Question 14 Marks
Shows a rod PQ of length 20.0cm and mass 200g suspended through a fixed point O by two threads of lengths 20.0cm each. A magnetic field of strength 0.500T exists in the vicinity of the wire PQ, as shown in the figure. The wires connecting PQ with the battery are loose and exert no force on PQ.
- Find the tension in the threads when the switch S is open.
- A current of 2.0A is established when the switch S is closed. Find the tension in the threads now.
Answer
$\Rightarrow\text{T}=\frac{\text{mg}}{2\cos30^\circ}$
$=\frac{200\times10^{-3}\times9.8}{2\sqrt{\frac{3}{2}}}=1.13$
$\Rightarrow2\text{T}\cos30^\circ=\text{mg}+\text{ilB}$
$=200\times10^{-3}9.8+2\times0.2\times0.5$
$=1.96+0.2=2.16$
$\Rightarrow2\text{T}=\frac{2.16\times2}{\sqrt{3}}=2.49$
$\Rightarrow\text{T}=\frac{2.49}{2}=1.245\approx1.25$
View full question & answer→
- When switch S is open
$\Rightarrow\text{T}=\frac{\text{mg}}{2\cos30^\circ}$
$=\frac{200\times10^{-3}\times9.8}{2\sqrt{\frac{3}{2}}}=1.13$
- When the switch is closed and a current passes through the circuit = 2A
$\Rightarrow2\text{T}\cos30^\circ=\text{mg}+\text{ilB}$
$=200\times10^{-3}9.8+2\times0.2\times0.5$
$=1.96+0.2=2.16$
$\Rightarrow2\text{T}=\frac{2.16\times2}{\sqrt{3}}=2.49$
$\Rightarrow\text{T}=\frac{2.49}{2}=1.245\approx1.25$