A wire of length $l$ is carrying a current $i$ is in uniform magnetic field $B$ perpendicular to wire. Torque due to magnetic field about its one end is :

\frac{i l^{2} B}{3}

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i l^{2} B

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\frac{i l^{2} B}{2}

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2 i l^{2} B

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Solution

The correct option is C $\frac{i l^{2} B}{2}$
$d F = B i . d x$
$\int d τ_{A} = \int_{0}^{l} B i . d x . x$
$τ_{A} = \frac{B i x^{2}}{2}$
$τ_{A} = \frac{B i l^{2}}{2}$

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A wire of length l is carrying a current i is in uniform magnetic field B perpendicular to wire. Torque due to magnetic field about its one end is :

The correct option is C il2B2dF=Bi.dx∫dτA=∫l0Bi.dx.xτA=Bix22τA=Bil22

A wire of length $l$ is carrying a current $i$ is in uniform magnetic field $B$ perpendicular to wire. Torque due to magnetic field about its one end is :

The correct option is C $\frac{i l^{2} B}{2}$
$d F = B i . d x$
$\int d τ_{A} = \int_{0}^{l} B i . d x . x$
$τ_{A} = \frac{B i x^{2}}{2}$
$τ_{A} = \frac{B i l^{2}}{2}$