Question

A square loop of uniform conducting wire is as shown in figure. A current $I$ (in
amperes) enters the loop from one end and leaves the loop from opposite end as
shown in figure. The length of one side of square loop is $l$ meter. The wire has
uniform cross section area and uniform linear mass density. In four situations of
column-I, the loop is subjected to four different uniform and constant magnetic field.
Under the conditions of column I, match the column-I with corresponding results of
column-II. (in column I is a positive non-zero constant).

• A. A-Q,R B-Q,R C-Q,R D-P,R
• B. A-R B-Q,S C-P D-P,R
• C. A-Q B-Q,R C-S D-Q
• D. A-Q B-Q C-Q,R D-P

Answer 1

Answer: (A)

. A-Q,R B-Q,R C-Q,R D-P,R

As we know $\overrightarrow{F}=I(\overrightarrow{l}\times \overrightarrow{B})$ , $where\overrightarrow{I}=currentintheconductor$

$\overrightarrow{l}=lengthofthecurrentcarryingcondutor\left(wire\right)$ , $\overrightarrow{B}=magneticfield$

(A) if $\overrightarrow{B}=B_0\hat{i}$ , ${\overrightarrow{F}}_{net}={\overrightarrow{F}}_{ab}+{\overrightarrow{F}}_{bc}+\overrightarrow{F_{ad}}+\overrightarrow{F_{dc}}$

${\overrightarrow{F}}_{net}=\overrightarrow{F_{ab}}+\overrightarrow{F_{dc}};\because \overrightarrow{F_{bc}}=0\&\overrightarrow{F_{ad}}=0$

$\overrightarrow{F_{net}}=\left(\frac{I}{2}\right)(l\hat{j}\times B_0\hat{i})+\left(\frac{I}{2}\right)(l\hat{j}\times B_0\hat{i})$

$\overrightarrow{F_{net}}=B_{0}l\frac{I}{2}(-\hat{k})+B_{0}l\frac{I}{2}(-\hat{k})$

$\overrightarrow{F_{net}}=B_{0}lI(-\hat{k})$

$\overrightarrow{F_{net}}=-B_{0}lI\left(\hat{k}\right)$

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