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 $v=5m/s$ along them. A magnetic field with $B=0.1T$ is directed out of the page. The absolute value of the emf around the triangle at the time $t=4s$ 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 $PQR,x=v\times t$
So, at $t=4s$, $x=5\times 4=20m$
Now in $△PRS$,
$\frac{SR}{SP}=\tan 45^{\circ }\Rightarrow SR=SP\left(\because \tan 45^{\circ }=1\right)$
or length $PQ$ of rod is $2\times SR=2x=40m$
i.e., area swept in $4s=$ area $\Delta PQR$
$=\frac{1}{2}\times$ base $\times$ height
$=\frac{1}{2}\times 2x\times x$
$=x^2=(20{)}^2=400m^2$
Area swept in $1s$
$=\frac{dA}{dt}=\frac{400}{4}=100m/s^2$
So, magnitude of emf induced in triangular loop is
$E=\frac{d\phi }{dt}=B\left(\frac{dA}{dt}\right)=0.1\times 100=10V$