0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

# Figure a shows a circuit consisting of an ideal battery with emf $$\mathscr{E}=6.00\muV$$, a resistance R, and a small wire loop of area $$5.0cm^2$$. For the time interval t = 10 s to t = 20 s, an external magnetic field is set up throughout the loop.The field is uniform, its direction is into the page in Figure a, and the field magnitude is given by $$B=at$$, where B is in teslas, a is a constant, and t is in seconds. Figure b gives the current $$i$$ in the circuit before, during, and after the external field is set up. The vertical axis scale is set by $$i_s=2.0mA$$. Find the constant a in the equation for the field magnitude.

Solution
Verified by Toppr

#### From the datum at $$t=0$$ in Fig. bwe see $$0.0015 \mathrm{A}=V_{\text {battery }} / R,$$ which implies that theresistance is$$R=(6.00 \mu \mathrm{V}) /(0.0015\mathrm{A})=0.0040 \Omega$$Now, the value of the current during$$10 \mathrm{s}<t<20 \mathrm{s}$$ leads us to equate$$\left(V_{\text {battery}}+\varepsilon_{\text {induced }}\right) / R=0.00050 \mathrm{A}$$This shows that the induced emf is $$\varepsilon_{\text{induced }}=-4.0 \mu \mathrm{V}$$. Now we use Faraday's law:$$\varepsilon=-\dfrac{d \Phi_{B}}{dt}=-A \dfrac{d B}{d t}=-A a$$Plugging in $$\varepsilon=-4.0\times 10^{-6} \mathrm{V}$$ and $$A=5.0 \times 10^{-4} \mathrm{m}^{2},$$ weobtain $$a=0.0080 \mathrm{T} / \mathrm{s}$$

3
Similar Questions
Q1

Figure a shows a circuit consisting of an ideal battery with emf $$\mathscr{E}=6.00\muV$$, a resistance R, and a small wire loop of area $$5.0cm^2$$. For the time interval t = 10 s to t = 20 s, an external magnetic field is set up throughout the loop.The field is uniform, its direction is into the page in Figure a, and the field magnitude is given by $$B=at$$, where B is in teslas, a is a constant, and t is in seconds. Figure b gives the current $$i$$ in the circuit before, during, and after the external field is set up. The vertical axis scale is set by $$i_s=2.0mA$$. Find the constant a in the equation for the field magnitude.

View Solution
Q2
$5.Figure6.14showsaaconductingloopconsistingofahalfcircleofradiusr=0.20m\phantom{\rule{0ex}{0ex}}andthreestraightsections.ThehalfcircleliesinuniformmagneticfieldBthatisdirectedoutofthapage\phantom{\rule{0ex}{0ex}}thefieldmagnitudeB=4{t}^{2}+2t+3,withBinteslaandtinseconds.Anidealbatterywithemf2Visconnectedtotheloop.Theresis\mathrm{tan}ceofloopis2ohm\text{'}\phantom{\rule{0ex}{0ex}}9a\right)WhatarethemagnitudeanddirectionofemfinducedaroundtheloopBatt=10s\phantom{\rule{0ex}{0ex}}\left(b\right)Whatiscurrentinloopatt=10s\phantom{\rule{0ex}{0ex}}$
View Solution
Q3

A square wire loop with 2.00 m sides is perpendicular to a uniform magnetic field, with half the area of the loop in the field as shown in Figure . The loop contains an ideal battery with emf $$\mathscr{E}_=20.0V$$. If the magnitude of the field varies with time according to $$B=0.0420-0.870t$$,with B in teslas and t in seconds, what are (a)the net emf in the circuit and (b) the direction of the (net) current around the loop?

View Solution
Q4

A circular region in an xy plane is penetrated by a uniform magnetic field in the positive direction of the z axis.The field’s magnitude B (in teslas) increases with time t (in seconds) according to $$B=at$$, where a is a constant. The magnitude E of the electric field set up by that increase in the magnetic field is given by Figure versus radial distance
r; the vertical axis scale is set by $$E_s=300\mu N/C$$, and the horizontal axis scale is set by $$r_s=4.00cm$$. Find a.

View Solution
Q5

As seen in Figure, a square loop of wire has sides of length 2.0 cm. A magnetic field is directed out of the page; its magnitude is given by $$B=4.0t^2y$$, where B is in teslas, t is in seconds, and y is in meters. At t = 2.5 s, what are the (a) magnitude and (b) direction of the emf induced in the loop?

View Solution