Question:medium

As shown in the figure, a uniform straight wire of length $30\sqrt{3}$ cm is bent in the form of an equilateral triangle ABC. A uniform magnetic field 2T is applied parallel to the side BC. If the current through the wire is 2A, the magnitude of the force on the side AC is

Show Hint

The magnetic force on a straight current-carrying wire segment is $\vec{F} = I(\vec{l} \times \vec{B})$. The magnitude $F = IlB\sin\theta$ is the key formula. In a closed loop in a uniform field, the net force is zero, but the force on an individual segment can be non-zero.

Updated On: Mar 30, 2026

$\frac{2}{\sqrt{3}}$ N
$0.2\sqrt{3}$ N
$1.2$ N
$0.6$ N

Show Solution

The Correct Option is D

Solution and Explanation

Was this answer helpful?

0

Top Questions on Moving charges and magnetism

Questions Asked in TS EAMCET exam

If $\omega \neq 1$ is a cube root of unity, then one root among the $7^{th}$ roots of $(1+\omega)$ is

TS EAMCET - 2025
Complex numbers

If $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 3 & 4 \\ 3 & 2 & 2 \\ 2 & 4 & 2 \end{pmatrix}$, then $\sqrt{|\text{Adj}(AB)|} =$

TS EAMCET - 2025
Matrices and Determinants

If \( p_1, p_2, p_3 \) are the altitudes and \( a=4, b=5, c=6 \) are the sides of a triangle ABC, then \( \frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2} = \)

TS EAMCET - 2025
Geometry

If \(\alpha, \beta, \gamma, \delta, \epsilon\) are the roots of the equation \(x^5 + x^4 - 13x^3 - 13x^2 + 36x + 36 = 0\) and \(\alpha<\beta<\gamma<\delta<\epsilon\) then \( \frac{\epsilon}{\alpha} + \frac{\delta}{\beta} + \frac{1}{\gamma} = \)

TS EAMCET - 2025
Algebra