SATHEE: Magnetics 6 Question 16

Magnetics 6 Question 16

16. Two infinitely long straight wires lie in the $x y$ -plane along the lines $x = \pm R$ . The wire located at $x = + R$ carries a constant current $I_{1}$ and the wire located at $x = - R$ carries a constant current $I_{2}$ . A circular loop of radius $R$ is suspended with its centre at $(0, 0, \sqrt{3} R)$ and in a plane parallel to the $x y$ -plane. This loop carries a constant current $I$ in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive, if it is in the $+ \hat{j}$ -direction. Which of the following statements regarding the magnetic field $B$ is (are) true?

(2018 Adv.)

(a) If $I_{1} = I_{2}$ , then $B$ cannot be equal to zero at the origin $(0, 0, 0)$

(b) If $I_{1} > 0$ and $I_{2} < 0$ , then $B$ can be equal to zero at the origin $(0, 0, 0)$

(d) If $I_{1} = I_{2}$ , then the $z$ -component of the magnetic field at the centre of the loop is $- \frac{μ_{o} I}{2 R}$

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Solution:

(a) At origin, $B = 0$ due to two wires if $I_{1} = I_{2}$ , hence $(B_{net})$ at origin is equal to $B$ due to ring. which is non-zero.

(b) If $I_{1} > 0$ and $I_{2} < 0, B$ at origin due to wires will be along $+ \hat{k}$ . Direction of $B$ due to ring is along $- \hat{k}$ direction and hence $B$ can be zero at origin.

(c) If $I_{1} < 0$ and $I_{2} > 0$ , B at origin due to wires is along $- \hat{k}$ and also along $- \hat{k}$ due to ring, hence $B$ cannot be zero.

(d)

At centre of ring, $B$ due to wires is along $x$ -axis.

Hence, $z$ -component is only because of ring which $B = \frac{μ_{0} i}{2 R} (- \hat{k})$ .

Magnetics 6 Question 16

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Magnetics 6 Question 16

Solution:

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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