Question

Two straight conducting rails form a right angle as shown below. A conducting bar in contact with the rails starts at the vertex at time $t =0$ and moves with constant velocity of $v=5 m / s$ along them. A magnetic field with $B=0.1 T$ is directed out of the page. The absolute value of the emf around the triangle at the time $t=4 s$ will be?

• A. 10 V
• B. 15 V
• C. 20 V
• D. 30 V

Answer 1

Answer: (A)

The given situation is shown below

For triangle $P Q R, x=v \times t$
So, at $t=4 s$, $x=5 \times 4=20 m$
Now in $\triangle P R S$,
$\frac{S R}{S P}=\tan 45^{\circ} \Rightarrow S R=S P \left(\because \tan 45^{\circ}=1\right)$
or length $P Q$ of rod is $2 \times S R=2 x=40 m$
i.e., area swept in $4 s=$ area $\Delta P Q R$
$=\frac{1}{2} \times$ base $\times$ height
$=\frac{1}{2} \times 2 x \times x$
$=x^{2}=(20)^{2}=400 m ^{2}$
Area swept in $1 s$
$=\frac{d A}{d t}=\frac{400}{4}=100 m / s ^{2}$
So, magnitude of emf induced in triangular loop is
$E=\frac{d \varphi}{d t}=B\left(\frac{d A}{d t}\right)=0.1 \times 100=10 V$