A square loop of uniform conducting wire is as shown in figure.A current $I$ (in ampere) enters the loop from one end and exits 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. The magnetic field is in Tesla and force is in newton.

Open in App

Solution

Case A:
$\to B = B_{0}^i$

$_{A C} = 0$ since $\to d l ∥ \to B$

$_{B D} = 0$ since $\to d l ∥ \to B$

$_{A B} = - I \frac{l}{2} B_{0}^k$ since $\to d l ⊥ \to B$

$_{C D} = - I \frac{l}{2} B_{0}^k$ since $\to d l ⊥ \to B$

Net force $_{l o o p} =_{A B} +_{C D} = - (I \frac{l}{2} B_{0}^k + I \frac{l}{2} B_{0}^k) = - I l B_{0}^k$

$C a s e B :$

$\to B = B_{0}^j$

$_{A B} = 0$ since $\to d l ∥ \to B$

$_{C D} = 0$ since $\to d l ∥ \to B$

$_{A C} = I \frac{l}{2} B_{0}^k$ since $\to d l ⊥ \to B$

$_{B D} = I \frac{l}{2} B_{0}^k$ since $\to d l ⊥ \to B$

Net force $_{l o o p} =_{A C} +_{B D} = (I \frac{l}{2} B_{0}^k + I \frac{l}{2} B_{0}^k) = I l B_{0}^k$

$C a s e C :$
This is combination of Case(A) and (B).
Since the forces on the loop ABDC are equal and opposite in the cases (A) and (B), they will cancel each other.
Net force $_{l o o p} = 0$

$C a s e D :$

$_{A B} =_{C D} = I \frac{l}{2} B_{0} (^j \times^k) = - I \frac{l}{2} B_{0}^i$

$_{A C} =_{B D} = I \frac{l}{2} B_{0} (^i \times^k) = - I \frac{l}{2} B_{0}^j$

$_{l o o p} =_{A B} +_{C D} +_{A C} +_{B D} = 2 \times (- I \frac{l}{2} B_{0}^i) + 2 \times (- I \frac{l}{2} B_{0}^j) = - \sqrt{2} I l B_{0} \frac{(^i +^j)}{\sqrt{2}}$

List 2 has magnitude of the forces shown only.

Suggest Corrections

Similar questions

Q. A square loop of uniform conducting wire is as shown in fig. A current I (in amperes) enters the loop from one end and exits the loop from opposite end as shown. The length of one side of the square loop is 1 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 magnetic fields. Under the conditions of column I, match the column I with corresponding results of column II

(B_{0}

in column I is a positive non-zero constant)

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).

Q. 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 metre. 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. (

B_{0}

in column I is a positive non-zero constant).

\begin{matrix} C o l u m n I & C o l u m n I I (A) \to B = B_{0}^i i n t e l s a & (p) m a g n i t u d e o f n e t f o r c e o n l o o p i s \sqrt{2} B_{0} I l n e w t o n (B) \to B = B_{0}^i i n t e l s a & (Q) m a g n i t u d e o f n e t f o r c e o n l o o p i s z e r o (C) \to B = B_{0} (^i +^j) i n t e s l a & (R) m a g n i t u d e o f n e t t o r q u e o n l o o p a b o u t i t s c e n t r e i s z e r o (D) \to B = B_{0}^k i n t e s l a & (S) N e t f o r c e p e r p e n d i c u l a r t o t h e p l a n e o f l o o p \end{matrix}

A plane square loop PQRS of side 'a' made of thin copper wire has 'n' turns and it carries a direct current 'I' ampere in the direction shown in the adjoining figure. This wire loop is placed in a magnetic field of flux density 'B' tesla, which is directed perpendicularly in to the plane of the loop. What is the torque acting on the loop?

Q. ABCD is a square loop made of a uniform conducting wire. A current enters the loop at A and leaves at D. The magnetic field is

Join BYJU'S Learning Program