dfrac{\mu_{0}}{2 \pi} \dfrac{I_{1} I_{2} l}{r}=\dfrac{\mu_{0}}{2 \pi} \dfrac{2 \times 3 \times 1}{0.05}=\dfrac{3 \mu_{0}}{\pi} \times 20 \\& =\dfrac{60 \times 4 \pi \times 10^{-7}}{\pi}=240 \times 10^{-7} \mathrm{~N}\end{aligned}\) Again force towards \(\begin{aligned}& B=\dfrac{\mu_{0}}{2 \pi} \times \dfrac{3 \times 4 \times 1}{0.08}=\dfrac{3 \mu_{0}}{\pi} \times 25 \\& =\dfrac{3 \times 4 \pi \times 10^{-7}}{\pi} \times 25=300 \times 10^{-7} \mathrm{~N}\end{aligned}\) \(\therefore\) Resultant force towards B \(=(300-240)\) \(10^{-7} \mathrm{~N}\) \(=0.6 \times 10^{-5} \mathrm{~N}\).
KCET - 2007
PHXII04:MOVING CHARGES AND MAGNETISM
362703
Three long, straight parallel wires carrying current, are arranged as shown in figure. The force experienced by a 25\(cm\) length of wire \(C\) is
1 \({10^3}\;N\)
2 \(2.5 \times {10^3}\;N\)
3 zero
4 \(1.5 \times {10^{ - 3}}\;N\)
Explanation:
Force on wire \(C\) due to wire \(D\) \({F_A} = {10^{ - 7}} \times \frac{{2 \times 30 \times 10}}{{3 \times {{10}^{ - 2}}}} \times 25 \times {10^{ - 2}} = 5 \times {10^{ - 4}}\;N\) (towards right) Force on wire \(C\) due to wire \(B\) \({F_B} = {10^{ - 7}} \times \frac{{2 \times 20 \times 10}}{{2 \times {{10}^{ - 2}}}} \times 25 \times {10^{ - 2}} = 5 \times {10^{ - 4}}\;N\) (towards left) \(\Rightarrow\) Net force on wire \(C\) is \(F_{n e t}=F_{A}-F_{B}=0\)
PHXII04:MOVING CHARGES AND MAGNETISM
362704
Two parallel wires 1\(m\) apart carry current of 1\(A\) and 3\(A\) respectively in opposite directions. The force per unit length acting between these two wires is
The force per unit length acting between two parallel wires carrying currents \(I_{1}\) and \(I_{2}\) and placed a distance \(d\) apart is \(f = \frac{{{\mu _0}{I_1}{I_2}}}{{2\pi d}}\) Here, \({I_1} = 1A,{I_2} = 3,A,d = 1\;m\) \({\mu _0} = 4\pi \times {10^{ - 7}}Tm{A^{ - 1}}\) \(\therefore f = \frac{{\left( {4\pi \times {{10}^{ - 7}}Tm{A^{ - 1}}} \right)(1\;A)(3\;A)}}{{2\pi (1\;m)}} = 6 \times {10^{ - 7}}\;N\;{m^{ - 1}}\) As the currents are in opposite directions, so \(\mathrm{f}\) is repulsive
KCET - 2015
PHXII04:MOVING CHARGES AND MAGNETISM
362705
Assertion : Two electron beams moving parallel come close each other Reason : Wires carrying current in opposite direction attracy each other
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Two parallel conductors carrying current in same direction attract each other, conductors carrying current in opposite direction repel each other.
PHXII04:MOVING CHARGES AND MAGNETISM
362706
An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current \(I\) along the same direction as shown in fig. Magnitude of force per unit length on the middle wire \(B'\) is given by :-
