Question

Figure shows a conducting loop $ABCDA$ placed in a uniform magnetic field (strength $B$) perpendicular to its plane. The part $ABC$ is the (three-fourth) portion of the square of side length $l$. The part $ADC$ is a circular arc of radius $R$. The points $A$ and $C$ are connected to a battery which supplies a current $I$ to the circuit. The magnetic force on the loop due to the field $B$ is:

A. $zero$
B. $BIl$
C. $2BIR$
D. ${\displaystyle \frac{BIlR}{I+R}}$

Answer 1

Hint: Since, the current through the loop $ABCDA$ splits into to part, one flows through $ABC$ and other through $ADC$. During the current flow, there will be some force generated. To find the total force generated in the loop $ABCDA$, the force generated on loops $ABC$ and $ADC$ are calculated. Since, the two currents travel the same distance in the magnetic field, the current gets added. Thus, the total force can be calculated.

Useful formula:
Relation between force, current and magnetic field will be given by,
$F=BIl$
Where, $F$ is the force on the conductor, $B$ is the magnetic field, $I$ is the current flows through the conductor and $l$ is the length of the conductor.

Complete Step by step solution:

Assume that,
The current through the loop $ABC$ is $I1$
The current through the loop $ADC$ is $I2$
The length where the current flow in loop $ABC$ is $l_1$
The length where the current flow in loop $ADC$ is $l_2$

The force generated on the loop $ABC$,
$F_{ABC}=B\times I1\times l_1......................................\left(1\right)$
Where, $F_{ABC}$ is the force generated on loop $ABC$, $I1$ is the current flows through the loop $ABC$ and $l_1$ is the length of the loop $ABC$.

The force generated on the loop $ADC$,
$F_{ADC}=B\times I2\times l_2......................................\left(2\right)$
Where, $F_{ADC}$ is the force generated on loop $ADC$, $I2$ is the current flows through the loop $ADC$ and $l_2$ is the length of the loop $ADC$.

Hence, the total force on loop $ABCDA$ is the sum of the force on loop $ABC$ and the force on loop $ADC$,
$F=F_{ABC}+F_{ADC}..........................................\left(3\right)$
Substitute the values of (1) and (2) equation (1),
$$\begin{gathered} F=\left(B\times I1\times l_1\right)+\left(B\times I2\times l_2\right) \\ F=B\left[\left(I1\times l_1\right)+\left(I2\times l_2\right)\right] \end{gathered}$$

Since, the current travels the same length in the loops $ABC$ and $ADC$ ,
 $l=l_1=l_2$
Hence,
$$\begin{gathered} F=B\left[\left(I1\times l\right)+\left(I2\times l\right)\right] \\ F=B\times l\left[I1+I2\right] \end{gathered}$$
The total current, $I=I1+I2$
Thus,
$$\begin{gathered} F=B\times l\left[I\right] \\ F=BIl \end{gathered}$$

Hence, the option (B) is correct.

Note: In the loop $ABCDA$, even the structure of the loops $ABC$ and $ADC$ may vary, but the current flows through the loops in the uniform magnetic field travels the same distance in both the loops. A force will be generated in the current carrying conductor placed in a uniform magnetic field, this is the statement of Faraday’s law of induction. The force generated in the loop $ABCDA$, is the sum of the force generated in loops $ABC$ and $ADC$.