Question

A jumper cable is used to start a car that has a dead battery. The cable barely reaches the car and is outstretched in a straight line. The electric starter motor on the car draws 134 AA . What is the magnetic field at a distance of 0.5 mm from the cable

Answer 1

Answer:

$2.68\times 10^{-5} T$

Explanation:

We are given that

Current,I=134 A

Distance,r=0.5m

We have to find the magnetic field at a distance of 0.5 m from the cable.

We know that magnetic filed ,

$B=\frac{2\mu_0 I}{4\pi r}$

Where $\frac{\mu_0}{4\pi}=10^{-7}$

Using the formula

$B=\frac{10^{-7}\times 134}{0.5}$

$B=2.68\times 10^{-5} T$

Hence, the magnetic field at a distance 0.5 m from the cable=$B=2.68\times 10^{-5} T$

Answer 2

Answer:

Explanation:

Current in the wire, i = 134 A

distance, r = 0.5 mm

The formula for the magnetic field due to straight conductor at a point is

$B = \frac{\mu _{0}}{4\pi }\frac{2i}{r}$

$B = 10^{-7}\frac{2\times 134}{0.5\times 10^{-3}}$

B = 0.0536 Tesla