Question

A straight wire of length 2 m carries a current of 10 A. If this wire is placed in a uniform magnetic field of 0.15 T making an angle of $ 45^\circ $ with the magnetic field, the applied force on the wire will be

• A. 1.5 N
• B. 3 N
• C. \( 3\sqrt{2} \) N
• D. \( 3/\sqrt{2} \) N

Hint:

When the angle between the wire and the magnetic field is \( 45^\circ \), use \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \) to calculate the force accurately.

Answer 1

The Correct Option is D

A straight wire carrying a current experiences a force when placed in a magnetic field due to the Lorentz force. The magnitude of this force can be calculated using the formula:

\( F = I \cdot L \cdot B \cdot \sin(\theta) \)

Where:

\( F \) = force on the wire (in Newtons)

\( I \) = current flowing through the wire (in Amperes)

\( L \) = length of the wire (in meters)

\( B \) = magnetic field strength (in Tesla)

\( \theta \) = angle between the wire and magnetic field (in degrees)

The given values are:

• \( I = 10 \, \text{A} \)
• \( L = 2 \, \text{m} \)
• \( B = 0.15 \, \text{T} \)
• \( \theta = 45^\circ \)

Plug these values into the formula:

\( F = 10 \cdot 2 \cdot 0.15 \cdot \sin(45^\circ) \)

Knowing that \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\), we find:

\( F = 10 \cdot 2 \cdot 0.15 \cdot \frac{\sqrt{2}}{2} \)

This simplifies to:

\( F = 3 \cdot \sqrt{2} \cdot \frac{1}{2} \)

\( F = \frac{3\sqrt{2}}{2} \)

Therefore, the force on the wire is:

\( \frac{3}{\sqrt{2}} \, \text{N} \)

Answer 2

The force on a current-carrying wire placed in a magnetic field is given by the formula: \[ F = BIL \sin \theta \] where:
- \( B \) is the magnetic field strength,
- \( I \) is the current in the wire,
- \( L \) is the length of the wire,
- \( \theta \) is the angle between the wire and the magnetic field. Given:
- \( B = 0.15 \, \text{T} \),
- \( I = 10 \, \text{A} \),
- \( L = 2 \, \text{m} \),
- \( \theta = 45^\circ \), Substitute these values into the formula: \[ F = (0.15)(10)(2) \sin(45^\circ) \] Since \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \), we get: \[ F = (0.15)(10)(2) \times \frac{\sqrt{2}}{2} \] \[ F = 3/\sqrt{2} \, \text{N} \]
Thus, the correct answer is: \[ \text{(4) } \frac{3}{\sqrt{2}} \, \text{N} \]