A uniform magnetic field of \( 2 \times 10^{-3} \, \text{T} \) acts along positive Y-direction. A rectangular loop of sides 20 cm and 10 cm with current of 5 A is in the Y-Z plane. The current is in anticlockwise sense with reference to the negative X-axis. Magnitude and direction of the torque is:
- \( 2 \times 10^{-4} \, \text{N} \, \text{m} \) along positive Z-direction
- \( 2 \times 10^{-4} \, \text{N} \, \text{m} \) along negative Z-direction
- \( 2 \times 10^{-4} \, \text{N} \, \text{m} \) along positive X-direction
- \( 2 \times 10^{-4} \, \text{N} \, \text{m} \) along positive Y-direction
The Correct Option is B
Approach Solution - 1
The problem involves calculating the torque on a current-carrying rectangular loop placed in a magnetic field. To solve this, we utilize the formula for torque \((\tau)\) on a current-carrying loop in a magnetic field, given by:
\(\tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta)\)
where:
- \(n\) is the number of loops (here it is 1).
- \(I\) is the current flowing through the loop (5 A).
- \(A\) is the area of the loop.
- \(B\) is the magnetic field (given as \(2 \times 10^{-3} \, \text{T}\)).
- \(\theta\) is the angle between the normal to the loop and the magnetic field direction.
The dimensions of the loop are given: 20 cm and 10 cm. Convert these to meters to calculate the area:
\(A = 0.2 \, \text{m} \times 0.1 \, \text{m} = 0.02 \, \text{m}^2\)
The loop is in the Y-Z plane, and the magnetic field is along the positive Y-direction. Therefore, the angle \(\theta = 90^\circ\) since the normal to the Y-Z plane is along the X-axis, which is perpendicular to the Y-direction.
Substituting the values into the torque formula:
\(\tau = 1 \times 5 \, \text{A} \times 0.02 \, \text{m}^2 \times 2 \times 10^{-3} \, \text{T} \times \sin(90^\circ)\)
\(\tau = 2 \times 10^{-4} \, \text{N} \cdot \text{m}\)
Now, to determine the direction of the torque, we use the right-hand rule for the cross product \(\mathbf{A} \times \mathbf{B}\). The fingers point in the direction of the area vector (normal to the loop, along the X-axis), and when curled towards the magnetic field (Y-direction), the thumb points in the result direction, which is the torque direction.
Given the current's anticlockwise direction concerning the negative X-axis, the torque direction will be opposite to the direction of the conventional curl from Y to Z direction, i.e., along the negative Z-direction.
Therefore, the correct answer is \(2 \times 10^{-4} \, \text{N} \, \text{m}\) along negative Z-direction.
Approach Solution -2
The magnetic moment \( \vec{M} \) is given by:
\[ \vec{M} = I \vec{A}. \]With \( I = 5 \, \text{A} \), \( A = 0.2 \times 0.1 = 0.02 \, \text{m}^2 \), and \( \vec{A} = 0.02 \, \hat{i} \), we get:
\[ \vec{M} = 5 \times 0.02 \, \hat{i} = 0.1 \, \hat{i}. \]The torque \( \vec{\tau} \) is given by:
\[ \vec{\tau} = \vec{M} \times \vec{B}. \]Substituting \( \vec{B} = 2 \times 10^{-3} \, \hat{j} \):
\[ \vec{\tau} = 0.1 \, \hat{i} \times 2 \times 10^{-3} \, \hat{j} = 2 \times 10^{-4} \, (-\hat{k}) = 2 \times 10^{-4} \, \text{Nm}. \]Therefore, the answer is:
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