SATHEE: Moving Charges and Magnetism

Moving Charges and Magnetism - Result Question 59

60. A square loop $A B C D$ carrying a current $i$ , is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be :

[2016]

(a) $\frac{2 μ_{0} I i}{3 π}$

(b) $\frac{μ_{0} I i}{2 π}$

(d) $\frac{μ_{0} I i L}{2 π}$

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Answer:

Correct Answer: 60. (a)

Solution:

(a) The direction of current in conductor

$X Y$ and $A B$ is same

$∴ F_{A B} = i ℓ B$ (attractive)

$F_{A B} = i (L) \cdot \frac{μ_{0} I}{2 π (\frac{L}{2})} (a r r o w) = \frac{μ_{0} i I}{π} (a r r o w)$

$F_{B C}$ opposite to $F_{A D}$

$F_{B C} (↑)$ and $F_{A D} (↓)$

$\Rightarrow$ cancels each other

$F_{C D} = i ℓ B$ (repulsive)

$F_{C D} = i (L) \frac{μ_{0} I}{2 π (\frac{3 L}{2})} (a r r o w) = \frac{μ_{0} i I}{3 π} (a r r o w)$

Therefore the net force on the loop

$F_{net} = F_{A B} + F_{B C} + F_{C D} + F_{A D}$

$\Rightarrow F_{net} = \frac{μ_{o} i I}{π} - \frac{μ_{o} i I}{3 π} = \frac{2 μ_{o} i I}{3 π}$

Moving Charges and Magnetism - Result Question 59

60. A square loop $A B C D$ carrying a current $i$ , is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be :

Answer:

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Moving Charges and Magnetism - Result Question 59

60. A square loop ABCD carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be :

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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60. A square loop $A B C D$ carrying a current $i$ , is placed near and coplanar with a long straight conductor XY carrying a current I, the net force on the loop will be :