A uniform magnetic field of 2 10^ -3 ~T acts along positive Y-dir… - Vidyadip

MCQ

A uniform magnetic field of $2 \times 10^{-3} \mathrm{~T}$ acts along positive $\mathrm{Y}$-direction. A rectangular loop of sides $20$ $\mathrm{cm}$ and $10 \mathrm{~cm}$ with current of $5 \mathrm{~A}$ is $\mathrm{Y}-\mathrm{Z}$ plane. The current is in anticlockwise sense with reference to negative $\mathrm{X}$ axis. Magnitude and direction of the torque is :

A
$2 \times 10^{-4} \mathrm{~N}-\mathrm{m}$ along positive $\mathrm{Z}$-direction
✓
$2 \times 10^{-4} \mathrm{~N}-\mathrm{m}$ along negative $Z$-direction
C
$2 \times 10^{-4} \mathrm{~N}-\mathrm{m}$ along positive $\mathrm{X}$-direction
D
$2 \times 10^{-4} \mathrm{~N}-\mathrm{m}$ along positive $\mathrm{Y}$-direction

✓

Answer

Correct option: B.

$2 \times 10^{-4} \mathrm{~N}-\mathrm{m}$ along negative $Z$-direction

b
$\overrightarrow{\mathrm{M}}=\mathrm{iA}$

$=5 \times(0.2) \times(0.1)(-\hat{\mathrm{i}})$

$=0.1(-\hat{\mathrm{i}})$

$\vec{\tau}=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}=0.1(-\hat{\mathrm{i}}) \times\left(2 \times 10^{-3}\right)(\hat{\mathrm{j}})$

$=2 \times 10^{-4}(-\hat{\mathrm{k}}) \mathrm{N}-\mathrm{m}$

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