A very long wire ABDMNDC is shown in figure carrying current I. AB and BC parts are straight, long a

Q: Why is the angle $\phi_2$ for wire ABN negative?

Because the wire segment ends at B, which is before the perpendicular point N. Both ends of the wire are on the same side of the perpendicular, so one angle must be negative.

Q: Why is the circular loop considered a full circle?

The text states "circular turn DMND", which implies the wire completes a full $360^\circ$ loop before continuing along the straight path.

Q: How do we determine the direction of the magnetic field?

Using the right-hand rule: point your thumb in the direction of the current, and your curling fingers will show the direction of the magnetic field.

Q: Why do the straight wires not cancel each other out?

Although they carry the same current, their geometry relative to the center O is different, and they produce magnetic fields in opposite directions with different magnitudes.

Q: What if the current flowed in the opposite direction?

The magnitude of the net magnetic field would remain exactly the same, but its direction would be reversed (into the page instead of out of the page).

Question

A very long wire ABDMNDC is shown in figure carrying current $I$. $AB$ and $BC$ parts are straight, long and at right angle. At $D$ wire forms a circular turn $DMND$ of radius $R$. $AB$, $BC$ parts are tangential to circular turn at $N$ and $D$. Magnetic field at the centre of circle is

Options:

(A) $\frac{\mu_0 I}{2\pi R}\left(\pi + \frac{1}{\sqrt{2}}\right)$

(B) $\frac{\mu_0 I}{2R}$

(C) $\frac{\mu_0 I}{2\pi R}(\pi + 1)$

(D) $\frac{\mu_0 I}{2\pi R}\left(\pi - \frac{1}{\sqrt{2}}\right)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

When faced with a complex current-carrying wire, the best strategy is to break it down into simpler, manageable segments. In this problem, the wire $ABDMNDC$ can be decomposed into three distinct parts: the straight wire segment $ABN$, the straight wire segment $BDC$, and the full circular loop $DMND$.

Let's establish a coordinate system to make things crystal clear. We place the corner $B$ at the origin $(0,0)$. The center of the circular loop, $O$, is located at $(R, R)$. The circle is tangent to the y-axis at $N(0, R)$ and tangent to the x-axis at $D(R, 0)$.

### The First Straight Wire: ABN

The wire comes from infinity along the y-axis, passing through $A$ and ending at the corner $B(0,0)$. To find the magnetic field $B_1$ at the center $O(R,R)$, we drop a perpendicular from $O$ to the y-axis, which meets at $N(0,R)$.

The segment goes from $y = \infty$ to $y = 0$. Notice carefully that both ends of this segment lie on the *same side* of the perpendicular point $N$. Therefore, the angles subtended at $O$ are $\phi_1 = 90^\circ$ (towards infinity) and $\phi_2 = -45^\circ$ (towards the origin $B$).

Using the Biot-Savart law for a finite straight wire:
$$B_1 = \frac{\mu_0 I}{4\pi R}(\sin\phi_1 + \sin\phi_2)$$
$$B_1 = \frac{\mu_0 I}{4\pi R}[\sin 90^\circ + \sin(-45^\circ)] = \frac{\mu_0 I}{4\pi R}\left(1 - \frac{1}{\sqrt{2}}\right)$$
By the right-hand rule, the current flows downwards, so the magnetic field at $O$ points **into the page** ($\otimes$).

### The Second Straight Wire: BDC

This segment starts at the corner $B(0,0)$ and extends to infinity along the x-axis. The perpendicular from $O(R,R)$ to the x-axis meets at $D(R,0)$.

This segment goes from $x = 0$ to $x = \infty$, crossing the perpendicular point $D$. Thus, the angles are on opposite sides of the perpendicular: $\phi_1 = 45^\circ$ (towards $B$) and $\phi_2 = 90^\circ$ (towards infinity).

$$B_2 = \frac{\mu_0 I}{4\pi R}(\sin 45^\circ + \sin 90^\circ) = \frac{\mu_0 I}{4\pi R}\left(\frac{1}{\sqrt{2}} + 1\right)$$
By the right-hand rule, the current flows rightwards, so the magnetic field at $O$ points **out of the page** ($\odot$).

### The Circular Loop: DMND

The text specifies a "circular turn DMND", which implies the wire completes a full $360^\circ$ loop. The current flows in a counter-clockwise direction.

The magnetic field at the center of a full circular loop is simply:
$$B_3 = \frac{\mu_0 I}{2R}$$
By the right-hand rule, the magnetic field points **out of the page** ($\odot$).

### Final Calculation

To find the net magnetic field at $O$, we sum the three contributions vectorially. Let's take the outward direction ($\odot$) as positive.

$$B_{net} = -B_1 + B_2 + B_3$$
$$B_{net} = -\frac{\mu_0 I}{4\pi R}\left(1 - \frac{1}{\sqrt{2}}\right) + \frac{\mu_0 I}{4\pi R}\left(\frac{1}{\sqrt{2}} + 1\right) + \frac{\mu_0 I}{2R}$$

Notice the beautiful cancellation that happens here. The $-1$ and $+1$ terms from the straight wires cancel each other out, while the $\frac{1}{\sqrt{2}}$ terms add up:

$$B_{net} = \frac{\mu_0 I}{4\pi R} \left( \frac{2}{\sqrt{2}} \right) + \frac{\mu_0 I}{2R} = \frac{\mu_0 I}{2\sqrt{2}\pi R} + \frac{\mu_0 I}{2R}$$

Factoring out the common terms, we arrive at our final elegant expression:

**$$B_{net} = \frac{\mu_0 I}{2\pi R}\left(\pi + \frac{1}{\sqrt{2}}\right)$$**

Select Answer:

Visualized Solution

$B_{n e t}$

$B_{1}$

$ϕ_{1}, ϕ_{2}$

$B_{1}$

$B_{2}$

$ϕ_{1}, ϕ_{2}$

$B_{2}$

$B_{3}$

$B_{n e t}$

$B_{n e t}$

The Sigma Insight: Biot-Savart Law

Analyzing the Setup

The First Straight Wire

The Second Straight Wire

The Circular Loop

Final Calculation

Similar Questions

A hairpin like shape as shown in figure is made by bending a long current carrying wire. What is the magnitude of a magnetic field at point $P$ which lies on the centre of the semicircle?

A non-planar loop of conducting wire carrying a current $I$ is placed as shown in the figure. Each of the straight sections of the loop is of length $2 a$ . The magnetic field due to this loop at the point $P (a, 0, a)$ points in the direction

A current loop, having two circular arcs joined by two radial lines as shown in the figure. It carries a current of $10 A$ . The magnetic field at point $O$ will be close to

A long straight wire along the z-axis carries a current $I$ in the negative z-direction. The magnetic vector field $B$ at a point having coordinate $(x, y)$ on the $z = 0$ plane is

A coil having $N$ turns is wound tightly in the form of a spiral with inner and outer radii $a$ and $b$ respectively. When a current $I$ passes through the coil, the magnetic field at the centre is

A coil having $N$ turns is wound tightly in the form of a spiral with inner and outer radii $a$ and $b$ , respectively. Find the magnetic field at centre, when a current $I$ passes through coil

A long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is $B$ . It is then bent into a circular loop of $n$ turns. The magnetic field at the centre of the coil will be

Find the magnetic field at point $P$ due to a straight line segment $A B$ of length $6 cm$ carrying a current of $5 A$ (See figure). (Take, $μ_{0} = 4 π \times 1 0^{- 7} N-A^{- 2}$ )

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bnet​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ϕ1​,ϕ2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ϕ1​,ϕ2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bnet​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bnet​

The Sigma Insight: Biot-Savart Law

Analyzing the Setup

The First Straight Wire

The Second Straight Wire

The Circular Loop

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A current loop, having two circular arcs joined by two radial lines as shown in the figure. It carries a current of 10 A. The magnetic field at point O will be close to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A long straight wire along the z-axis carries a current I in the negative z-direction. The magnetic vector field B at a point having coordinate (x,y) on the z=0 plane is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5 A (See figure). (Take, μ0​=4π×10−7 N-A−2)

$B_{n e t}$

$B_{1}$

$ϕ_{1}, ϕ_{2}$

$B_{1}$

$B_{2}$

$ϕ_{1}, ϕ_{2}$

$B_{2}$

$B_{3}$

$B_{n e t}$

$B_{n e t}$

A current loop, having two circular arcs joined by two radial lines as shown in the figure. It carries a current of $10 A$ . The magnetic field at point $O$ will be close to

A long straight wire along the z-axis carries a current $I$ in the negative z-direction. The magnetic vector field $B$ at a point having coordinate $(x, y)$ on the $z = 0$ plane is

Find the magnetic field at point $P$ due to a straight line segment $A B$ of length $6 cm$ carrying a current of $5 A$ (See figure). (Take, $μ_{0} = 4 π \times 1 0^{- 7} N-A^{- 2}$ )