Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
154 views
in Physics by (98.2k points)
closed by
Infinite number of straight wires each carrying current I are equally placed as shown in the figure Adjacent wires have current in opposite direction Net magnetic field at point `P` is
image .
A. `(mu_(0)I)/(4pi)(In2)/(sqrt3a)hatk`
B. `(mu_(0)I)/(4pi)(In4)/(sqrt3a)hatk`
C. `(mu_(0)I)/(4pi)(In4)/(sqrt3a)(-hatk)`
D. zero

1 Answer

0 votes
by (97.8k points)
selected by
 
Best answer
Correct Answer - B
`B_(1)=(mu_(0))/(4pi) (I)/(a) (sinalpha +sinbeta) =(mu_(0))/(4pi) (Ixx2 sin 30^(@))/(a cos 30^(@))`
`oversetrarr(B) = B_(1) hatk - B_(2)hatk +B_(3)hatk -B_(4)hatk+......`
`= (mu_0)/(4pi)(Ixx2 sin 30^@)/(a cos 30^(@))hatk - (mu_0)/(4pi) (I xx 2 sin 30^(@))/(2a cos 30^(@))hatk + `
`(mu_(0))/(4pi) (I xx 2 sin30^(@))/(3a cos 30^(@))hatk+......`
`=(mu_(0))/(4pi)(2I)/(sqrt3a) hatk - (mu_0)/(4pi) (2I)/(sqrt3 2a) hatk + mu_(0)/(4pi)(2I)/(sqrt33a) hatk+....`
` =(mu_0)/(4pi) (2I)/(sqrt3a) [1 - (1)/(2) + (1)/(3) - (1)/(4) + .... ]hatk`
`=(mu_0)/(4pi) (2I)/(sqrt3a) log_(e) (1+1)hatk=(mu_(0)I)/(4pi sqrt3a )log_(e) 4hatk` .

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...