Question:
An isosceles triangular current carrying loop is kept perpendicularly in a uniform magnetic field \(B_0\), as shown in figure. The force on the loop due to magnetic field is
An isosceles triangular current carrying loop is kept perpendicularly in a uniform magnetic field \(B_0\), as shown in figure. The force on the loop due to magnetic field is
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The net force on any closed current-carrying loop placed in a uniform magnetic field is zero, although the loop may experience torque.
Updated On: Jun 24, 2026
- \(ILB_0\cos\theta\)
- \(2ILB_0\cos\theta\)
- \(0\)
- \(ILB_0\sin\theta\)
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The Correct Option is C
Solution and Explanation
Step 1: Understand the magnetic force on a current-carrying wire.
The magnetic force on a current-carrying wire placed in a magnetic field is given by \[ \vec{F}=I(\vec{L}\times \vec{B}) \] where \(I\) is the current, \(\vec{L}\) is the length vector of the conductor, and \(\vec{B}\) is the magnetic field.
Step 2: Apply the result for a closed current loop in a uniform magnetic field.
For a closed current loop placed in a uniform magnetic field, the net magnetic force on the complete loop is always zero.
This is because the force on one part of the loop is balanced by the force on another part of the loop.
Mathematically, \[ \vec{F}_{net}=I\oint d\vec{l}\times \vec{B} \] Since \(\vec{B}\) is uniform, \[ \vec{F}_{net}=I\left(\oint d\vec{l}\right)\times \vec{B} \] For a closed loop, \[ \oint d\vec{l}=0 \] Therefore, \[ \vec{F}_{net}=I(0)\times \vec{B} \] \[ \vec{F}_{net}=0 \]
Step 3: Interpret for the triangular loop.
The given loop is triangular and closed.
Although individual sides of the triangle may experience magnetic forces, the vector sum of all these forces becomes zero because the magnetic field is uniform.
Hence, the net force on the current-carrying triangular loop is \[ 0 \]
Step 4: Final conclusion.
Therefore, the force on the loop due to the magnetic field is \[ \boxed{0} \]
The magnetic force on a current-carrying wire placed in a magnetic field is given by \[ \vec{F}=I(\vec{L}\times \vec{B}) \] where \(I\) is the current, \(\vec{L}\) is the length vector of the conductor, and \(\vec{B}\) is the magnetic field.
Step 2: Apply the result for a closed current loop in a uniform magnetic field.
For a closed current loop placed in a uniform magnetic field, the net magnetic force on the complete loop is always zero.
This is because the force on one part of the loop is balanced by the force on another part of the loop.
Mathematically, \[ \vec{F}_{net}=I\oint d\vec{l}\times \vec{B} \] Since \(\vec{B}\) is uniform, \[ \vec{F}_{net}=I\left(\oint d\vec{l}\right)\times \vec{B} \] For a closed loop, \[ \oint d\vec{l}=0 \] Therefore, \[ \vec{F}_{net}=I(0)\times \vec{B} \] \[ \vec{F}_{net}=0 \]
Step 3: Interpret for the triangular loop.
The given loop is triangular and closed.
Although individual sides of the triangle may experience magnetic forces, the vector sum of all these forces becomes zero because the magnetic field is uniform.
Hence, the net force on the current-carrying triangular loop is \[ 0 \]
Step 4: Final conclusion.
Therefore, the force on the loop due to the magnetic field is \[ \boxed{0} \]
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