Question Number 59842 by rahul 19 last updated on 15/May/19 Commented by tanmay last updated on 15/May/19 Commented by tanmay last updated on 15/May/19 $${i}\:{am}\:{unable}\:{to}\:{find}\:\mathrm{2}\:{factor}\:{in}\:{answer}\:{and}\:{cause}…
Question Number 55725 by Tinkutara last updated on 03/Mar/19 Answered by ajfour last updated on 03/Mar/19 $$\:\:\:{short}-{circuited}\:{case} \\ $$$$\:\:\frac{{LdI}}{{dt}}=−{IR} \\ $$$$\Rightarrow\:\int_{{I}_{\mathrm{0}} } ^{\:{I}} \frac{{dI}}{{I}}\:=\:−\frac{{R}}{{L}}\int_{\mathrm{0}} ^{\:\:{t}}…
Question Number 53686 by ajfour last updated on 25/Jan/19 Commented by ajfour last updated on 25/Jan/19 $${Charged}\:{particle}\:\left({q},{m}\right)\:{released} \\ $$$${at}\:{origin}.\:{Find}\:{velocity}\:\bar {{v}}\:{as}\:{a} \\ $$$${function}\:{of}\:{the}\:{z}\:{coordinate}. \\ $$$${Electric}\:{field},\:{E}_{\mathrm{0}} \hat…
Question Number 53342 by aseerimad last updated on 20/Jan/19 Commented by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19 $${pls}\:{clarify}\:{direction}\:{of}\: \\ $$$$\overset{\rightarrow} {{E}}={electric}\:{field}\:\:\left({is}\:{it}\:+{ve}\:{x}\:{axis}\:\right) \\ $$$$\overset{\rightarrow} {{V}}={velocity}\:{proton}\left({is}\:{it}\:+{ve}\:{x}\:{axis}\right. \\ $$$$\overset{\rightarrow}…
Question Number 48310 by rahul 19 last updated on 22/Nov/18 Answered by tanmay.chaudhury50@gmail.com last updated on 22/Nov/18 $${current}\:{I}=\frac{{dq}}{{dt}} \\ $$$${E}=\frac{{q}}{{c}}\:\:{so}\:\frac{{dE}}{{dt}}=\frac{{I}}{{c}}\:\:\:{I}={c}\frac{{dE}}{{dt}} \\ $$$${force}\:{on}\:{coducting}\:{slider}\:{F}={IBl} \\ $$$${F}={m}\frac{{dV}}{{dt}}={IBl} \\…
Question Number 42422 by ajfour last updated on 25/Aug/18 Commented by ajfour last updated on 25/Aug/18 $${Q}.\mathrm{42278}\:\:{solution}\:\:\left({rishav}'{s}\:{question}\right) \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on…
Question Number 41434 by rahul 19 last updated on 07/Aug/18 Commented by rahul 19 last updated on 07/Aug/18 $$\mathrm{If}\:\mathrm{the}\:\mathrm{ammeter}\:\mathrm{A}_{\mathrm{2}\:} \:\mathrm{reads}\:\mathrm{1}.\mathrm{6A}\:\mathrm{and} \\ $$$$\mathrm{ammeter}\:\mathrm{A}_{\mathrm{3}} \:\mathrm{reads}\:\mathrm{0}.\mathrm{4}\:\mathrm{A}\:. \\ $$$$\mathrm{Find}\:\mathrm{reading}\:\mathrm{of}\:\mathrm{ammeter}\:\mathrm{A}_{\mathrm{1}}…
Question Number 40975 by rahul 19 last updated on 30/Jul/18 Answered by ajfour last updated on 30/Jul/18 $$\left({b}\right)\:\mathrm{10}{A}/{s} \\ $$ Commented by rahul 19 last…
Question Number 40946 by Necxx last updated on 29/Jul/18 $${Each}\:{of}\:{two}\:{long}\:{straight}\:{wires}\:\mathrm{4}{cm} \\ $$$${apart}\:{carry}\:{equal}\:{electric}\:{currents} \\ $$$${and}\:{experience}\:{a}\:{force}\:{of}\:\mathrm{2}×\mathrm{10}^{−\mathrm{4}} {N}/{m}. \\ $$$${What}\:{is}\:{the}\:{magnitude}\:{of}\:{the} \\ $$$${electric}\:{current}\:{in}\:{each}? \\ $$ Answered by tanmay.chaudhury50@gmail.com last…
Question Number 40945 by Necxx last updated on 29/Jul/18 $${Two}\:{parallel}\:{plate}\:{conductors}\:\mathrm{1}{m} \\ $$$${from}\:{each}\:{other}\:{carry}\:{an}\:{electric} \\ $$$${current}\:{of}\:\mathrm{2}{A}\:{each}.{Find}\:{the}\:{magnetic} \\ $$$${force}\:{per}\:{metre}\:{on}\:{each}\:{wire}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com