Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{e^{x}}{\sqrt{16-e^{2 x}}} d x$ Solution: Let $\mathrm{e}^{x}=\mathrm{t}$ Then we have, $\mathrm{e}^{x} \mathrm{dx}=\mathrm{dt}$ Therefore, $\int \frac{e^{x}}{\sqrt{16-e^{2 x}}} d x=\int \frac{d t}{\sqrt{4^{2}-t^{2}}}$ Since we have, $\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)+c$ Hence, $\int \frac{d t}{\sqrt{4^{2}-t^{2}}}=\sin ^{-1}\left(\frac{e^{x}}{a}\right)+c$...
Read More →An electron and a positron are released from (0, 0, 0)
Question: An electron and a positron are released from (0, 0, 0) and (0, 0, 1.5R) respectively, in a uniform magnetic field $B=B_{0} \hat{i}$ each with an equal momentum of magnitude $\mathrm{p}=\mathrm{eBR}$. Under what conditions on the direction of momentum will the orbits be non-intersecting circles? Solution: When the centres are greater than 2R, then the circular orbits of electron and positron shall not overlap. Let d be the distance between Cp and Ce Then d2= 4R2+ 9/4R2 6R2cos As d is gr...
Read More →A rectangular conducting loop consists of two wires
Question: A rectangular conducting loop consists of two wires on two opposite sides of length l joined together by rods of length d. The wires are each of the same material but with cross-sections differing by a factor of 2. The thicker wire has a resistance R and the rods are of low resistance, which in turn are connected to a constant voltage source Vo. The loop is placed in uniform a magnetic field B at 45oto its plane. Find , the torque exerted by the magnetic field on the loop about an axis...
Read More →Use contradiction method to prove that
Question: Use contradiction method to prove that $p: \sqrt{3}$ is irrational is a true statement. Solution: Let us assume that $\sqrt{3}$ is a rational number. For a number to be rational, it must be able to express it in the form $p / q$ where $p$ and $q$ do not have any common factor, i.e. they are co-prime in nature. Since $\sqrt{3}$ is rational, we can write it as $\sqrt{3}=\frac{p}{q}$ $\rightarrow \frac{p}{\sqrt{3}}=q$ [ squaring both sides ] $\rightarrow \frac{p^{2}}{3}=q^{2}$ Thus, $\mat...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{\sec ^{2} x}{\sqrt{4+\tan ^{2} x}} d x$ Solution: Let $\tan x=t$ Then $\mathrm{dt}=\sec ^{2} \mathrm{x} \mathrm{dx}$ Therefore, $\int \frac{\sec ^{2} x}{\sqrt{4+\tan ^{2} x}} d x=\int \frac{d t}{\sqrt{2^{2}+t^{2}}}$ Since, $\int \frac{1}{\sqrt{\left(x^{2}+a^{2}\right)}} d x=\log \left[x+\sqrt{\left(x^{2}\right.}+a^{2}\right)+c$ Hence, $\int \frac{d t}{\sqrt{2}^{2}+t^{2}}=\log \left[t+\sqrt{t^{2}+2^{2}}\right]+c$ $=\log \left[\tan x+\sqrt{\t...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{x}{\sqrt{x^{4}+a^{4}}} d x$ Solution: $\int \frac{\mathrm{x}}{\sqrt{\mathrm{x}^{4}+\mathrm{a}^{4}}} \mathrm{dx}=\int \frac{\mathrm{x}}{\sqrt{\left(\mathrm{x}^{2}\right)^{2}+\left(\mathrm{a}^{2}\right)^{2}}} \mathrm{dx}$ Let $x^{2}=t$, so $2 x d x=d t$ Or, $x d x=d t / 2$ Hence, $\int \frac{\mathrm{x}}{\sqrt{\left(\mathrm{x}^{2}\right)^{2}+\left(\mathrm{a}^{2}\right)^{2}}} \mathrm{dx}=\int \frac{1}{\sqrt{\mathrm{t}^{2}+\left(\mathrm{a}^{2}\r...
Read More →A 100 turn rectangular coil ABCD is hung from one arm of a balance.
Question: A 100 turn rectangular coil ABCD is hung from one arm of a balance. A mass 500 g is added to the other arm to balance the weight of the coil. A current 4.9 A passes through the coil and a constant magnetic field of 0.2 T acting inward is switched on such that only arm CD of length 1 cm lies in the field. How much additional mass m must be added to regain the balance? Solution: When t = 0, the external magnetic field is off. Mgl = Wcoill 0.5 gl = Wcoill Wcoil = 0.5 9.8 N Let m be the ma...
Read More →A long straight wire carrying a current of 25 A
Question: A long straight wire carrying a current of 25 A rests on a table as shown in the figure. Another wire PQ of length 1 m, mass 2.5 g carries the same current but in the opposite direction. The wire PQ is free to slide up and down. To what height will PQ rise? Solution: The magnetic field produced by a long straight current-carrying wire is given as B = 0I/2h Magnetic force on the small conductor is F = BIl sin = BIl F = mg = 0I2l/2h h = 0.51 cm...
Read More →A multirange voltmeter can be constructed by using
Question: A multirange voltmeter can be constructed by using a galvanometer circuit as shown in the figure. We want to construct a voltmeter that can measure 2V, 20V, and 200V using galvanometer of resistance 10Ω and that produces maximum deflection for a current of 1 mA. Find R1, R2, and R3 that have to be used. Solution: iG(G+R1) = 2 for 2V range iG(G+R1+R2) = 20 for 20V range iG(G+R1+R2+R3) = 200 for 200V range Solving the above, we get R1 = 1990 Ω R2 = 18kΩ R3 = 180 kΩ...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{\sqrt{5 x^{2}-2 x}} d x$ Solution: We have $\int \frac{\mathrm{dx}}{\sqrt{5 \mathrm{x}^{2}-2 \mathrm{x}}}=\int \frac{\mathrm{dx}}{\sqrt{5\left(\mathrm{x}^{2}-\frac{2 \mathrm{x}}{5}\right)}}$ $=\frac{1}{\sqrt{5}} \int \frac{d x}{\sqrt{\left(x-\frac{1}{5}\right)^{2}-\left(\frac{1}{5}\right)^{2}}}$ completing the square Put $x-1 / 5=t$ then $d x=d t$ Therefore $\int \frac{d x}{\sqrt{5 x^{2}-2 x}}=\frac{1}{\sqrt{5}} \int \frac{d x}{\sqrt{(t)...
Read More →Do magnetic forces obey Newton’s third law.
Question: Do magnetic forces obey Newtons third law. Verify for two current elements $d l_{1}=d l \hat{i} \quad$ located at the origin and $d l_{2}=d l \hat{j} \quad$ located at $(0, \mathrm{R}, 0) .$ Both carry current $\mathrm{I}$. Solution: The magnetic forces do not obey Newtons third law if there is no current flowing in the conductor which is placed parallel to each other....
Read More →An electron enters with a velocity v = v0i into a cubical
Question: An electron enters with a velocity v = v0i into a cubical region in which there are uniform electric and magnetic fields. The orbit of the electron is found to spiral down inside the cube in the plane parallel to the x-y plane. Suggest a configuration of fields E and B that can lead to it. Solution: The configuration of the fields E and B are the spiral path....
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{\sqrt{7-6 x-x^{2}}} d x$ Solution: $7-6 x-x^{2}$ can be written as $7-\left(x^{2}+6 x+9-9\right)$ Therefore $7-\left(x^{2}+6 x+9-9\right)$ $=16-\left(x^{2}+6 x+9\right)$ $=16-(x+3)^{2}$ $=(4)^{2}-(x+3)^{2}$ $\int \frac{1}{\sqrt{7-6 x-x^{2}}} d x=\int \frac{1}{\sqrt{(4)^{2}-(x+3)^{2}}} d x$ Let $x+3=t$ $\mathrm{d} \mathrm{x}=\mathrm{dt}$ $\int \frac{1}{\sqrt{(4)^{2}-(x+3)^{2}}} d x=\int \frac{1}{\sqrt{(4)^{2}-(t)^{2}}} d t$ $=\sin ^{-1}\l...
Read More →A charged particle of charge e and mass m is moving
Question: A charged particle of charge e and mass m is moving in an electric field E and magnetic field B. Construct dimensionless quantities and quantities of dimension [T]-1. Solution: mv2/R = evB eB/m = v/R = ꞷ B = F/ev = [MA-1T-2] [ꞷ] = [eB/m]=[v/R] = [T-1]...
Read More →By giving a counter-example, show that the statement is false :
Question: By giving a counter-example, show that the statement is false : p : If n is an odd positive integer, then n is prime. Solution: Let us take a odd positive integer n = +9 Even though $(+9)$ is an odd positive integer, It is divisible by 3 . To be a prime number, a number must only have itself and 1 as its factors. Since 9 has 3 as its factor too, it is not a prime number in spite of being an odd positive integer. $\therefore$ The given statement $p$ is false....
Read More →A current-carrying loop consists of 3 identical
Question: A current-carrying loop consists of 3 identical quarter circles of radius R, lying in the positive quadrants of the x-y, y-z, and z-x planes with their centres at the origin, joined together. Find the direction and magnitude of B at the origin. Solution: The vector sum of the magnetic field at the origin due to the quarter is given as $\vec{B}_{n e t}=\frac{1}{4}\left(\frac{\mu_{0} I}{2 R}\right)(\hat{i}+\hat{j}+\hat{k})$...
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{\sqrt{16-6 x-x^{2}}} d x$ Solution: let I $=\int \frac{1}{\sqrt{16-6 x-x^{2}}} d x$ $=\int \frac{1}{\sqrt{-\left[x^{2}+6 x-16\right]}} d x$ $=\int \frac{1}{\sqrt{-\left[x^{2}+2 x(3)+(3)^{2}-(3)^{2}-16\right]}} d x$ $=\int \frac{1}{\sqrt{-\left[(x-3)^{2}-25\right]}} d x$ $=\int \frac{1}{\sqrt{25-(x+3)^{2}}} d x$ $\operatorname{let}(x+3)=t$ $\mathrm{d} \mathrm{x}=\mathrm{dt}$ $I=\int \frac{1}{\sqrt{5^{2}-t^{2}}} d t$ $=\sin ^{-1}\left(\fra...
Read More →Consider the statement :
Question: Consider the statement : $q$ : For any real numbers $a$ and $b, a^{2}=b^{2} \Rightarrow a=b$ By giving a counter-example, prove that $q$ is false. Solution: Let us take the numbers a= +5 and b= -5. $2=(+5)^{2}=25$ $b^{2}=(-5)^{2}=25$ $\therefore a^{2}=b^{2}$ But, $+5 \neq-5$, thus $a \neq b$. $\therefore \mathrm{q}$ is false....
Read More →Two long wires carrying current I1 and I2 are arranged
Question: Two long wires carrying current I1 and I2 are arranged as shown in the figure. The one carrying I1 is along is the x-axis. The other carrying current I2 is along a line parallel to the y-axis given by x = 0 and z = d. Find the force exerted at O2 because of the wire along the x-axis. Solution: The magnetic field B on a current-carrying conductor is given as F = I(LB) = ILB sin O2and I1are parallel to the y-axis and are in the direction of Y I2is parallel to the y-axis and is along Y-ax...
Read More →Describe the motion of a charged particle
Question: Describe the motion of a charged particle in a cyclotron if the frequency of the radio frequency (rf) field were doubled. Solution: When the frequency of the radio frequency field was doubled, the time period of the radio frequency is halved which results in half revolution of the charges....
Read More →Let p : If x is an integer and x2 is even, then x is even,
Question: Let $p$ : If $x$ is an integer and $x^{2}$ is even, then $x$ is even, Using the method of contrapositive, prove that p is true. Solution: Let $p: x$ is an integer and $x^{2}$ is even. q: x is even For contrapositive, $\sim p=x$ is an integer and $x^{2}$ is not even. $\sim \mathrm{q}=\mathrm{x}$ is not even. The contrapositive statement is: If $x$ is an integer and $x^{2}$ is not even, then $x$ is not even. Proof; Let $x$ be an odd/not even integer x = 2n + 1 $\{2 n$ must be an even int...
Read More →The magnetic force depends on v which depends
Question: The magnetic force depends on v which depends on the inertial frame of reference. Does then the magnetic force differ from inertial frame to frame? Is it reasonable that the net acceleration has a different value in different frames of reference? Solution: The net acceleration can have a different value in different frames of reference as velocity depends on frame of reference....
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{1}{\sqrt{7-3 x-2 x^{2}}} d x$ Solution: let $I=\int \frac{1}{\sqrt{7-3 x-2 x^{2}}} d x$ $=\int \frac{1}{\sqrt{-2\left[x^{2}+\frac{3}{2} x-\frac{7}{2}\right]}} d x$ $=\frac{1}{\sqrt{2}} \int \frac{1}{\left.\sqrt{-\left[x^{2}+2 x\left(\frac{3}{4}\right)+\left(\frac{3}{4}\right)^{2}-\left(\frac{3}{4}\right)^{2}-\frac{7}{2}\right.}\right]} d x$ $=\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{-\left[\left(x-\frac{3}{4}\right)^{2}-\frac{65}{16}\right]}}...
Read More →Show that a force that does no work
Question: Show that a force that does no work must be a velocity dependent force. Solution: $d W=\vec{F} \cdot \overrightarrow{d l}=0$ $\vec{F} \cdot \vec{v} d t=0$ $\vec{F} \cdot \vec{v}=0$...
Read More →Verify that the cyclotron frequency
Question: Verify that the cyclotron frequency ꞷ = eB/m has the correct dimensions of [T]-1. Solution: The path traced by the particle in a cyclotron is a circular path in which magnetic force acts as a centripetal force mv2/R = evB eB/m = v/R = ꞷ B = F/ev = [MLT-2]/[AT][LT-1] = [MA-1T-2] [ꞷ] = [eB/m] = [v/R] = [T]-1...
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