Question

A horizontal rod 0.2 m long is mounted on a balnce and carries a current. At the location of the rod a uniform horizontal magnetic field has magnitude 0.067 T and direction perpendicular to the rod. The magnetic force on the rod is measured by the balance and is found to be 0.13N. What is the current?

Answer 1

Answer:

0.13 = BIL

0.13 = 0.067 ×2 × I

I = 1 A approx

Explanation:

hope the answer will help you

Answer 2

Given :

Length of rod l = 0.2 m

Magnetic field b = 0.067 T

Angle between Rod and Magnetic field a = 90°

Magnetic force f= 0.13 N

To find :

Current I

Solution :

The force on rod containing current f =

$f = i \: l \times \: b \\ \\ f = ilb \sin \: a \\ \\ f = ilb \sin90 \\ f = ilb$

where

f = force applied

i = current aplied

l = length of rod

and b = magnetic field.

Putting all the values of l ,b and f in above equation,

We get the magnitude of force applied on wire.

So,

$0.13 \: = \: i \times(0.2) \times(0.067)$

$\textbf{ \large \: i \: = } \frac{ \textbf{ \large \: \: 0.13}}{ \textbf{ \large \:0.134} }$

So, The current flowing through rod is 0.970 Ampere.