Physics Test - 23

\(\begin{aligned} & \vec{\tau}=\vec{p} \times \vec{E} \\ & \vec{p}=q \vec{\ell} \\ & \vec{E}=0.2 \frac{V}{\mathrm{~cm}}=20 \frac{V}{m} \\ & \vec{p}=4 \times(\hat{i}-\hat{j}+\hat{k}) \\ & =(4 \hat{i}-4 \hat{j}+4 \hat{k}) \mu C-m \\ & \vec{\tau}=(4 \hat{i}-4 \hat{j}+4 \hat{k}) \times(20 \hat{i}) \times 10^{-6} \mathrm{Nm} \\ & =(8 \hat{k}+8 \hat{j}) \times 10^{-5}=8 \sqrt{2} \times 10^{-5} \\ & \alpha=2\end{aligned}\)

A Zener diode is always used in the reverse bias to get the property of breakdown voltage. We use a Zener diode to regulate the voltage of the alternating input signal in various places because of this property. Zener diode is used as the ac voltage stabilizer.

It is a special purpose semiconductor diode, named after its inventorC. Zener. It is designed to operate under reverse bias in the breakdownregion and used as a voltage regulator.

Zener diode is fabricated by heavily doping both p-, and n- sides of the junction. Due to this, depletion region formed is very thin \(\left(<10^{-6} \mathrm{~m}\right)\) and the electric field of the junction is extremely high \(\left(\sim 5 \times 10^{6} \mathrm{~V} / \mathrm{m}\right)\) even for a small reverse bias voltage of about \(5 \mathrm{~V}\). The \(\mathrm{I}-\mathrm{V}\) characteristics of a Zener diode is shown in Fig. (b). It is seen that when the applied reverse bias voltage \((V)\) reaches the breakdown voltage \(\left(V_{x}\right)\) of the Zener diode, there is a large change in the current. Note that after the breakdown voltage \((V_{z})\), a large change in the current can be produced by almost insignificant change in the reverse bias voltage. In other words, Zener voltage remains constant, even though current through the Zener diode varies over a wide range. This property of the Zener diode is used for regulating supply voltages so that they are constant.

Transducer is a device used for converting one form of energy into another. Therefore, the term transducer is applied to gramophone pickups, microphones and loudspeakers but not for the amplifier.

Under uniform magnetic field, force evB acts on proton and provides the necessary centripetal force \(\frac{ mv ^2}{ a }\)

\(\frac{ mv ^2}{ a }= evB \)

\(v =\frac{ aeB }{ m }\)

Now, angular momentum

\(J=r \times p\)

\(= a \times mv\)

J\( = a \times m \left(\frac{ aeB }{ m }\right) \)

\(J = a ^2 eB\)

A ball balanced on a vertical rod is an example of unstable equilibrium. A system is in unstable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in the same direction as the displacement from equilibrium.

If a ball is placed on vertical rod, it is in unstable equilibrium because once it is displaced from its place, it will experience the net force in the direction of displacement and never come back to its original position. The potential energy of the ball is maximum at this point.

Magnetic moment associated with electron \(M=\frac{e ~h}{4 ~\pi ~m_e}\)

where \(e\) is charge on electron, \(h\) is plank constant, \(m_e \) is mass of electron

\(\theta=30^{\circ}\)

\(\tau=M B \sin \theta\) where \(\tau\) is Torque.

\(\tau=\frac{e ~h}{4 ~\pi~ m_e} B \times \sin 30^{\circ}\)

\(=\frac{e ~h~B}{8 ~\pi ~m_e}\), which is the torque.

Length of a solenoid, \(L=600 mm\)

Radius of the solenoid, \(d=7.5 mm =7.5 \times 10^{-3} m\) Current, \(I=3 A\)

Total number of turns, \(N=50\)

Step 2: Find Magnetic flux

Magnetic flux through one loop is

\(\phi=B A\)

\(\phi=\frac{\mu_r \mu_0 N^2 A I}{L}\)

\(\phi=\frac{600 \times 4 \pi \times 10^{-7} \times 50^2 \times \pi \times\left(7.5 \times 10^{-3}\right)^2 \times 3}{600 \times 10^{-3}}\)

\(\phi=1.66 \times 10^{-3} Wb\)

\(\phi=1.66 mWb\)

As we already know that Energy of electromagnetic radiation depends upon the frequency of radiation.

Energy of a photon of EM radiation with frequency is \(E = h \nu\)

Frequency and wave length are related by \(\lambda v= c\)

\(E = h \nu=\frac{ hc }{\lambda}\)

\( \lambda=\frac{ hc }{ E } \)

\( =\frac{1240 ev - nm }{13.2 kev }\)

\( =\frac{1240 ev - nm }{13.2 \times 1000 ev } \)

\(=\frac{1240}{13200}=0.093 nm\)\( =0.093 nm\)

As X - ray wavelength is in the Range of \(0.01 nm\) to \(1 nm\).

Hence answer is X-rays.

Microwaves are used for telecommunication.

• It modulates the mechanical waves emitted by our vocal cords. Great physicists such as Hertz, Tesla, Branly, and Marconi understood that it was possible to use electromagnetic waves to convey information through the air.
• Many different types of EMF are used to convey information over a distance, the most common being radio waves, also called radio frequencies.
• This is a wide spectrum ranging from tens of kilohertz to 300 gigahertz.