Why do we use $\sin\theta$ instead of $\cos\theta$?

Torque is defined by the cross product $\vec{\tau} = \vec{M} \times \vec{B}$. The magnitude of a cross product inherently involves the sine of the angle between the two vectors.

JEE Main 2019

LEVELJEE Main

Animated Solution for Physics - Magnetic Effects of Current: A rectangular coil (dimension $5 cm \times 2.5 cm$ ) with 100 turns, carrying a current of 3A in the clockwise direction, is kept centred at the origin and in the X-Z plane. A magnetic field of 1 T is applied along X-axis. If the coil is tilted through $4 5^{\circ}$ about Z-axis, then the torque on the coil is

Select Answer:

Visualized Solution

The Physical Setup

Coil dimensions: $5 cm \times 2.5 cm$
Turns: $N = 100$
Current: $I = 3 A$
Magnetic Field: $B = 1 T \hat{i}$

Area Vector and Angle

Area vector $A$ is perpendicular to the plane of the coil.
Angle between $A$ and $B$ is $θ = 4 5^{\circ}$ .

Torque Formula

Magnetic Moment: $M = N I A$
Torque: $τ = ∣ M \times B ∣ = M B sin θ$
$τ = N I A B sin θ$

Calculating Area

$A = 5 cm \times 2.5 cm$
$A = 12.5 cm^{2}$
$A = 12.5 \times 1 0^{- 4} m^{2}$

Substitution

$τ = (100) \times (3) \times (12.5 \times 1 0^{- 4}) \times (1) \times sin (4 5^{\circ})$

Final Calculation

$sin (4 5^{\circ}) = \frac{1}{2} \approx 0.707$
$τ = 300 \times 12.5 \times 1 0^{- 4} \times 0.707$
$τ = 3750 \times 1 0^{- 4} \times 0.707$
$τ \approx 0.265 N-m \approx 0.27 N-m$

The Way Forward

What if the coil was tilted about the X-axis instead?
The angle $θ$ would remain $9 0^{\circ}$ , and torque would be maximum!

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The Sigma Insight: Magnetic Moment of Current Loop

Solution Diagram

Analyzing the Setup

Imagine a rectangular coil perfectly aligned in the X-Z plane. The dimensions are given as

5 cm

2.5 cm

, and it consists of

100

turns of wire. A current of

3 A

flows through it in a clockwise direction.

Suddenly, a uniform magnetic field of

1 T

is switched on, pointing directly along the positive X-axis. If the coil remained in the X-Z plane, its area vector (pointing along the Y-axis) would be perfectly perpendicular to the magnetic field, resulting in maximum torque. However, the problem states that the coil is tilted by

4 5^{\circ}

about the Z-axis.

The Master Equation

To find the torque on a current-carrying loop in a magnetic field, we use the fundamental cross-product relationship:

τ = M \times B

Here,

M

is the magnetic dipole moment of the coil, defined as

M = N I A

, where

N

is the number of turns,

I

is the current, and

A

is the area vector.

The magnitude of the torque can be written as:

τ = N I A B sin θ

where

θ

is the angle between the area vector

A

and the magnetic field

B

Finding the Angle and Area

Because the coil is tilted by

4 5^{\circ}

about the Z-axis, its area vector (which was originally along the Y-axis) also tilts by

4 5^{\circ}

. Since the magnetic field is along the X-axis, the angle

θ

between the area vector and the magnetic field becomes exactly

4 5^{\circ}

Before substituting the values, let's calculate the area of the coil in standard SI units to avoid any silly mistakes:

A = 5 cm \times 2.5 cm = 12.5 cm^{2}

A = 12.5 \times 1 0^{- 4} m^{2}

Final Calculation

Now, we substitute all our known values into the master equation:

τ = (100) \times (3 A) \times (12.5 \times 1 0^{- 4} m^{2}) \times (1 T) \times sin (4 5^{\circ})

We know that

sin (4 5^{\circ}) = \frac{1}{2} \approx 0.707

. Let's compute the numerical part:

τ = 300 \times 12.5 \times 1 0^{- 4} \times 0.707

τ = 3750 \times 1 0^{- 4} \times 0.707

τ \approx 0.375 \times 0.707 \approx 0.265 N-m

Rounding off to two decimal places, we get our final answer:

$τ \approx 0.27 N-m$

Common Traps & FAQs

Why is the area vector initially along the Y-axis?

The area vector is always perpendicular to the plane of the coil. Since the coil lies in the X-Z plane, the only direction perpendicular to both the X and Z axes is the Y-axis.

Does the direction of the current matter for the magnitude of the torque?

No, the direction of the current (clockwise vs. counterclockwise) flips the area vector by

18 0^{\circ}

. This changes the sign of the torque vector, but the magnitude

∣ τ ∣ = N I A B sin θ

remains exactly the same.

Why do we use

sin θ

instead of

cos θ

Torque is defined by the cross product

τ = M \times B

. The magnitude of a cross product inherently involves the sine of the angle between the two vectors.

What if the coil was tilted about the X-axis instead?

If tilted about the X-axis, the area vector would rotate in the Y-Z plane. It would remain perfectly perpendicular to the magnetic field (which is along the X-axis), meaning

θ

would stay

9 0^{\circ}

, and the torque would be maximum.

Why did we convert the area to square meters?

To get the final torque in standard SI units (Newton-meters), all input variables must be in SI units. Since

1 cm = 1 0^{- 2} m

1 cm^{2} = 1 0^{- 4} m^{2}

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