A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b ) and DA (radius =a ) of the loop are joined by two straight wires AB and CD . A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30^ \circ . Another straight thin wire steady current I_1 flowing out of the plane of the paper is kept at the origin. Due to the presence of the current I_1 at the origin:

The forces on AD are BC are zero.

The magnitude of the net force on the loop is given by I_1I \over 4\pi \mu _0[ 2( b - a ) + \frac\pi 3( a + b ) ].

The magnitude of the net force on the loop is given by \mu _0II_1 \over 24ab( b - a ).

The forces on AB and DC are zero.

Magnetics PYQ — JEE Main 2009 · Shift 0 · Q57 with Solution

View written solutionFree

Correct answer: A

Magnetic field due to the wire at the origin

A long straight wire carrying current $I_1$ out of the plane produces magnetic field lines that are concentric circles centered at $O$ .

At a distance $r$ from the origin,

B(r)=\frac{\mu_0 I_1}{2\pi r}

and the direction is tangential to the circle centered at $O$ (anticlockwise by right-hand rule).

Force on a current element

For a small element $d\vec l$ of the loop carrying current $I$ ,

d\vec F = I\, d\vec l \times \vec B.

We now examine each part of the loop.

Force on the circular arcs $AD$ and $BC$

Both $AD$ and $BC$ are arcs centered at $O$ .

Along an arc centered at $O$ , the current element $d\vec l$ is tangential to the circle.
The magnetic field $\vec B$ due to the central wire is also tangential to the same circle.

Hence, for every point on each arc,

d\vec l \parallel \vec B

or antiparallel, so

d\vec l \times \vec B = 0.

Therefore,

\vec F_{AD}=0, \qquad \vec F_{BC}=0.

So Option A is correct.

Force on the straight segments $AB$ and $CD$

The straight wires $AB$ and $CD$ lie along radial lines from the origin because they join the two concentric arcs at fixed angular positions.

Thus, along these segments:

$d\vec l$ is radial,
$\vec B$ is tangential.

Therefore,

d\vec l \perp \vec B,

so the force is not zero in general.

Its magnitude on a radial segment from $r=a$ to $r=b$ is

F = I\int_a^b B(r)\,dr = I\int_a^b \frac{\mu_0 I_1}{2\pi r}\,dr = \frac{\mu_0 I I_1}{2\pi}\ln\!\left(\frac{b}{a}\right).

So each straight segment experiences a nonzero force. Hence Option D is false.

Also, since options B and C give specific net-force expressions inconsistent with the above logarithmic dependence, they are false.

Check the remaining options

A: True, because force on each arc is zero.
B: False; net force does not have that form.
C: False; dimensions/form are incorrect, and actual force on radial segments involves $\ln(b/a)$ .
D: False; forces on $AB$ and $CD$ are not zero.

Final answer

The correct option is:

\boxed{A}

Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.

More from Magnetics

View written solutionFree

Correct answer: A

Magnetic field due to the wire at the origin

A long straight wire carrying current $I_1$ out of the plane produces magnetic field lines that are concentric circles centered at $O$ .

At a distance $r$ from the origin,

B(r)=\frac{\mu_0 I_1}{2\pi r}

and the direction is tangential to the circle centered at $O$ (anticlockwise by right-hand rule).

Force on a current element

For a small element $d\vec l$ of the loop carrying current $I$ ,

d\vec F = I\, d\vec l \times \vec B.

We now examine each part of the loop.

Force on the circular arcs $AD$ and $BC$

Both $AD$ and $BC$ are arcs centered at $O$ .

Along an arc centered at $O$ , the current element $d\vec l$ is tangential to the circle.
The magnetic field $\vec B$ due to the central wire is also tangential to the same circle.

Hence, for every point on each arc,

d\vec l \parallel \vec B

or antiparallel, so

d\vec l \times \vec B = 0.

Therefore,

\vec F_{AD}=0, \qquad \vec F_{BC}=0.

So Option A is correct.

Force on the straight segments $AB$ and $CD$

The straight wires $AB$ and $CD$ lie along radial lines from the origin because they join the two concentric arcs at fixed angular positions.

Thus, along these segments:

$d\vec l$ is radial,
$\vec B$ is tangential.

Therefore,

d\vec l \perp \vec B,

so the force is not zero in general.

Its magnitude on a radial segment from $r=a$ to $r=b$ is

F = I\int_a^b B(r)\,dr = I\int_a^b \frac{\mu_0 I_1}{2\pi r}\,dr = \frac{\mu_0 I I_1}{2\pi}\ln\!\left(\frac{b}{a}\right).

So each straight segment experiences a nonzero force. Hence Option D is false.

Also, since options B and C give specific net-force expressions inconsistent with the above logarithmic dependence, they are false.

Check the remaining options

A: True, because force on each arc is zero.
B: False; net force does not have that form.
C: False; dimensions/form are incorrect, and actual force on radial segments involves $\ln(b/a)$ .
D: False; forces on $AB$ and $CD$ are not zero.

Final answer

The correct option is:

\boxed{A}

Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.

Magnetics question

Magnetics question

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Magnetics question

Magnetics question

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