Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
gOlU g3n|[0]uS Grade: Upto college level

6 years ago

510 Points

drop perpendicular from P to line which is below OP ...let this perpendicular cut this line at Q ....

now we have right angled  triangle POQ in which ,

POQ = @ , PQO = 90o &  OPQ = 90-@

length of this perpendicular = xsin@

now , we have formula for magnetic field due to current carrieng straight wire (B) = (uoI/4pia)[sin@1+sin@2]

here , a = perpendicular distance

@1 , @2 are angles substended with perpendicular line to end points of wire ...

@1 = @    (from left end it is making @)

@2 = 90       (right end is at infinity )

now , B = (uoI/4pi(xsin@) ) .  [sin@+1]

direction of this field is in -ve Z axis ...

in same manner , B due to another straight wire is same ...

total magnetic field = 2B

= (uoI/2pi(xsin@))[1+sin@] (-k)

this is the required ans

6 years ago
510 Points

@1 = 90-@ , by mistake i have written @

correct this , then ans will be final ans ..

6 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Magnetism

View all Questions »
• Complete Physics Course - Class 12
• OFFERED PRICE: Rs. 2,756
• View Details
Get extra Rs. 220 off
USE CODE: Neerajwin
• Complete Physics Course - Class 11
• OFFERED PRICE: Rs. 2,968
• View Details
Get extra Rs. 237 off
USE CODE: Neerajwin