The magnetic field existing in a region is given by \vecB=0.2(1+2 x) \hatk . A square loop of edge 50 ~cm carrying 0.5 A current is placed in x - y plane with its edges parallel to the x - y axes, as shown in figure. The magnitude of the net magnetic force experienced by the loop is \underline\hspace2cm mN .

Magnetics PYQ — JEE Main 2024 · 4 Apr · Shift 1 · Q86 with Solution

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Correct answer: 50

Given magnetic field

The field is

\vec B = 0.2(1+2x)\,\hat k

So the magnetic field is along the $z$ -direction and varies with $x$ .

Loop details

Square loop side:

a=50\text{ cm}=0.5\text{ m}

Current:

I=0.5\text{ A}

The loop lies in the $xy$ -plane, with sides parallel to the axes.

Magnetic force on a current element

Force on a wire segment is

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

Since $\vec B$ depends only on $x$ , the two vertical sides (parallel to $y$ -axis) experience forces of different magnitudes because they are at different $x$ positions.

The two horizontal sides (parallel to $x$ -axis) experience equal and opposite forces, so they cancel out.

Thus, net force comes only from the two sides parallel to the $y$ -axis.

Force on a vertical side at position $x$

For a side of length $a$ parallel to $y$ -axis, $B$ is constant along that side because $x$ is constant there.

Hence magnitude of force on that side is

F = I a B(x)

If the left side is at $x=x_1$ and the right side at $x=x_2=x_1+a$ , then net force magnitude is

F_{\text{net}} = I a\,[B(x_2)-B(x_1)]

Now,

B(x)=0.2(1+2x)

So,

B(x_2)-B(x_1)=0.2\big[(1+2x_2)-(1+2x_1)\big]

=0.2\cdot 2(x_2-x_1)

=0.4a

Since $a=0.5$ m,

B(x_2)-B(x_1)=0.4\times 0.5=0.2\text{ T}

Therefore,

F_{\text{net}} = I a (0.2)

=0.5\times 0.5\times 0.2

=0.05\text{ N}

Convert to mN

0.05\text{ N}=50\text{ mN}

Final answer

The magnitude of the net magnetic force on the loop is

\boxed{50\text{ mN}}

More from Magnetics

View written solutionFree

Correct answer: 50

Given magnetic field

The field is

\vec B = 0.2(1+2x)\,\hat k

So the magnetic field is along the $z$ -direction and varies with $x$ .

Loop details

Square loop side:

a=50\text{ cm}=0.5\text{ m}

Current:

I=0.5\text{ A}

The loop lies in the $xy$ -plane, with sides parallel to the axes.

Magnetic force on a current element

Force on a wire segment is

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

Since $\vec B$ depends only on $x$ , the two vertical sides (parallel to $y$ -axis) experience forces of different magnitudes because they are at different $x$ positions.

The two horizontal sides (parallel to $x$ -axis) experience equal and opposite forces, so they cancel out.

Thus, net force comes only from the two sides parallel to the $y$ -axis.

Force on a vertical side at position $x$

For a side of length $a$ parallel to $y$ -axis, $B$ is constant along that side because $x$ is constant there.

Hence magnitude of force on that side is

F = I a B(x)

If the left side is at $x=x_1$ and the right side at $x=x_2=x_1+a$ , then net force magnitude is

F_{\text{net}} = I a\,[B(x_2)-B(x_1)]

Now,

B(x)=0.2(1+2x)

So,

B(x_2)-B(x_1)=0.2\big[(1+2x_2)-(1+2x_1)\big]

=0.2\cdot 2(x_2-x_1)

=0.4a

Since $a=0.5$ m,

B(x_2)-B(x_1)=0.4\times 0.5=0.2\text{ T}

Therefore,

F_{\text{net}} = I a (0.2)

=0.5\times 0.5\times 0.2

=0.05\text{ N}

Convert to mN

0.05\text{ N}=50\text{ mN}

Final answer

The magnitude of the net magnetic force on the loop is

\boxed{50\text{ mN}}

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